[STPM] Physics

[Daniel]

I do not understand what does it mean by "The velocity of the ball V2 after collision has the same magnitude as its velocity before collision V1, but the direction of V2 is at right angles to V1". Let said the V2 is 5m/s, so the V1 would be 5m/s too?

[chiunlin]

I do not understand what does it mean by "The velocity of the ball V2 after collision has the same magnitude as its velocity before collision V1, but the direction of V2 is at right angles to V1". Let said the V2 is 5m/s, so the V1 would be 5m/s too?

Yes. When we refer to the magnitude of the velocity, we are actually referring to the speed of the object and disregard its direction.

[Daniel]

I do not understand what does it mean by "The velocity of the ball V2 after collision has the same magnitude as its velocity before collision V1, but the direction of V2 is at right angles to V1". Let said the V2 is 5m/s, so the V1 would be 5m/s too?

Yes. When we refer to the magnitude of the velocity, we are actually referring to the speed of the object and disregard its direction.

So is there any change of velocity in the above statement? Because the V2 and V1 has the same velocity, so it should be 0 ??? But how come the question ask me to find the change in Velocity ?

[david_david]

change in speed will be 0

change in velocity=v2-v1
=v2+(-v1)

Draw vector diagram.should be 45 degrees , magnitude surd 2 of V1

[Daniel]

change in speed will be 0

change in velocity=v2-v1
=v2+(-v1)

Draw vector diagram.should be 45 degrees , magnitude surd 2 of V1

You got it correct!! But i am still confused. If the v1 = 5m/s and v2 = 5 m/s. How come there is still change in velocity??

[david_david]

Velocity is a vector.
It has direction in addition to magnitude.
5 km north plus 5km east won't give you 10km.( just an example)
the same if you subtract.( subtraction is actually adding the negative of the vector)

[Daniel]

Thanx David.. i could barely understand it

[chenchow]

the v1 and v2 you show is speed, which is scalar, i.e. non-directional. You just look at how fast it moves, but don't care where it moves to.

Whereas velocity cares about how fast it moves and which direction it moves to.

[USSDefiantNX74205]

Okay, since we're talking about STPM physics here, I'm gonna ask a couple of questions I didn't really get from my just concluded mid-year exams.

From paper II:

A rocket is fixed vertically on its launching pad. Before launching is started, the mass of the rocket and its fuel is 1.9 X 10^3 kg. During launching, gas is emitted from the rocket with a velocity of 2.5 X 10^3 m/s relative to the rocket. The fuel is used at a constant rate of of 7.4 kg/s. Find the thrust of the rocket.

From paper I:

A helicopter of mass 3.0 X 10^3 kg rises vertically with a constant velocity of 25 m/s. By taking the acceleration of free fall as 10 m/s^2, calculate the resultant force acting on the helicopter.

A) zero
B) 3.0 X 10^4 N downwards
C) 4.5 X 10^4 N upwards
D) 7.5 X 10^4 N upwards

Any help or explanation on answering thse two questions from ya all would be greatly appreciated...

[david_david]

1)Thrust refers to the force exerted by the rocket, not considering it's weight right?

then it should be the velocity of the gas emitted times the rate of change of mass( the rate the fuel is burned).....from newton's 2nd law.(F=vdm/dt+mdv/dt)... but in this case dv/dt is zero

2)since it has uniform velocity, resultant force should be zero.when there is a net force there is always acceleration.

I hope it is correct.......

[Daniel]

Here is the questions that took me few hours to think about but still cound not find the solution.

A rocket is launched vertically an it moves upwards with a uniform acceleration of 19.6 for 60s. At that point the rocket stops burning fuel and continues to move as a free object under the influence of gravity.
a)What is the maximum height reached by the rocket?
b)What is the total time of travel from the moment the rocket is launched until it crashed to the ground???

[Daniel]

Here is the questions that took me few hours to think about but still cound not find the solution.

A rocket is launched vertically an it moves upwards with a uniform acceleration of 19.6 for 60s. At that point the rocket stops burning fuel and continues to move as a free object under the influence of gravity.
a)What is the maximum height reached by the rocket?
b)What is the total time of travel from the moment the rocket is launched until it crashed to the ground???

I knew that the V=0 but can said that U = 0 too ?

[david_david]

the rocket starts from rest, so u=o

after accelerating for 19.6 for 60 s it should reach a new velocity of
1176 m/s( v=u+at)
during this period it moves 35280m( s=ut+1/2at2)

thats the first phase. after that, it is like a projectile influenced by gravity only( if there's no air resistance)

now we must find how high more it will go.we take 1176m/s as the initial velocity. at max height the velocity will be zero.

v2=u2+2as. so we can find s, but g is negative( downwards).from calculation you should get s=70488m( if you take g=9.81)

so the max height will be 35280+70488= 105768( i'm ignoring sig fig rule)


to find the time , we break it up into the part when it's accelerating and the part when its influenced by gravity.for the first part it's 60s

for the second part:

s=ut+1/2at2

since it's 35280m above the ground, when it crashes , it's -35280m
<taking upwards as positive>

-35280=1176t-4.905t2. it's quadratic so solve for t( ignore negative value as it refers to the time before that)

i got t=236s

so the total time should be 60+236=296s


I hope it's right..........

[Daniel]

Hey David! you got it exactly the same answer in the book. Thankx a lot :D .Perhaps i should ask you some tougher questions. hehhee... By the way, are you in form 6 too?

[Daniel]

If i throw a ball upward vertically, the a=-g ?
What if a rocket launch with 15 m/s^2 from the land to the sky , will the a=-15 too?
If an apple fall from a tree, the a=g, the V is positve or negative?

The questions seems stupid but sometime i do get confused.

[david_david]

Actually studying in S'pore( A level) but thinking of terminating and going for JPA France..Haven't really decided

It all depends which direction you want to consider as positive

If upwards is positive, then a=-g because gravity acts downwards/

the converse is true if you take downwards as positive

If rocket launced and you consider upwards as positive, then a=+15

[gohweihan]

This tip came from my physics teacher last time.

For acceleration, velocity and distance relative to the earth surface, always take the direction upwards as positive and downwards as negative. That way, it will be easier to avoid confusion.

[Daniel]

I just read the chapter Projectiles on my own. Why the range R is maximum when:-
sin 2Q = 1 ?? what it must be 1? why not 2,3 and so on?

[Daniel]

A ball is thrown with a velocity U at an angle Q to the horizontal.
a) Find in terms of U,Q and g the maximum height reached by the ball
b) At what angle should the ball be thrown so that its maximum height reached is the same as its horizontal range?

I dont get the B)

[USSDefiantNX74205]

I just read the chapter Projectiles on my own. Why the range R is maximum when:-
sin 2Q = 1 ?? what it must be 1? why not 2,3 and so on?

The maximum value for sin is 1 and the minimum value is -1. That means, the value of sin for any number will always be between 1 and -1 with these two being the maximum and minimum limits respectively.

[solace]

er..coz when u throw a ball in a curve ( parabola?) the biggest range it can achieve is when the ball is thrown at 45 degrees to the ground.. so..sin 2(45) = 1? i kinda cracked my head to figure it when i was readin d chapter..hehe..but..i lazy to go back f6 la..haha........

[blumarine]

hey guys!Is there any methods that can easily to understand and remember the text?I can easily understand the text but when i came to exercise,i would be very confuse and just dunno how to answer the question..thats my biggest problem in physics...

[knkh_87]

blumarine, I suppose that your problem arises as a result of you following the text too closely. If you are using a Malaysian Physics book as reference, you would notice that they cite a lot of examples. I feel that these are very beneficial, but they inculcate a sense of being overly reliant on these questions (my personal opinion is that we feel that these questions are THE kind of questions in the real exam, but it may not apply to everyone).

For STPM, I propose that before you venture into any one chapter, do relevant questions from this book: Pertandingan Fizik Olimpiad (Pelangi). It is of SPM level, but the questions really require in-depth analysis and total understanding of your basic concepts of Physics first. It will prove to be of immense help to you in the aspect of regathering all your F5 Physics and strengthening your understanding of the chapter first. Then, you may start reading on the Form 6 theory of the chapter.

Hope this proposal helps.

[knkh_87]

Daniel's question:A ball is thrown with a velocity U at an angle Q to the horizontal.
a) Find in terms of U,Q and g the maximum height reached by the ball
b) At what angle should the ball be thrown so that its maximum height reached is the same as its horizontal range?

For part b)

These are the two derived equations:
Range (horizontal) = (U^2 sin 2Q)/g
Maximum height (vertical) = (U^2 sin^2 Q) / 2g, g is gravitational acceleration

In the case where range=maximum height reached,
(u^2 sin 2Q)/g = (u^2 sin^Q)/2g

sin 2Q=(sin^2 Q)/2
4 sin Q cos Q= sin^2 Q
sin Q (sin Q- 4 cos Q)=0

sin Q=0, Q=0 or 180

sin Q-4 cos Q=0
sin Q= 4 cos Q
tan Q=4
Q=75.96 (4 SF) or 76 (2 SF)- Answer

Assuming that Q is acute, the angle Q should be 75.96 degrees or zero (if acceptable).

P.S. I can't insert square symbols, so the calculations may look a little messy. Sorry about that.

[ElansarGelmir]

A ball is thrown with a velocity U at an angle Q to the horizontal.
a) Find in terms of U,Q and g the maximum height reached by the ball
b) At what angle should the ball be thrown so that its maximum height reached is the same as its horizontal range?

I dont get the B)

hey, i think i've seen and done this question before... but mine's about canon...

oh yeah, my test, once i didn't insert a negative (velocity) in my calculation... wah, that bloody teacher sucked 4/7 marks away from me lar...

[Daniel]

How to excel in ProjectileS??Do i need to memorise the important formulae like Max Height, Max. Range ?? i just managed to finish 1 questions correctly out of 5 questions... hmm.. i am dying
When a rifle is fired horizontally at a target P on a screen at a range of 50m, the bullet strikes the screen at a point 20.0mm below p. The screen is now moved to a distance of 100m and the rifle is fired again horizontally at P in its new position.
Assuming that air resistance may be nelected, what is the new distance below P which the screen would now be struck.

[knkh_87]

When a rifle is fired horizontally at a target P on a screen at a range of 50m, the bullet strikes the screen at a point 20.0mm below p. The screen is now moved to a distance of 100m and the rifle is fired again horizontally at P in its new position.
Assuming that air resistance may be nelected, what is the new distance below P which the screen would now be struck.

For the initial range (50m)
Horizontal distance travelled, sx=50m
Assuming that the velocity of the bullet when it is fired from the rifle is u, 50=ut1, t1=50/u.

Vertical distance travelled, sy=2.00x10^-2m
When the bullet was fired from the rifle, the (vertical) velocity was zero. The vertical acceleration of the bullet is g (9.81 m/s^2)

Using s=ut + 1/2 at^2
2.00x10^-2=0(t1) + 1/2(g)(50/u)^2...(1)

With the screen now 100m from the rifle, the time of travel, t2=100/u

Using s=ut + 1/2at^2,
new distance below P, h=0+1/2g(100/u)^2....(2)

Dividing equation (2) by equation (1),

h/2.00x10^-2 = (100/50)^2 = 4
h= 0.08m=8.0 cm below P

I would advise against memorising formulae for projectiles, though they are not provided in the formula sheet. You should understand the basic kinematic equations and rederive the projectile equations in the exam. It is tedious, but it is safer and provides a better understanding.

[ElansarGelmir]

How to excel in ProjectileS??Do i need to memorise the important formulae like Max Height, Max. Range ?? i just managed to finish 1 questions correctly out of 5 questions... hmm.. i am dying
When a rifle is fired horizontally at a target P on a screen at a range of 50m, the bullet strikes the screen at a point 20.0mm below p. The screen is now moved to a distance of 100m and the rifle is fired again horizontally at P in its new position.
Assuming that air resistance may be nelected, what is the new distance below P which the screen would now be struck.

Daniel, you got the questions from the past year A Level papers? No wonder all the questions seem to be familiar...

my tip is, for the projectile question, draw the diagram of the problem... Try to insert as many informations into the diagram that will help u to find the Sx and Sy (range and height)... hope this helps...

[david_david]

Daniel, for your projectile question, I don't think calculations are required

The horizontal distance is doubled, and the horizontal velocity is constant.

implies that the time to hit target is doubled.

vertical displacement=1/2gt^2

displacement is directly proportional to time squared

time doubled, distance quadrupled(x4)

so it would be 8.0cm(80mm)

[Daniel]

A ball is thrown horizontally with a velocity of 20m/s from a cliff. Assuming that air resistance may be neglected, what is the speed of the ball 5 seconds later?

[david_david]

find new vertical velocity.(v=u+at), u=0

horizontal velocity remains the same

use pythagoras theorem to find resultant .

since it speed, no need to find angle

[gohweihan]

To Daniel's question,

A ball is thrown horizontally with a velocity of 20m/s from a cliff. Assuming that air resistance may be neglected, what is the speed of the ball 5 seconds later?

Declaration of constants and given information:

Gravitational acceleration, g = -9.81msˉ?
Vertical initial velocity, u = 0 msˉ?
Time, t = 5s

To find the vertical end velocity, V(v) use the formula

v = u + at

where a = g, because acceleration in this system comes from gravity, and V(v) = v.

So, with V(v) = u + gt

V(v) = (0) + (-9.81)(5)
V(v) = -49.05msˉ?

Negative sign shows the direction of velocity, that is downwards (relative to earth surface). After 5 seconds, the horizontal velocity remains the same, therefore

Horizontal velocity, V(h) = 20msˉ?
Vertical velocity, V(v) = -49.05msˉ?

Use the phytagoras theorem, like david_david said, and you should get

Magnitude of velocity squared, v? = [V(h)]? + [V(v)]?
Magnitude of velocity squared, v? = (20)? + (-49.05)?
Magnitude of velocity squared, v? = 2805.9025m?sˉ?

Speed = magnitude of velocity, v
Speed = √2805.9025
Speed = 52.9708msˉ?

[Daniel]

Thanx all of you. I have a better understanding right now :D

[Daniel]

A tractor of mass 1000kg tows a trailer of mass 1000kg. The total resistance is 4000N and is constant. One quarter of the total resistance acts on the trailer. Initially the acceleration of both the tractor and trailer is 2 m/s/s. Finally both of them move with the same uniform speed of 6 m/s.
What is the force on the tractor along the tow-bar when:-
a)The acceleration is 2m/s
b)The tractor and trailer move at 6m/s

[ElansarGelmir]

A tractor of mass 1000kg tows a trailer of mass 1000kg. The total resistance is 4000N and is constant. One quarter of the total resistance acts on the trailer. Initially the acceleration of both the tractor and trailer is 2 m/s/s. Finally both of them move with the same uniform speed of 6 m/s.
What is the force on the tractor along the tow-bar when:-
a)The acceleration is 2m/s
b)The tractor and trailer move at 6m/s

Hmm... i am not a physics student, and sux in it, but i would like to give it a try... Daniel, dun follow my solution until it's confirmed correct by the pihak-pihak berkuasa (like gohweihan and knkh)...


Total resistance R = 4000N
0.75 R (3000N) on tractor,

a)
so, to accelerate at 2m/s/s,
net ma=Force-resistance
1000(2)=Force-3000N
Force = 5000 N

b)
at uniform speed, a=0 (what's the use of velocity anyway?)

1000(0)=Force-3000N
Force = 3000N



Hmmm... Recommers, is this correct? Not sure, about it. Just take it as a practice.

[david_david]

Lets consider the trailer only

the question tests whether you understand newtons third law.Don't consider the tractor as you dunno the force of the engine driving it foward.

question wants the force on the tractor along the tow bar( connected to the trailer).However, this force is equal to the force the tractor exerts on the trailer to pull it foward( newtons third law)

resistance on trailer=.25x4000=1000N

ma=net force-resistance
1000(2)=f-1000

F=3000N


For the constant velocity, the force equals resistance,=1000N

[Daniel]

David is correct. David you are in form 6 too? your seems like a physics expert.

[chenchow]

For discussion about STPM, please join the Guidance for SPM and STPM Special Interest Group, this thread will be moved in on Tuesday 1am Malaysia time. You won't see this thread if you don't join in.

http://recom.homelinux.org:8000/~recom/modules.php?name=SIG&id=SPM

Click on the top right, Join, to join the SIG.

[Daniel]

A boy throws a ball vertically upwards and later catches it again. The system consists of the ball and the earth. The boy is part of the earth. Before the ball is thrown, the total linear momentum is zero. The gravitational force between the ball and the Earth is an internal force . Therefore, there is no external force and momentum is conserved.... When the ball moves upwards, the Earth moves downwards. Since the mass of the Earth M is very large compared to m, the mass of the ball, the velocity of V the Earth can be neglected. When the ball moves towards the Earth, the Earth too moves towards the ball so that the total linear momentum remains zero.

Sorry for the long texts. The only think which i could hardly understand is what is the relationship the ball with the EARTH? Why suddenly earth involved in this situation?

[wawa]

The only think which i could hardly understand is what is the relationship the ball with the EARTH? Why suddenly earth involved in this situation?

You're viewing the Earth as part of the system. As your text said. the boy is part of Earth.

[Daniel]

Is it got difference between m1v1 + m2v2 = m1u1 + m2u2 and mv + mu = 0 ??

[Daniel]

An object of mass M is projected with a velocity of v from the point P to Q ( 45 degrees from PQ ) Negelcting air resistance, what is the change in momentum of the object when it moves from P to Q?

[knkh_87]

Momentum is a vector. You will notice that the magnitudes at the start and at the end of the projection are equal. The horizontal direction is similar, but the vertical one is different. (When you start, the vertical velocity is upwards, whilst the vertical velocity at the end is downwards).

Try drawing the vector diagram. Your answer should be the (square root of 2) mv.

[Daniel]

Momentum is a vector. You will notice that the magnitudes at the start and at the end of the projection are equal. The horizontal direction is similar, but the vertical one is different. (When you start, the vertical velocity is upwards, whilst the vertical velocity at the end is downwards).

Try drawing the vector diagram. Your answer should be the (square root of 2) mv.

Thanks a lot. Only those vector quantity can used vector diagram to solve its problems?

[Daniel]

1.A car of mass 3000 kg travelling with a velocity of 90km/h collides with a stationary car of mass 2000kg. What is the total kinetic energy of both cars after collision?

2.A radioactive atom which is initially at rest decays into two particles of mass m and M that move with velocity u and v respectively. Calculate the ratio of the kinetic energy of the particle of mass m to the kinetic energy of the particles of mass M.

[Daniel]

Nobody answer me?

[blumarine]

Please help me to solve the question below...

A car is travelling at 13ms-1 towards some traffic lights.When the driver sees the lights change to red,the car is at a distance of 25m from the stop line.The reaction time of the driver is 0.7s and the condition of the road does not permit the car to decelerate greater than 4.5ms-2.with the brakes fully applied,how far from the stop line is the car when it stops?On which side of the line is the car?

[chenchow]

I would say that you should draw a timeline, and then picture the situation.

Initial state: 13ms-1 and 25meter.. 0.7s reaction time, so, that would be 9.1m gone before he reacts. So, he has 15.9 meter to react right?

Max deaccelerate is 4.5ms-2 right? so, each sec, he can deaccelerate 4.5ms-2 right?

so, to go from 13ms-1 to 0ms-1, you can see, he would need nearly 3 second (do the full calculation).

So, with that time and the velocity at each point, you can see whether the 15.9 meter would be sufficient to react...

Try and see, and also let everyone knows what you know and what you don't know. This will help your thinking thoughts.

[Daniel]

Please help me to solve the question below...

A car is travelling at 13ms-1 towards some traffic lights.When the driver sees the lights change to red,the car is at a distance of 25m from the stop line.The reaction time of the driver is 0.7s and the condition of the road does not permit the car to decelerate greater than 4.5ms-2.with the brakes fully applied,how far from the stop line is the car when it stops?On which side of the line is the car?

Are you using Fajar bakti book too?

[chenchow]

You guys can learn together...The more people are learning, the easier it will be. So, bring your friends in.

But I hope that you guys when post questions, can post say what you know and what you don't know...So, you won't be directly spoon-fed. This is more for you guys to collaborate and learn together, kay?

Often, many of you will face the same kind of problems in those topics that are tougher...

[chiunlin]

1.A car of mass 3000 kg travelling with a velocity of 90km/h collides with a stationary car of mass 2000kg. What is the total kinetic energy of both cars after collision?

2.A radioactive atom which is initially at rest decays into two particles of mass m and M that move with velocity u and v respectively. Calculate the ratio of the kinetic energy of the particle of mass m to the kinetic energy of the particles of mass M.

1. Apply Mu=(M+m)v then calculate what is .5(M+m)v^2
2. Apply mu=Mv then calculate (.5mu^2)/(.5mv^2)

[Daniel]

1. An object of mass m falls from rest from the top of a buildings of height h. What is its momentum just before it hits the ground?

2. A body of mass m has kinetic energy E. what is its momentum in terms of m and E.?

[Daniel]

1.A car of mass 3000 kg travelling with a velocity of 90km/h collides with a stationary car of mass 2000kg. What is the total kinetic energy of both cars after collision?

2.A radioactive atom which is initially at rest decays into two particles of mass m and M that move with velocity u and v respectively. Calculate the ratio of the kinetic energy of the particle of mass m to the kinetic energy of the particles of mass M.

1. Apply Mu=(M+m)v then calculate what is .5(M+m)v^2
2. Apply mu=Mv then calculate (.5mu^2)/(.5mv^2)

Thanks chiulin

[gohweihan]

1.A car of mass 3000 kg travelling with a velocity of 90km/h collides with a stationary car of mass 2000kg. What is the total kinetic energy of both cars after collision?

Actually, unless we know whether the collision is elastic or inelastic, it would be hard to come up with a definite answer, as the two would yield different results.

[chenchow]

1. An object of mass m falls from rest from the top of a buildings of height h. What is its momentum just before it hits the ground?

2. A body of mass m has kinetic energy E. what is its momentum in terms of m and E.?

Q1. Conservation of energy. Total Energy (Potential) = Total Kinetic Energy (assumption that no friction). With that you can find your velocity and you can find your momentum = mv

Q2. E = 0.5 mv^2. so, what is v = in terms of E? just flip it.
and Momentum = mv

Hopefully you guys can try yourself and mention what you know and what you don't know... This will enhance your learning...

[chiunlin]

1. An object of mass m falls from rest from the top of a buildings of height h. What is its momentum just before it hits the ground?

2. A body of mass m has kinetic energy E. what is its momentum in terms of m and E.?

1. mgh=.5mv^2. Understand why?
2. Algebraic manipulation. Express v in terms of E and m then substitute it into mv.

[gohweihan]

Please help me to solve the question below...

A car is travelling at 13ms-1 towards some traffic lights.When the driver sees the lights change to red,the car is at a distance of 25m from the stop line.The reaction time of the driver is 0.7s and the condition of the road does not permit the car to decelerate greater than 4.5ms-2.with the brakes fully applied,how far from the stop line is the car when it stops?On which side of the line is the car?

Declare information

Initial velocity, u = 13msˉ?
End velocity, v = 0msˉ?
Maximum deceleration, a = -4.5msˉ?
Distance from line, D = 25m
Reaction time, t' = 0.7s

Note the negative sign for deceleration, as it will be useful during calculation. Also, we derive the end velocity as 0msˉ? because logically, a stopped car has no velocity.

Firstly, consider the distance covered from the time the light changes to the time the driver hammers the brakes

Distance to reaction, x = ut'
Distance to reaction, x = (13)(0.7)
Distance to reaction, x = 9.1m

Secondly, find the distance covered from the time the driver hammers the brakes to the time the car comes to a halt. Consider this formula

v? = u? + 2as

Reverse engineer this formula, and you get

s = (v? - u?)/2a

Distance to stop, s = (v? - u?)/2a
Distance to stop, s = [(0)? - (13)?] / 2(-4.5)
Distance to stop, s = -169 / -9
Distance to stop, s = 18.7778m

Total distance covered to stop, y = x + s
Total distance covered to stop, y = 9.1 + 18.7778
Total distance covered to stop, y = 27.8778m

Distance from line, S = D - y
Distance from line, S = 25 - 27.8778
Distance from line, S = -2.8778m

A negative sign shows that the car stops after the line.

Therefore, the car stops 2.8778 metres after the line.

[Daniel]

1. An object of mass m falls from rest from the top of a buildings of height h. What is its momentum just before it hits the ground?

2. A body of mass m has kinetic energy E. what is its momentum in terms of m and E.?

1. mgh=.5mv^2. Understand why?
2. Algebraic manipulation. Express v in terms of E and m then substitute it into mv.

Actually i am just confused with the g after reading projectiles. I was confused whether to use -g or g. Since questions 1 is a linear motion and consist of one dimension so it would be g instead of -g? am i right?

[Daniel]

Thankx chenchow and chiunlin again

[chiunlin]

1.A car of mass 3000 kg travelling with a velocity of 90km/h collides with a stationary car of mass 2000kg. What is the total kinetic energy of both cars after collision?

Actually, unless we know whether the collision is elastic or inelastic, it would be hard to come up with a definite answer, as the two would yield different results.

Collision with a stationary object should be inelastic. No assumption needed.

To simplify matters, remember that energy is a scalar quantity and there's no need to determine the sign.

[kevinkhoo1986]

1. An object of mass m falls from rest from the top of a buildings of height h. What is its momentum just before it hits the ground?

2. A body of mass m has kinetic energy E. what is its momentum in terms of m and E.?

1. mgh=.5mv^2. Understand why?
2. Algebraic manipulation. Express v in terms of E and m then substitute it into mv.

I understand what daniel is facing. I am also confused with the g too. So the v = (2gh)^0.5 and the answer is m(2gh)^0.5 , but the g is always constant with -10ms^-2 right? if i replace the g with -10, it would be m(-20h)^0.5 and i could not find the answer bcoz the calculator will show sign of "error".

[blumarine]

thnks weihan..i will think again the question then try to solve it before i look at your explanation...
Thanks to chen chow also...

[Daniel]

M1(V1-U1) = M2(U2-V2) -> Momentum conservation
M1(V1^2 - U1^2) = M2(U2^2 - V2^2) -> Total kinectic energy

How did they get V1 + U1 = U2 + V2 from the above equations? The book did not show the step.

[blumarine]

the gravity is -9.81ms^-2 rite?My teacher say we must use the correct measure rite?He say if we use -10ms^-2 in Stpm we wont get mark..Is it correct?

Besides why we have to use -9.81ms^-2 when the object fall free and use 9.81ms^-2 when is in normal situation?Please explain to me....Thanks.... :lol:

[USSDefiantNX74205]

Looks like there are lotsa questions around here and since I'm a student, I'll try to answer these questions as a student. All you physics experts out there please be kind and check to see if my answers are correct ok? Appreciate it....

1.A car of mass 3000 kg travelling with a velocity of 90km/h collides with a stationary car of mass 2000kg. What is the total kinetic energy of both cars after collision?

Velocity of first car before collision = 25m/s
Velocity of second car before collision = 0m/s
Mass of first car = 3000kg
Mass of second car = 2000kg

Since kinetic energy is conserved in an elastic collision (assuming that it is an elastic collision), then:

1/2 X 3000 X 25? + 1/2 X 2000 X 0 = 937500J

2. An object of mass m falls from rest from the top of a buildings of height h. What is its momentum just before it hits the ground?

To find its momentum just before it hits the ground, one must find its velocity just before hitting the ground.

Initial velocity, u = 0
Velocity just before hitting the ground = v (unknown)
Acceleration, a = 9.81m/s?
Displacement, s = h

Using the equation v? = u? + 2as,
v? = 0 + 2(9.81)h
v? = 19.62h
v = 4.43√h m/s

Momentum = mv = 4.43m√h kgm/s

3. A body of mass m has kinetic energy E. what is its momentum in terms of m and E.?

Formula for Kinetic Energy = 1/2 X mv?

Therefore,
E = 1/2 X mv?
v? = 2E/m
v = √(2E/m)

Momentum = mv
Therefore, momentum of mass = m√(2E/m)

[USSDefiantNX74205]

the gravity is -9.81ms^-2 rite?My teacher say we must use the correct measure rite?He say if we use -10ms^-2 in Stpm we wont get mark..Is it correct?

Besides why we have to use -9.81ms^-2 when the object fall free and use 9.81ms^-2 when is in normal situation?Please explain to me....Thanks....

I think we have to use 9.81m/s? because it is more accurate, and your answer must not vary much from the ones given to the examiners. If you use 10m/s?, the answer will be slightly off. In my physics book, it was stated that the value of g anywhere in the book is 9.81m/s? unless otherwise stated.

To your second question, I believe normal situation means under the influence of earth's pull of gravity (9.81m/s?) which is the acceleration of an object in freefall close to the surface of the earth.

[Daniel]

the gravity is -9.81ms^-2 rite?My teacher say we must use the correct measure rite?He say if we use -10ms^-2 in Stpm we wont get mark..Is it correct?

Besides why we have to use -9.81ms^-2 when the object fall free and use 9.81ms^-2 when is in normal situation?Please explain to me....Thanks....

I think we have to use 9.81m/s? because it is more accurate, and your answer must not vary much from the ones given to the examiners. If you use 10m/s?, the answer will be slightly off. In my physics book, it was stated that the value of g anywhere in the book is 9.81m/s? unless otherwise stated.

To your second question, I believe normal situation means under the influence of earth's pull of gravity (9.81m/s?) which is the acceleration of an object in freefall close to the surface of the earth.

But according to my book and my friend told me that the acceleration is constant no matter the object moving upwards or falling down which is = -9.81 m/s/s.

[Daniel]

A students measures the speed of sound as 327.66m/s. He estimates the uncertainty in the result as [+][-] 3%. Which of the following gives the result to the correct number of significant figures?
a) 327 m/s b) 327.7 m/s c) 328 m/s d) 330 m/s

How am i going to calculate it? :roll:

[chiunlin]

M1(V1-U1) = M2(U2-V2) -> Momentum conservation
M1(V1^2 - U1^2) = M2(U2^2 - V2^2) -> Total kinectic energy

How did they get V1 + U1 = U2 + V2 from the above equations? The book did not show the step.

M1(V1-U1) = M2(U2-V2) - (1)
M1(V1^2 - U1^2) = M2(U2^2 - V2^2) -(2)
(1)/(2): V1+U1=U2+V2

[chiunlin]

When calculating questions regarding collisions, apply conservation of linear momentum first before conservation of energy.

As for g, let's go back to the basic.
We learn that force between two objects,
F=G(m1)(m2)/r^2 where G is the gravitational constant, m1 is the mass of the first object, m2 is the mass of the second object and r is the distance between the two objects.

Take m1 as the mass of earth,M=6x10^24kg(if i'm not mistaken)
r as the radius of the earth.
If you calculate GM/r^2, you will get g which is about 9.81m s^-2.
Observe that if any of the variable, M or r changes, then g also changes. If you measure g at Mount Himalaya, it will actually be smaller than the g on sea level.

[chiunlin]

A students measures the speed of sound as 327.66m/s. He estimates the uncertainty in the result as [+][-] 3%. Which of the following gives the result to the correct number of significant figures?
a) 327 m/s b) 327.7 m/s c) 328 m/s d) 330 m/s

How am i going to calculate it? :roll:

327.66x3% approx 10
This implies that the measurement is 327.66+-10.
From the margin of error, the result should be given as 330m/s.

Btw, let me say that physics requires a lot of common sense to master it, you only need to comprehend totally the concepts. If any of you are wondering, my physics is only SPM level.

[Daniel]

A students measures the speed of sound as 327.66m/s. He estimates the uncertainty in the result as [+][-] 3%. Which of the following gives the result to the correct number of significant figures?
a) 327 m/s b) 327.7 m/s c) 328 m/s d) 330 m/s

How am i going to calculate it? :roll:

327.66x3% approx 10
This implies that the measurement is 327.66+-10.
From the margin of error, the result should be given as 330m/s.

Btw, let me say that physics requires a lot of common sense to master it, you only need to comprehend totally the concepts. If any of you are wondering, my physics is only SPM level.

Thank you for replying me. You answer is correct. By the way, i am still do not understand why you use 10 (2 Significant Figure) instead of 9.8298 or 9.83 or 9.8 ? Margin of error?

[Daniel]

And why the answer should be 330 m/s with 2 significant number?

[knkh_87]

Thank you for replying me. You answer is correct. By the way, i am still do not understand why you use 10 (2 Significant Figure) instead of 9.8298 or 9.83 or 9.8 ? Margin of error?

Maximum uncertainties in physical quantities are only calculated to one significant figure. With regard to the example above, I would consider 10 to be one significant figure (it can be considered as one or two significant figures though).

As to why 330m/s is the answer has hardly any relation to the fact of it having 2 significant figures. Uncertainties occur in tenths ('puluhs'-sorry,forgot the English term for it), therefore, it is sufficient to write the value accurate to the 'puluh' place.[/quote]

[knkh_87]

Got these questions from a book on Pre-University Physics. I don't have any definite answers to these, so I hope to gather some views on them.

1. Two men want to break a cord. They first pull against each other. Then they tie one end to a wall, and pull together. Is either procedure better than the other?

2. Explain carefully what is meant by potential energy. Why is it considered to be a property of a system rather than a body?

3. A man on a sheet of smooth ice sets himself in motion by throwing successively his two boots, each of mass m, in the same horizontal direction with velocity v relative to himself. Calculate the man's final velocity if his mass without his boots is M. ( I think you can express this in terms of m,v, and M)

[chiunlin]

1. The second procedure is better. Think of the tension in the cord.
2.Definition of potential energy: Energy of a system because of its state or position.
3. Apply conservation of momentum and do some algebraic manipulation to get the answers.

[Daniel]

1)A block of mass m is pushed off with an initial speed u so that it moves up a rough plane inclined at an angle Q to the horizontal. The frictin between the block and the plane is R. Deduce an expression for the velocity v of the block after it has moved a distance d a long the inclined plane. Explain any additional symbols in your expression.

2)The initial kinectic energy of an object moving on a horizontal surface is K. Friction between the object and the surface causes the velocity of the object to decrease uniformly to zero in time t. What is the kinetic energy of the object at time = t/2

Thank you in advance

[chiunlin]

1) Apply v^2=u^2+2ad. Now the problem here is to calculate a.
Total force on the block=Friction+weight of block along the plane.

2.v=initial speed. Calculate what is the velocity in terms of v when time=t/2. Calculate the kinetic energy and expressed it in terms of K.

[Daniel]

I dont get the second one.. :?:

[gohweihan]

What chiunlin said for the second is to use v to symbolize the initial velocity. Since you know the end velocity is 0m/s, and it takes the time of t to slow from v to 0m/s, you can derive the deceleration from them. From that deceleration, you can then derive the velocity of the object at t/2 expressed in v.

Using this velocity at t/2, you can then get the kinetic energy of the object at that time and express it in K.

[Daniel]

heheheh.... Thankx gohwehan and chiulin, i got the answer already. I feel so stupid surrounded by so many experts in here.

[Daniel]

This is math T question, a^2 = b^2
can i eliminated the square for both of it and become a = b , my friends insisted it was a wrong method but i think it should be alright.

[gohweihan]

This is math T question, a^2 = b^2
can i eliminated the square for both of it and become a = b , my friends insisted it was a wrong method but i think it should be alright.

Firstly, perhaps should start another thread on Mathematics T.

As for your question, it may not seem obvious at first, but you would not get a complete set of answers if you cancel out the square from both ends of the equation.

If you cancel out the square, you would be doing something like

a? = b?
a = b

And you only get one answer, that is a = b.

But now, consider this

a? = b?

Shift b? over to the left, and

a? - b? = 0

Expand the equation, and you get

(a + b)(a - b) = 0

? a = b, a = -b

Therefore, your final answer would be a = b and a = -b.

You see, by cancelling out the square, you are eliminating another set of possible answer. Any power numbers should not be eliminated at first sight, as it may (or at times may not) come into play at a later stage of the question. In this case, it is proven that you cannot eliminate the square in the beginning of the question.

[Daniel]

The initial velocity of a particle is 15 m/s a long the ox direction. An instant later, its velocity is 15 m/s at an angle of 60 degree to the ox direction. Find the magnitude and directin of the change in velocity of the particle.

I know the answer would be 15 since it would be triangle form with 60 degrees, but i do not know how to do the calculation. (langkah kerja)

[chiunlin]

To understand the question better, draw a scaled vector diagram.
Change in velocity=v-u=v+(-u).
Then draw the resultant vector.

[david_david]

1)A block of mass m is pushed off with an initial speed u so that it moves up a rough plane inclined at an angle Q to the horizontal. The frictin between the block and the plane is R. Deduce an expression for the velocity v of the block after it has moved a distance d a long the inclined plane. Explain any additional symbols in your expression.




can you consider the energy and work involved :?: :?:

initial kinetic energy=potential energy gained+work done against friction+final kinetic energy

[Daniel]

Help me to solve these three questions. Thank you :wink:
http://server6.uploadit.org/files/danielseliong-PhysicsQ.JPG

[Daniel]

http://server6.uploadit.org/files/danielseliong-physciss.JPG
How to solve these two questions?

[Daniel]

What does Coefficient mean in Physics???
hence, what is the meaning with this phrase "the coefficient of friction between the block and the table is 0.4" . There is no unit for coefficient?

[pandaboy]

What does Coefficient mean in Physics???
hence, what is the meaning with this phrase "the coefficient of friction between the block and the table is 0.4" . There is no unit for coefficient?

The coefficient of friction has no unit, it is called the "miu", rite? Correct me if I'm wrong. My brain kinda karat already after one month not touching any book.

Erm..coefficient....wat is it called in BM? Suddenly I cant remember. Lets see what the dictionary has to say about this.

Main Entry: co?ef?fi?cient
Pronunciation: "kO-&-'fi-sh&nt
Function: noun
Etymology: New Latin coefficient-, coefficiens, from Latin co- + efficient-, efficiens efficient
1 : any of the factors of a product considered in relation to a specific factor; especially : a constant factor of a term as distinguished from a variable
2 a : a number that serves as a measure of some property or characteristic (as of a substance, device, or process)

Is it pemalar in BM? Correct me if I'm wrong again. :wink: I seriously think I'm wrong here. Cant find the word.

Wow, I really need to revise soon. I forgot everything I learnt in the past 1 and a half years... OMG...

[pandaboy]

Question 4:
Since it's an elastic collision, so the total kinetic energy is conserved.

Taking u2 and v2 as final velocity for P and Q respectively,

(1/2) m (u + v)^2 = (1/2) m (u2 + v2)^2

This gives :
(u + v)^2 = (u2 + v2)^2

when P comes to rest , u2 = 0
(u + v)^2 = (v2)^2

So v2 is equals to the positive and negative values of the square roots of (u + v)^2, which gives

(Ans) = -(u +v) or (u +v)

Seriously, correct me if I'm wrong! hehe....

:oops: Pai seh la if wrong....

[teazo_darren]

help!!!
I am STRUGGLING with physics calculation parts....dunno how to deal with it....wat cos, sin and there and that......aihzzz!.HELP!.who can enlightening me.plszzz........~.thanks!

[caffe_mocha]

physic seems to be easy to understand but it's very difficult when it comes to the ques....i always look at the ques but dunno how to solve it........hai......

[wawa]

try to find the relations between those questions you are answering. Don't view them seperately.
Then you'll find out that they'll all asking the same thing. It takes some time to arrive there.
IF you can't answer the questions, it means you have not grasp or understand properly the essence of concepts. There's no such thing as understand but cannot answer questions.

[chiunlin]

If you can't solve the questions, that means you haven't grasp completely the concepts. However, there are times when the questions involved a higher level of mathematics and you may not know how to progress. Anyway, before solving any questions, ask yourself these questions:
i) What does the question asked for?
ii)What concepts do i need to know in order to answer this question?
iii) What are the information given what is missing in order to arrive at the answer?
iv) What do i need to do in order to obtain the missing information?
v)Lastly, how to proceed with all the information to get required answer
Not to forget that you need to check your answers.

[Daniel]

I have a very lousy physics teacher which just gives us endless notes without explaining it. How to solve both of this questions?
1)Particles of mass 0.2kg and 0.5 kg are at the 20 cm and 66 cm mark respectively on a uniform metre rule of mass 0.3 kg. Find the position of the centre of mass of the system.

2)A uniform metal plate which is made up of a square of sides 20 cm and an equilateral triangle. Fine the position of the centre of mass of the metal plate.

[yekban81]

I have a very lousy physics teacher which just gives us endless notes without explaining it. How to solve both of this questions?
1)Particles of mass 0.2kg and 0.5 kg are at the 20 cm and 66 cm mark respectively on a uniform metre rule of mass 0.3 kg. Find the position of the centre of mass of the system.


The rule for this question: The sum of momen at the centre of the mass must be equal to zero. Imagine the line is the ruler and each masses includes the ruler mass(at the centre of the line) is located as follow. Assume x as the center of the mass. Based on this diagram, the sum of momen at x is equal to zero if x is the center of the mass.

___0.2g______0.3g__0.5g________x

0=SUM Momen at x
0=0.2g(x-0.20) + 0.3g(x-0.50) + 0.5g (x-0.66)



2)A uniform metal plate which is made up of a square of sides 20 cm and an equilateral triangle. Fine the position of the centre of mass of the metal plate.


Use the same concept but with different approach; use area (m2) as mass(kg).

[Daniel]

:lol: Gosh.... i finally get it. thanx a lot yekban81. By the way, are you in form 6 too?

[yekban81]

I never take STPM. I am pursuing master by research in UTM, Skudai.

[Daniel]

just 23 years old already pursuing master? I am lucky indeed to have a master teaching me right now :)

[Daniel]

A spring with a load moving up and down. Why we say that the system are performing SHM? And why we need to relate it with w, that is the angular velocity? The spring is just moving up and down and does not perform any circular motion right? But why we need to relate it with angular velocity? Can someone explain it further?

[yekban81]

SHM-Single Harmonic System..I forget liao the definition. Is there any other type of harmonic system that you learn in your STPM Physics?

The motion of spring with load is characterised by using the displacement and time. Let us assume that this motion is a perfect harmonic motion; no damp effect, no friction. Hence this spring moves up and down at a similar amplitude all the time in a certain frequency.

When you plot the displacement and time in a single graph, you get a curve which is very similar to the sin/cos curve. In fact it is in the form of equation with sin/cos function. w is the value of theta per time in radian per second, thus it is named as "angular velocity".
From the equation:
S=Asin(wt) where wt=theta

To me, it is just a constant value in a sin/cos equation which represent the motion of a particular spring with a particular load. w is a parameter created to express theta in terms of time, t.

[cooldownguy86]

i have a question guys. hope you can help.

State Newton's second law of motion. (no problem)
When an object moves through a fluid it experiences a retarding force due to turbulence.
For a sphere of radius r moving with a speed v in a fluid of density p, the retarding force is given by F = kpr^2v^2, where k is a constant.

By relating the retarding force to the transfer of momentum between the sphere and the fluid, explain why the force F is directly propotional to pr^2v^2

it's hard...i can't understand it. thanks in advance

[doodle]

Let me try...

The movement of the sphere in fluid displaces the fluid. Force is defined as the rate of change of momentum, F = d(mv)/dt. Since velocity is constant, then F = v.dm/dt. The mass of fluid displaced when the sphere moves x distance is m = p.V = p.pi.r^2.x. Thus F = v.p.dV/dt = v.p.pi.r^2.dx/dt = pi.p.r^2.v^2 since v = dx/dt.

[bush]

A spring with a load moving up and down. Why we say that the system are performing SHM? And why we need to relate it with w, that is the angular velocity? The spring is just moving up and down and does not perform any circular motion right? But why we need to relate it with angular velocity? Can someone explain it further?

SHM is an occillation or motion such that the acceleration of the system is directly proportional and oppositely directed to its displacement.( acceleration is always directed to its equilabrium point)

for omega w of a SHM, it is called angular frequency, and not angular velocity though they have the same units and dimensions, angular velocity is used in circular motion separately. its just terms......terms to screw up the life of a student

[Mark]

A car is travelling at 13 m/s towards some traffic lights..when the driver sees the lights change to red,the car distance is at a distance 25m from the stop line. The reaction time of the driver is 0.7s and the condition of the road does not permit the car to decelerate greater than 4.5m/s^2. With the brakes fully applied,how far from the stop line is the car when it stops? On which side of the line is the car?


Answer:2.9m after the stop line

[littlebigone]

i used a velocity-time graph to solve this question.

velocity times time gives you distance, therefore area under the curve that plots velocity will give you the distance travelled.

in this problem, i assumed constant deceleration, therefore time taken to get from 13 to 0 is 2.888 ~ 2.9 secs (13/.5)

Since we assume constant deceleration, we can easily imagine that a straight line graph with negative slope, with y axis plotting velocity and x axis plotting time. the line cuts the axis at (0,13) and (2.9,0), so we have right angle triangle with sides 13 and 2.9. The area of this triangle is 18.85 = 0.5 x 13 x 2.9.

But the driver takes 0.7 secs to react and so travels 9.1 m before starting to brake.

The total distance travelled is 27.95 m so the car stops 2.95 meters after the stop line. I think the errors are the result of rounding.

Let me know if this solution works for you.

[Mark]

thx littlebigone...i think you have done it correctly..

[yangyang]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

[lyt87]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

1a) v=4m/s s=6
s=vt, 4t=6
t = 1.5

b)no.

[bush]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

1a) v=4m/s s=6
s=vt, 4t=6
t = 1.5

b)no.

i don't think it's correct. The bus is also moving, accelerating to be precise.

[lyt87]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

1a) v=4m/s s=6
s=vt, 4t=6
t = 1.5

b)no.

i don't think it's correct. The bus is also moving, accelerating to be precise.

so, u mean the door is the bus door?
isee, so is it vt= 6 + 0.5 at^2

[bush]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

1a) v=4m/s s=6
s=vt, 4t=6
t = 1.5

b)no.

i don't think it's correct. The bus is also moving, accelerating to be precise.

so, u mean the door is the bus door?
isee, so is it vt= 6 + 0.5 at^2

thats what the question says

[bush]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

1a) v=4m/s s=6
s=vt, 4t=6
t = 1.5

b)no.

i don't think it's correct. The bus is also moving, accelerating to be precise.

so, u mean the door is the bus door?
isee, so is it vt= 6 + 0.5 at^2

thats what the question says

[Nelson]

hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx :)

1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2
a.) How long does it take for the man to gain the door?
b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up?


If possible, pls show da working 4 me...thx u...

I use some mathematical methods to solve it:

For (a),
the displacement for the man to gain the door,
s=vt-6
Substituting v=4,
s=4t-6 ----------(1)
the displacement of the door,
s=(1/2)(at^2)
Substituting a=1.2
s=0.6t^2--------------(2)
Comparing (1) and (2),
4t-6=0.6t^2
20t-30=3t^2
3t^2-20t+30=0
t={-(-20)?√[(-20)^2-4(3)(30)]}/2(3)
t=(20?6.32)/6
t=(20+6.32)/6 or t=(20-6.32)/6
t=4.39 or 2.28
the time is 2.28 or 4.39

For (b)
displacing 6 with 10 in (1),
s=4t-10---------(3)
Comparing (2) and (3),
4t-10=0.6t^2
20t-50=3t^2
3t^2-20t+50=0
b^2-4ac
=(-20)^2-4(3)(50)
=-200≤0
and the equation 3t^2-20t+50=0 does not have roots
Hence the man will not catch up the bus.

[miracle]

i'm new member in this group...nice to meet all of the physics expert...i have 1 question wanna ask....

an aeroplane which is flying with a horizontal velocity of 50m/s at a height of 300m releases a parcel when it is vertically above a point X on the earth's surface.
a)the time of the flight of the parcel
b)velocity of the parcel just before strikingthe earth's surface
c)the distance between the point X and the point where the parcel strikes the earth's surface

tq....

[DecentMerson]

i'm new member in this group...nice to meet all of the physics expert...i have 1 question wanna ask....

an aeroplane which is flying with a horizontal velocity of 50m/s at a height of 300m releases a parcel when it is vertically above a point X on the earth's surface.
a)the time of the flight of the parcel
b)velocity of the parcel just before strikingthe earth's surface
c)the distance between the point X and the point where the parcel strikes the earth's surface

tq....

i don't think this is a trick question rite??
just keep in mind that to find the vertical velocity, u can ignore the given horizontal velocity...

for a) we need to find t, given s, u, and a...

so, by using s = ut + .5at^2, where s = 300, u =0, and a = 9.81 (or 10), solve for t.

for b), this is a bit tricky... since velocity is a vector, i assume that u need to find the resultant velocity...
horizontal component is given = 50,
vertical component is v = u + at = at from part a (u is zero for part a)

take the square root of the sum of 50^2 and (vertical velocity)^2...
U'll need to find the angle of the velocity too... use tangent to find the angle...

for c) this is just (50 X t) because again, horizontal velocity and vertical velocity can be look at separately...

wow... can't believe that I still can solve Physics question(hopefully i didn't mess up)... it's been at least 15 months since I touch any Physics stuff...

[Nelson]

i'm new member in this group...nice to meet all of the physics expert...i have 1 question wanna ask....

an aeroplane which is flying with a horizontal velocity of 50m/s at a height of 300m releases a parcel when it is vertically above a point X on the earth's surface.
a)the time of the flight of the parcel
b)velocity of the parcel just before strikingthe earth's surface
c)the distance between the point X and the point where the parcel strikes the earth's surface

tq....

I think the solution is as follows:

For a),
we use s=(1/2)(at^2) where s=300, a=g=9.81
300=(1/2)(9.81t^2)
t^2=61.16
t=7.82s

For b),
given that horizontal velocity=50m/s
Let vertical velocity=v
v^2=u^2+2as where u=0, a=g=9.81, s=300
v^2=0^+2(9.81)(300)
v^2=5886
v=76.72m/s
velocity of the parcel
=√(76.72^2+50^2)
=√8368
=91.58m/s
the angle, θ
=tanˉ?(91.58/50)
=tanˉ?1.83
=61.37?
the velocity is 91.58m/s which is 61.37? from the horizontal line.

For c),
said that when the plane is vertically above point X, it releases the parcel
the parcel needs 7.82s to strike the ground
before that the distance it passes
=(horizontal velocity) x (time needed to strike the ground)
=50x7.82
=391
Hence the distance between point X and the point where the parcel strikes the earth's surface
=the distance which the parcel passes
=391m

Is this the solution?

[miracle]

i have 1 question wanna to ask....

a hose ejects water at a speed of 20cm/s through a hole of area 100cm^2.if the water strikes a wall normally,calculate the force on the wall,assuming the velocity of water normal to the wall is zero after collision. (ans=0.4N)

[miracle]

still 1 question wanna asked...

a wind travelling at a uniform velocity of 50m/s strikes a rigid wall which is perpendicular to the direction of the wind.estimate the pressure exerted by the wind on the wall. (density of wind=1.25kg/m^3) (ans=3125Pa)

i got questions to ask. in centripetal force.\
we know that, R cos (angle)= mg
then, can mg cos (angle)= R?

[kucingbiru]

no, mg/cos(angle) = R

[kucingbiru]

i have 1 question wanna to ask....

a hose ejects water at a speed of 20cm/s through a hole of area 100cm^2.if the water strikes a wall normally,calculate the force on the wall,assuming the velocity of water normal to the wall is zero after collision. (ans=0.4N)

i'm not sure if u've seen this equation.

F=(m_dot)*(V) where m_dot is the rate of mass of the fluid passing through the exit.

Volume_dot is the rate of volume of the fluid passing through the exit.

Volume_dot = V * A
= 0.2m/s * 0.01m^2

Volume_dot = 0.002 m^3/s

but we need mass_dot. mass is volume X rho (density of water).

let rho= 1000kg/m^3

thus, mass_dot = Volume_dot * rho
= 0.002m^3/s * 1000 kg/m^3

mass_dot = 2kg/s

F = mass_dot * V
= 2 kg/s * 0.2 m/s = 0.4 kg.m/s^2

F=0.4 N

[kucingbiru]

still 1 question wanna asked...

a wind travelling at a uniform velocity of 50m/s strikes a rigid wall which is perpendicular to the direction of the wind.estimate the pressure exerted by the wind on the wall. (density of wind=1.25kg/m^3) (ans=3125Pa)

similar to the previous question, but this time we donno the area of the front of the wind. so we also donno the area of the wall that got hit by the wind. but the areas should be the same.

let the area be A.

F= (mass_dot)* V and pressure, P = F/A

mass_dot = rho*Volume_dot

Volume_dot = A * V = AV

so mass_dot = rho*AV

so F=rho*AV * V = rho *A*V^2

P = F/A = (rho*A*V^2)/A

P= rho*V^2

= 1.25kg/m^3 * 50m/s * 50m/s = 3125 kg/m.s^2


P= 3125 Pa.

[youngyew]

From a redundant thread (http://recom.org/modules.php?name=Forums&file=viewtopic&t=2891).

hie.. im in lower six now.. i have difficulty in physics especially the topic that my teacher is teaching now.. projectile.. it seems that i can understand wut is being explained in the book..but when it comes to trying out questions.. i can barely apply wut i knew bout that topic.. i cant even start.. ne advice? thanks..
post a hw problem. maybe we can discuss it together.

[kucingbiru]

mod: you might want to rename this thread to "Physics SPM & STPM" or create one dedicated to spm physics.

[littlepiglet]

i got problems with a pyhsics' question.

A tow-truck uses a rope to pull a car of mass 1000kg. the tow-truck accelerates slowly from rest with the two-rope slack. when the rope becomes taut the car starts to accelerate and the tow-truck moves with a constant speed of 0.8m/s until the car achieves the same speed.

a) what is the change in momentum of the towed car?
b) during acceleration of the towed car, the tension in the rope is 4000N. How far does the towed car travel from rest before it achieves the speed of 0.8 m/s?
c) what is the distance travelled by the tow-truck during the time?

can anyone help me to solve te qeustion? thank you

[kucingbiru]

i got problems with a pyhsics' question.

A tow-truck uses a rope to pull a car of mass 1000kg. the tow-truck accelerates slowly from rest with the two-rope slack. when the rope becomes taut the car starts to accelerate and the tow-truck moves with a constant speed of 0.8m/s until the car achieves the same speed.

a) what is the change in momentum of the towed car?
b) during acceleration of the towed car, the tension in the rope is 4000N. How far does the towed car travel from rest before it achieves the speed of 0.8 m/s?
c) what is the distance travelled by the tow-truck during the time?

can anyone help me to solve te qeustion? thank you


a) m=1000kg, V1=0, V2=0.8m/s
change of momentum, dp = p2 - p1 (p= momentum, mv)

p1 = mV1, p2 = mV2

so dp = 800 kg m/s




b) the pulling force, T=4000N
mass of car, m =1000kg

so, the acceleration of the car is, a=T/m = 4000N/1000kg = 4m/s^2

V1=0m/s
V2=0.8m/s <--initial and final velocities of the car

you should know this function.
V2^2 = V1^2 + 2as
where V1 is initial velocity, V2 final velocity, a acceleration and s displacement.

rearranging this funciton, we can get s as a function of those.

s = (V2^2 - V1^2 ) / 2a

plug in all values u know. u'll get s= 0.08m i guess.


c) the tow-truck starts at rest. it starts moving and at some point it moves at a constant velocity, so at the time its acceleration A = 0.

when A=0, the tow truck moves some more at 0.8m/s for some distance until the car reaches 0.8m/s.

so, the tow truck moves for some distance D at a constant speed for some time t.

D= Vt <--both D and t are unknowns. but the time the truck takes to travel at constant speed should be the same as the time the car takes to accelerate to 0.8m/s.

now go back to the car to get the time t. we know the acceleration of the car, a=4m/s^2, we know the initial and final velocities V1 and V2.

u should know this function too. V2 = V1 + at

rearrange this equation, t= (V2-V1) / a

t = (0.8 - 0)/4

t= 0.2s


plug it in to D=Vt to get the distance the truck traveled,

D = 0.8 m/s* 0.2s

D=0.16m.




p/s: i'm not sure if any of these is correct. i'm just trying to make myself sleepy.

[littlepiglet]

thank you so much, kucingbiru. now i know why the distance for c is 0.16m

[kucingbiru]

a trolley move at constant v 3ms-1 on a smooth horizontal surface.if sand is dropped perpendicular on the trolley at the rate of 0 kg per minute,what is the horizontal force required to maintain the movement of the trolley at constant v 3 ms-1

Variable mass problem (if it's really variable)***.

This is the force equation for variable mass

Sigma F = mv' + m' u ...(1)

where m is the mass of the object, v' is its acceleration, m' is the rate of change of mass of the object and u is is the relative velocity (relative to the object) of the particle that's being added/removed.

Please let me know if you're not familiar with this equation. I dont think i learnt it in elementary physics.

So, we want horizontal force. let's just rewrite equation (1) for x component (let x be horizontal).

Sigma F_x = m (v_x)' + m'(u_x) ..... (2)
where (v_x)' is the x-component of the acceleration of the trolley, and (u_x) is the x-component of the relative velocity of the particle added/ejected.

now, we donno the mass of the trolley, so we donno m. BUT, it says that we're trying to keep the velocity of the trolley constant, thus its acceleration is 0. so,
(v_x)' = 0 so we can cancel m from the equation.

we can now reduce equation (2) to

Sigma F_x = m' (u_x)

Recall that u_x is the x-component of the velocity of the sand relative to the trolley. the sand is dropped vertically (let's just say it's vertical relative to a non moving non rotating coordinate system). Let the absolute velocity (again, relative to that static coordinate system) of the sand is S. S only has a y-component. the x component of S, S_x =0.

equation of relative velocity,
V_b = V_a + V_b/a
where V_b and V_a are absolute velocities of b and a respectively, and V_b/a is the velocity of b relative to a.

since u_x is the velocity of the sand relative to the trolley,
u_x = v - S_x where v is the velocity of the trolley.

recall that S_x = 0. so

u_x = v - 0

u_x = v = 3 m/s

Sigma F_x = m' (u_x)

plug in the value of m' to get the answer.

***[if m' is 0 (as you stated in your question), thn Sigma F_x is 0. or in other words, if u dont change the mass, the trolley will move at a constant velocity without having to apply any force (simple Newton's first law interpretation, you dont need to go this far to get the answer :P).]]

[Daniel]

geometrical optics is killing me! i could barely understand the whole topic. The -ve and +ve sign is the sole killer.

1)1/u+1/v=1/f
when the U is -ve, what does it actually mean? it mean it's the object on the left hand side the P? or? What about when v is -ve?

2)How to determine whether the image form from the first refraction is virtual object or real object?

3)How to determine an image is "nyata or maya"?

Please enlighten me.... Tq

[chiunlin]

geometrical optics is killing me! i could barely understand the whole topic. The -ve and +ve sign is the sole killer.

1)1/u+1/v=1/f
when the U is -ve, what does it actually mean? it mean it's the object on the left hand side the P? or? What about when v is -ve?

2)How to determine whether the image form from the first refraction is virtual object or real object?

3)How to determine an image is "nyata or maya"?

Please enlighten me.... Tq

The sign rule for geometric optics
i) U or distance of the object is +ve when it is at the same side as the incoming ray. -ve otherwise.

ii) V or distance of image is +ve when it is at the
same side as the outgoing ray.

iii) f, focal length, same as V.

So to answer your questions, U can only be -ve in a compound system(a system with more than one lenses/mirrors). Example: Imagine a system with two lenses 20 cm apart. Let's say the object is to the left of the first lens, and the image of the object after passing through the first lens is 30 cm to the right of the first lens(10cm to the right of the second lens). The image serves as the object of the second lens, but since it is at the opposite side of the incoming rays of the second lens, we take U to be -ve.

Similarly, when V is -ve, it means that the image is formed at the opposite side of the outgoing rays(diverging), and the image formed is always virtual( V +ve means real image, -ve means virtual image)

[kucingbiru]

The sign rule for geometric optics

that depends on the convention adopted by the book too right?

[chiunlin]

The sign rule for geometric optics

that depends on the convention adopted by the book too right?
Yeah,I forgot about that, different authors have different rules. The stated rules in my previous posts are those I find to be the easiest to remember.

[Confused]

Can anybody explain to me when should we consider Conservation of Momentum and Conservation of Kinetic energy in the calculation?

[youngyew]

From a redundant thread (http://recom.org/modules.php?name=Forums&file=viewtopic&p=68014#68014).

momemtum is always conserved in all collisions.

KE is only conserved when the collision is elastic.

a free body force diagram is a diagram which shows all forces that act on that body.......

[youngyew]

There are very few examples of totally elastic collisions in macroscopic scale. The only perfect elastic collisions are the ones that happen in vacuum (so that the energy doesn't get transferred to other non-involving molecules) and in which the involved bodies doesn't get deformed in the process. The trivial example of totally elastic collision would be the collision between air molecules. Snooker ball collisions, although may be near elastic, is still far from perfect because deformity does happen to the ball at the instance of contact, and sound energy (which actually is the vibration of air molecules) is produced during the collision.

Therefore, you have to be sure that all forms of energy is accounted for before you apply the law of conservation of energy. Or else, it's very probably that we will end up with ridiculous answers due to the "missing energy".

[bush]

i doubt we will need to consider all these as quantitative questions asked at STPM/A-levels require u to assume such things as insignificant.

[kucingbiru]

i doubt we will need to consider all these as quantitative questions asked at STPM/A-levels require u to assume such things as insignificant.

yeah, but one should state his assumptions clearly.

[Sam]

A ball is thrown with a velocity U at an angle Q to the horizontal.
a) Find in terms of U,Q and g the maximum height reached by the ball
b) At what angle should the ball be thrown so that its maximum height reached is the same as its horizontal range?

I dont get the B)

It just means that the resultant force is in equlibrium

[Daniel]

Hmm..... what is the difference between:
1)Tensile force, tensile strength, and tensile stress and what is the meaning of tensile itself??

2)I could not differenciate elastic limit and proportional limit...

please enlighten me. TQ

[bush]

Hmm..... what is the difference between:
1)Tensile force, tensile strength, and tensile stress and what is the meaning of tensile itself??

2)I could not differenciate elastic limit and proportional limit...

please enlighten me. TQ

Fancy names when dealing with materials.....

didn't like this option, chose nuclear and particle physics.

[kucingbiru]

Hmm..... what is the difference between:
1)Tensile force, tensile strength, and tensile stress and what is the meaning of tensile itself??

2)I could not differenciate elastic limit and proportional limit...

please enlighten me. TQ

compression to compressive stress is tension to tensile stress. get it?

[Nelson]

Hello, I need some help here!

A turntable rotates at 45 revolutions per minute and then at 40/3 revolutions per minute. What is the ratio of the accelerations of a point on the rim of the turntable? (ans:1.82)

Could anyone please help me.

[Daniel]

A solid sphere of mass m is launched so that the sphere will slide along a flat surface with an initial velocity of u. After a short lapse of time, the sphere starts rolling without slipping on the surface. At the instant when the sphere starts rolling without slipping:
a) Show that the velocity of the sphere is 5u/7.

I just don?t know why the energy is not conserved and the angular momentum is conserved (shown by the answer). The frictional forces on the surface will produced a torque on the sphere since it is not acting on the rotational axle right? Then the angular momentum should not be conserved. Please enlighten me.

[windrunner]

Hmm..... what is the difference between:
1)Tensile force, tensile strength, and tensile stress and what is the meaning of tensile itself??

2)I could not differenciate elastic limit and proportional limit...

please enlighten me. TQ

Tensile force is a force, a STRONG thingy has a high max tensile force before a thing breaks, stress is the ratio of force over cross sectional area, tensile means pull, or expand. (something like compress to push)

Elastic limit is the max force the object returns to original size and shape.
It might not be a straight line!!!! There would be a slight curve after proportional limit.

Proportional limit is like hooke's law, it is the limit before elastic limit. It is a straight line in the Force-extension graph.

[kucingbiru]

a STRONG thingy has a high max tensile force before a thing breaks

it's usually called tensile strength

[confused-freaker]

I got a question to ask...its from phy past year stpm.....

question (b)......i compared answers from longman stpm 5 year series book and the pre-u stpm phy...the answer n working is different...and i dont understand the working for both...here are the question n answers:

http://server3.uploadit.org/files/chickens-phyQ.jpg

http://server3.uploadit.org/files/chickens-ans1.jpg

http://server2.uploadit.org/files/chickens-ans2.jpg

[reign226]

The correct way to do this is the first method. If you see the second method, the answer is nonsensical because it's 1.07m, while the length of the air wedge is only one tenth of that figure.

Anyway, there are two ways to do this.

a) Using the formula 2t + 0.5[lambda] = n[lambda]

rearrange to form:

2t= n[lambda] - 0.5[lambda]

Now we have two equations, one with Na for the green light and Nb for the red light. Since thickness is the same, we can equate the two formulas getting:

Na x [lambda for a] - 0.5[lambda for a] = Nb x [lambda for b] - 0.5[lambda for b]

With me so far? Okay we need another equation so we can solve them. Since the wavelength for A is smaller for B, we can expect there to be a larger number of fringes for A (smaller wavelength = closer separation for fringes = more fringes will be formed).

And since yellow light is the superposition of maxima from both lights, we can assume that:

Na - Nb = 1.

That is to say, there is only a difference of ONE maxima for the two light's maximas to coincide on the same spot. This is the hardest leap of logic IMO that you have to make, so try to understand and visualize this in your brain.

Anyway once we have this two formulas just replace the second one into the first to get rid of either Na or Nb (your choice), then do the math and you'll find that Na = 5.0007 (something like tat).

Okay, now the final thing to do. How far is the 5th bright fringe from O? There's actually only 4 bright fringes if N = 5 because the bright fringe starts with N = 1 (page 355 of the blue book). Also, we cannot ignore the fact that there is 0.5 of a dark fringe between the first bright fringe and O. Therefore, we have a distance of 4.5 'strips' or 4.5n.

To get N simply use the formula [lambda] / [2 tan(theta)] then multiply by 4.5 and you'll get the (sensible) answer.

Phew. The only leaps of logic that I had to take was that bright fringes start counting from 1. I kept multiplying by 5.5 and not getting the answer as a result until I consulted the book. I would say this question is quite easy if you know how to correctly get the first two formulas (which in itself isn't hard. those are the formulas u have to memorize anyway) and to realize that for bright fringes they start from N = 1 and that there is half a dark fringe in front.

[reign226]

I don't think the answer is right for the turntable question. You sure you got the numbers right? One spins at 45RPM, the other at 13.33 RPM and the ratios of their acceleration is only a measly 1.82??

the formula used is a = r[omega]^2

since radius is the same, if I divide the two equations i get:

a1/a2 = [omega for 1]^2 / [omega for 2]^2

1 is 45RPM, 2 is 13.33RPM.

[omega] = 2[pi]/[period]

Try as i might I keep getting 11.39.

[Nelson]

I don't think the answer is right for the turntable question. You sure you got the numbers right? One spins at 45RPM, the other at 13.33 RPM and the ratios of their acceleration is only a measly 1.82??

the formula used is a = r[omega]^2

since radius is the same, if I divide the two equations i get:

a1/a2 = [omega for 1]^2 / [omega for 2]^2

1 is 45RPM, 2 is 13.33RPM.

[omega] = 2[pi]/[period]

Try as i might I keep getting 11.39.

Thank you reign226 for reminding me this question. Actually the value is not 40/3 but is 100/3. It's a typing error. Feeling very sorry for everyone who has tried this question.

[knkh_87]

Okay, brain teaser. :D Not an STPM question, but more of an application of STPM work.

For those who have visited Sunway Lagoon, you will understand these better:

(1) When you are on top of a flume ride or in a position right before you slide down a steep slope on a roller coaster, you will momentarily feel weightless. Using a free body diagram, explain this situation.

(2) How would a person in the situation in (1) prove that he/she is weightless.

(3) Sunway Lagoon has the longest suspension bridge (in a theme park) in the world. Why is it very difficult to estimate the tension in its main cables?

[reign226]

Lemme have a go:

1) On a flume ride going up, you're basically being thrown upwards like a ball. You move up with a constant velocity (let's say) so that means there is an acceleration opposite of the earth's gravity. At the top, this acceleration disappear but just before it does, you basically have no resultant acceleration hence the weightlessness.

2) Uhh... carry an accelerometer? Dunno this one. But I know all the males will know it instantly in ther, uhh, bebolas. ^^

3) Hmm, interesting, I didn't know that. I assume using Young's Modulus, I get F = EAe/l and since L is so goddamned big, the % of error of the final answer becomes significant. Not to mention it's hard to measure the elongation of such a bridge.

[bush]

the suspension bridge.......people walking around.....tension won't be constant.......

[Tasslehoff]

1) concept of Inertia.
2) Have a weighing scale.. perhaps seated on it, but it'll be dangerous..
3)no idea about the height and the location of the suspension bridge, but perhaps the wind could play a factor. *remembers a particular clip of a suspension bridge being ripped and heaved a few hundred meters by a hurricane* depending on where the direction of the wind, the tension in the cables varies as the wind rocks the bridge.

[okaywhy]

1. Inertia is the resistence to change the state of motion. The greater the mass of a body, the greater the resistence.
So, does it means that the fatter the guy in a car, the greater effect of moving forward when a car is to decelerate?

2. Why electric current, I is a base quantity? Why not charge, Q instead? Since electric current, I=Q/t.

3. If a rubber band is pulled by the same magnitude of forces in the right and the left hand side of a body, resultant force = 0, the rubber band had been extended for a length of 2cm.
work done, W=Fs, since F=0, W=(0N)(0.02m)=0. But how come there is no work done since the rubber band is extended???

4. Energy enables a body to work. Kinetic Energy and Potential Energy both give rise to work done, but how can Heat Energy result in work done?
Say a ball falls because of potential energy, the moving ball strikes another ball and the ball striked is moved because of kinetic energy, but how can heat do work? even the ball has got heat energy, it still cant result in work done.

[bush]

yes.

simply because Electric current is selected to be one of the base units, although its definition uses units which are not base units themselves.

look at it this way. Fix the end of the band and stretch the other end so that it gives an extension of 2cm. Can you see work done?

Heat energy is defined as the energy transferred during heating....in microscopic levels, heat energy is actually the kinetic energy of molecules of a system. The higher the temperature, the more KE the molecules will possess.

[okaywhy]

in microscopic levels, heat energy is actually the kinetic energy of molecules of a system. The higher the temperature, the more KE the molecules will possess.


So, since average KE of molecules=temperature, is it implies that a highly accelerated cold water beam could have a higher temperature than a calm but hot water? Given that both samples have the same volume.

Do yuu know why the temperature in the highlands is lower? Don't the molecules in greater height have got more kinetic energy?

[reign226]

The way I see it, higher molecules have more potential energy than kinetic energy, while molecules closer to earth have less potential and more kinetic. Ergo, etc etc.

[bush]

I was speaking strictly on the lines of ideal gases, that is why the generalisation became Thermodynamic temperature=Random Kinetic energy of molecules. This is not really the case since what we were doing is to find a relationship between macroscopic properties and microscopic properties. Therefore, we cannot see them as equal terms though both are closely related.

In a more accurate sense, Temperature is a quantity that acts as an indicator of internal energy of a system. Some people define temperature as a measure of random agitation of matter, some define it as an indicator or guide which determine which direction heat will flow.

By viewing from an ideal correlation between KE and temperature with all other physical quantities being negligible, I don't see anything wrong with that statement. However, the same cannot be said for the system as qouted by you since there are many other physical factors which must be considered like molecular potential and others.

[fly]

hi, i'hv few questions here.....
conditions:
1. if the tangential velocity,v>sqr[GM(mass of the earth)/r], then the orbit of the satellite is an ellipse wif the Earth as the near focus of the ellipse.
2. if the tengential velocity,v<sqrt[GM(mass of the earth)/r],then the orbit of the satellite is an ellipse wif the Earth as the far focus of the ellipse.

Is thr juz the difference btw v>GM/r & v<GM/r as state above?
any uses if the satellite orbiting as the conditions above?

Find the tengential velocity,v required 4 the satellite 2 orbit 500km above the earth's surface

since, v=sqrt(GM/r) but also =sqrt(gR2/r)
after calculated, i found out tat thr is a difference...

the 1st 1 is
sqrt(6.67E-11x6.02E24/(6.87E6)=7645ms-1

the 2nd is
sqrt[9.81x(6.37E6)2/6.87E6]=7612ms-1

so which 1 is the correct answer? y?

Thx..... :)

[okaywhy]

forgive my stupidity.... my basics are very lousy.
What's the difference between force and energy?
What's force and what's energy?

Force is defined as something that causes a mass to accelerate.
Unit for force:kgms^-2
Energy is defined as something that enables work done.
Unit for energy: kgm^2s^-2

Although mathematically, I see the apparent difference between them, but, qualitatively, I really can't totally fathom this, can't energy causes mass to accelerate s well? Similarly, can't force enables work done?

[wawa]

orgive my stupidity.... my basics are very lousy.
What's the difference between force and energy?
What's force and what's energy?

Force is defined as something that causes a mass to accelerate.
Unit for force:kgms^-2
Energy is defined as something that enables work done.
Unit for energy: kgm^2s^-2

Although mathematically, I see the apparent difference between them, but, qualitatively, I really can't totally fathom this, can't energy causes mass to accelerate s well? Similarly, can't force enables work done?

Don't think absolutely mathematically when trying to understand physics. Physics is a language trying to describe how things work around us in a certain manner, and it would help if you'd understand how it works, and not just maths. (Count QM out at the moment)

Anyways, the idea is that any object without any force acting on it will always remain stationary if originally stationary, or in constant velocity if originally that way. In other words, if I dont' "disturb" it, it will remain that way. Something like if you throw your clothes on the floor and everybody refuses to pick it up, it will remain on the floor! BUT, if there is a wind, or your mother finally decides she has enough and picks it up, the wind or your mother will be changing the "state" of the clothes. Similarly, a force is needed to change the state of an object, ie you change the direction of a moving ball, you stomp on a pingpong ball and "deform" the ball (the shape is changed), you pick up the ball (from stationary to moving). Remember, maths is just a compact way of saying physics. F= ma? The force that you apply will cause an alteration in the state of an object of a certain mass by providing a certain acceleration.

Okay, this is one way of looking at the world around you. Gravitational forces push the planets around so that you have sunrise and dusk, your computer you are looking at now is there because there is frictional force stopping it from sliding off you table, you type words because you are exerting mechanical force one the keyboard is allows electricity and electrons to move around. THE FORCE is everywhere!!! And you thought Star Wars was fiction! :wink:

Energy is another important way to view the world and physics. Some of you may have seen the picture whereby it looks both like a beautiful girl's face and a crooked old lady's face by viewing it differently. A bit same here. Energy is an alternate way of viewing the world. You guys know the definitions of potential, kinetic blah blah blah. I think Feynman's explanation is that best. Any of you who has Feynman's lectures in physics go look up volume one.

It is an idea that all objects in this universe has this mystical thing "energy", you cannot create nor destroy it, but thankfully for us, we can transfer it to other objects! So when you pick up your clothes (after your mother has nagged you) and put it into your drawer 10m's from the floor, you give "energy" to the clothes so that they can remain in the drawer. But wait! Physics is also about measurement, so how do you measure this mysterious object "energy"? Easy, we decide! We, humans, define it to be .... (check you textbooks E = mgh etc etc ). To make it even easier, we even decided to distinguish the ways "energy" is being possessed by an object , ie kinetic (movement), Potential (location) etc etc....

So, now we have two very different yet legitimate way of looking at the world, force moves everything around us or everything around us has "energy" and what we are doing is just transferring them around. How do we reconcile these two things? Let's say you are kicking the ball. Yup, for the force point of view, you are exerting a force on it (lets say 10N), and the ball changes its state from stationary to moving. If you look from energy point of view, you are merely providing energy to the ball, so that it will have kinetic energy, which when going higher, the ball converts it to potential energy, and then.... blah blah blah. So, how does a certain amount of force change this funny thing "energy"? Both view are right, but how to get information of view 2 (energy) when all you know about is information from view 1? Remember your W=FL stuff?

So...clearer now? Well, I hope you'd be able to deduce the answers to your last few questions now.

[okaywhy]

Wow, :D brilliant explanation... thank you! Ya, physics should be taught in this way.

Another question, haha

When a bat of mass M hits a ball of mass m,
the ball experiences a force of F1, and it exerts a force F2 on the bat.
According to Newton's 3rd law, F1=F2=F

Using F=ma,
the acceleration of the ball, a1=F/m
the acceleration of the bat, a2=F/M

since M>>>>m, a2<<a1


ok, now the question is, what if the mass of the bat is the same as the mass of the ball? i.e, M=m

Supposably, the bat will experiences the same acceleration as the ball, i.e, a1=a2, right?

But, in real life experience, err, take snooker, all the balls are having the same mass, but why the ball didn't bounce back after it collides with another ball since their acceleration should be the same?

why Newton 3rd Law seems to contradict the momentum way of explaining collisions?

[kimsiang]

Wow, :D brilliant explanation... thank you! Ya, physics should be taught in this way.

Another question, haha

When a bat of mass M hits a ball of mass m,
the ball experiences a force of F1, and it exerts a force F2 on the bat.
According to Newton's 3rd law, F1=F2=F

Using F=ma,
the acceleration of the ball, a1=F/m
the acceleration of the bat, a2=F/M

since M>>>>m, a2<<a1


ok, now the question is, what if the mass of the bat is the same as the mass of the ball? i.e, M=m

Supposably, the bat will experiences the same acceleration as the ball, i.e, a1=a2, right?

But, in real life experience, err, take snooker, all the balls are having the same mass, but why the ball didn't bounce back after it collides with another ball since their acceleration should be the same?

why Newton 3rd Law seems to contradict the momentum way of explaining collisions?

Good questions.But then,the actual situation is more complicated.
If we consider the easiest case,for example,a block with velocity v collides with another block with the same mass on a smooth surface[without friction],the 1st block will stop and the 2nd block will continue to move with velocity v.{you can try to work out the solution using the conservation of momentum and energy}

But then,when it goes to snooker,you have to consider about rotational motion and also friction.[Actually i alwaz play billiard in my dormitary here.]
The motion of the 2 balls(in snooker) is decided by how you hit the 1st ball.You can let the ball stop,go forward,go inverse by hitting the ball at different height.[you can search for the further details n calculation online, :D ]
Try to play it and then you will understand, :D .

[wawa]

Remember, Physics is HUMAN's way of trying to understand and describe the world. We try to be as near as possible, but fact is....we aren't always right! Next time if do continue studying, you'd find that in order for us scientists and engineers to understand our surroundings, we repeatedly use a very very very important tool -- approximation.

Let's say, you want to model the behavior of traffic moving on a highway. Very difficult huh? There's so many factors and variables to consider, and it'll take a whole horde of mathematicians, engineers and computers to completely describe how it is going. But we don't have all the time! Guess what? We use approximation -- it acts something like that on the whole, we might not get the smallest details correct, but GENERALLY is it all right.

When you write of M>>>>m etc, you are using that important tool -- approximation! So essentially you are saying, M is sooo soo sooo much "heavier" than m, I can say that a2 is soo soo soo much smaller than a1. But what does that mean when you are saying that? I mean that the "effect" of the force on M is soo soo soo much smaller than than the force on m, in fact, so small that sometimes i can ignore it in M! It is an approximation!

You can even try this at home! Try a bowling ball and a ping pong ball, where the mass difference is huge. Roll the bowling ball into the pingpong, and you'd find out YOURSELF that on the whole this is true -- the bowling ball's velocity doesn't change approximately. Of course if you look closely, it ain't , there's a small difference! Kinda like some people if you look really really closely, they have all this zits and blackheads and whatever gooey idea you can think of on their nose, but if you look a distance away, it looks ok! :wink:

BTW, if might be good if you look up on elastic and inelastic collisions.

For your bouncing back, again....you can try my real life, rather painful experience. If you ram straight into a person, face into face, chances you'd fall backwards. What happened if you rammed the shoulder of that person instead straight on? Remember, F=ma, we are approximation that its a point source...remeber what i said for the past two posts...and put two and two together. :wink: you can consider if you are soo soo huge and you ram into a small kid, what would happen?

BTW, newton third law cannot contradict "momentum way". If my memory's correct, 3rd law is the law which your "momentum way" is derived from.

Let me know if you're still lost.

[okaywhy]

thank you wawa! :D

I've completed reading all chapters regarding mechanics for STPM, i.e linear motion, circular motion and gravitation. But yet, I still can't answer the question, why doesn't the moon fall? Simlarly, why the earth doesn't fall into the Sun?

Some books give me the answer that it's the result of the moon's velocity. Ok, how does the gravity results the moon to move in such velocity?

And since the gravitational forces exerted by the earth and the moon are the same, why doesn't earth revolves around the moon?

Can anyone kindly explain it both mathematically and qualitativly?

[youngyew]

thank you wawa! :D

I've completed reading all chapters regarding mechanics for STPM, i.e linear motion, circular motion and gravitation. But yet, I still can't answer the question, why doesn't the moon fall? Simlarly, why the earth doesn't fall into the Sun?

Some books give me the answer that it's the result of the moon's velocity. Ok, how does the gravity results the moon to move in such velocity?

And since the gravitational forces exerted by the earth and the moon are the same, why doesn't earth revolves around the moon?

Can anyone kindly explain it both mathematically and qualitativly?
I don't have enough time here, so I would just try to give an outline to the answer.

Why does the moon move with such a velocity, it's due to its initial motion. But how did the moon start moving, or where did the moon come from, nobody knows for sure. The prevalent theory (if im not mistaken) is that a huge comet or interstellar object hit the earth some billion years ago, and the moon was one of the debris which maintained a momentum and stayed in a good orbit by chance.

As for the reason moon doesn't fall onto earth, there is an alternative way to see it - the moon is always falling, but it never hits the ground! We all know that all projectile follows a parabola track before hitting the earth. As you increase the horizontal component of the initial velocity, the curve of the projectile changes in such a way that it draws a larger curve with larger horizontal velocity. When you have sufficient velocity, the projectile will then draw a curve which is more convex (meaning more "pointed outwards) or equally convex with the earth, and that is when you have an object travelling in an orbit. Remember that the object is always falling, but it just won't hit the ground because the earth is round.

For the third question, the earth does revolve around the moon. In fact, if you want to be precise, neither the moon revolves around the earth nor the earth revolves around the moon. Both of them are actually revolving around a virtual point in the space which is the centre of mass of both the moon and the earth (barycenter).

Earth and Moon orbit about their barycenter, or common center of mass, which lies about 4700 km from Earth's center (about 3/4 of the way to the surface). Since the barycenter is located below the Earth's surface, Earth's motion is more commonly described as a "wobble". When viewed from Earth's North pole, Earth and Moon rotate counter-clockwise about their axes; the Moon orbits Earth counter-clockwise and Earth orbits the Sun counter-clockwise.

For more information, refer to http://en.wikipedia.org/wiki/Moon http://en.wikipedia.org/wiki/Orbit and http://en.wikipedia.org/wiki/Barycenter There are some comprehensive info, and it makes a good starting point to learn about orbits and satellites.

[okaywhy]

But,what happened if the mass is not fixed?
We know that
F=(dp/dt)
F=(d[mv]/dt)

if we let m=m(t)[mass changes with time,or as function of time],
F =(dm(t)/dt)v + m(t)(dv/dt)
F =(dm(t)/dt)v + m(t)a

Hence from above ,
F ~ dm(t)/dt [rate of changes of mass]
F ~ a .
CLear??Maybe a bit difficult.....


If I'm not mistaken, the rockets are using this concept... my teacher told me that its about the system of variable mass. But the rocket is accelerating as well, so, it can't be just F=(dm/dt)v since the v is changing as well, isn't it?

What's m(t)[mass changes with time,or as function of time]?

[okaywhy]

Thank you youngyew, it is very helpful indeed.

It's also said that Sun and earth are revolving around a common centre of mass/barycentre.

But, I'm still confused, we've got 9 planets in the solar system, thus, the Sun may have 9 different centres of mass with respect to 9 different planets with different mass. How does it works???

[wawa]

kimsiang wrote:

But,what happened if the mass is not fixed?
We know that
F=(dp/dt)
F=(d[mv]/dt)

if we let m=m(t)[mass changes with time,or as function of time],
F =(dm(t)/dt)v + m(t)(dv/dt)
F =(dm(t)/dt)v + m(t)a

Hence from above ,
F ~ dm(t)/dt [rate of changes of mass]
F ~ a .
CLear??Maybe a bit difficult.....


If I'm not mistaken, the rockets are using this concept... my teacher told me that its about the system of variable mass. But the rocket is accelerating as well, so, it can't be just F=(dm/dt)v since the v is changing as well, isn't it?

What's m(t)[mass changes with time,or as function of time]?

Remember that your rocket is burning up fuel as it goes. Obviously, the lesser and lesser fuel would mean a lighter and lighter aircraft.
So...mass changes with time loh.

It's also said that Sun and earth are revolving around a common centre of mass/barycentre.

But, I'm still confused, we've got 9 planets in the solar system, thus, the Sun may have 9 different centres of mass with respect to 9 different planets with different mass. How does it works???

Remember...Principle of Superposition....

[okaywhy]

But the rocket is accelerating as well,
so, it can't be just F=(dm/dt)v,
as the v is changing as well,
isn't it?


My main question is this. I feel weird that...
F=(dm/dt)v, the v is constant. That means the rocket is not accelerating?

[okaywhy]

How does the formation of bonds between molecules results in an exothermic reaction?

Physically, we know that bonds means potential energy between molecules,
now since, internal energy, U = potential energy + kinetic energy,
when the potential energy of the molecules increases, the kinetic energy of the molecules will decreases, right?
Using 1/2 m<c^2>=f/2 KT,
we have, Kinetic energy, KE ~ Temperature, T
Hence, when bond formed, KE decreases, temperature decreases.

But, lo and behold, how can a decrease in temperature results in an exothermic reaction? Doesn't it violates the Zeroth Law of Thermodynamics which states that heat must flow from high temperature to low temperature?

[bush]

your internal energy should be constant if we take your statement,'when the potential energy of the molecules increases, the kinetic energy of the molecules will decreases' to be true....

well, for the final part,I think we look at the system, it loses energy or it radiates heat, hence is temperature decreases.

I doubt that the zeroth law is violated, it just shows that the system has a higher temperature than whatever that is touching the system.

Put it this way, the temperature is not low enough to be called a low temperature.....does it make sense?

[okaywhy]

your internal energy should be constant if we take your statement,'when the potential energy of the molecules increases, the kinetic energy of the molecules will decreases' to be true....


Oh... ! Thanks for your explanation!

So, if internal energy is not constant, does it make sense if I were to put it this way that:

Bonds formed, i.e. PE of molecules increases,
KE of molecules increases because degree of freedom increases,
hence, temperature increases,
heat loss by the system, i.e. exorthermic?

[bush]

your internal energy should be constant if we take your statement,'when the potential energy of the molecules increases, the kinetic energy of the molecules will decreases' to be true....


Oh... ! Thanks for your explanation!

So, if internal energy is not constant, does it make sense if I were to put it this way that:

Bonds formed, i.e. PE of molecules increases,
KE of molecules increases because degree of freedom increases,
hence, temperature increases,
heat loss by the system, i.e. exorthermic?

I guess so, in this case U is not constant, ie increases. Hence the temperature of the system increases. Zeroth law comes in,heat flows out from the system assuming that the temperature of that system is higher than its surroundings.

[littlepiglet]

Do anyone have some tips how to remember all the formula inside the books? So many formula and i almost mixed all of them together. Does the STPM give the list of formula? (So that i don't need to remember all of them.)

Do we need to know how to proved all the formula? At first I think that we don't need to remember how the formula come. But when I roughly see the past year paper, the essay question need us to proved the formula. Which formula is most probably come out at the exam?

[victorvandort]

hi , can anyone tell me what's gradient and how's it used in physics?

and also, any good books for physics stpm? i'm not taking tuition at the moment and am rather worried. any books even a levels is ok...

or even, any good tuition centers around kl? other than kasturi..

thanks

[kimsiang]

hi , can anyone tell me what's gradient and how's it used in physics?


There are 2 definations for it:
1.Gradient of a graph at 1 point - the slope of the tangent to the graph at that point.

2.gradient of a scalar field,f(x,y,z) at a point - the vector pointing in the direction of the greastest increase in the scalar with distance.It has components along the coordinate axes that are the partial derivatives,fx,fy,fz of the function with respect to each variable:

grad f = (fx)i + (fy)j + (fz)k

i,j,k are unit vectors along thex-,y-, and z-axis.

For example,the force experienced by a particle inside the potential V(x,y,z) is given by F = - grad V,
and,Electric field is the negative gradient of electrical potential.

[ohmygod]

and also, any good books for physics stpm? i'm not taking tuition at the moment and am rather worried. any books even a levels is ok...


thanks

getting university books is good, I think. You can understand the concept better.

[okaywhy]

Electric current, I is defined as the rate of flow of charge, i.e. dQ/dt.

Now, my question is, isn't that the amount of charges contained in the wire is always constant? Say the wire is made of copper, regardless of whether the wire is connected to the battery or not, the amount of free electrons (i.e charge carriers) is the same, right? Then, what is dQ after all?

[bush]

Would it be easier if you consider it as, the flow of charge passing through a point per unit time?

[okaywhy]

Where is the point actually?

http://img373.imageshack.us/img373/8517/simplecircuit7hl.gif

What I don't understand is the "dQ" (change of charge) in I=dQ/dt. Normally, in other formulae such as a=dv/dt, the v changes. But what is actually happening in a flow of charge carriers while the charge is always constant throughout the whole wire?

[bush]

Don't you know that when you have infinite amount of points joined together, you'll get 'a line'?

Anyways, the point is arbitrary, it can be anywhere on the circuit. That is to say, the rate of flow of current at any one point in the circuit.

However, a point in a circuit when translated into 3D sense will become a plane which cuts the conductor into two. The current, IIRC can be defined as the electric charge passing through that plane per unit time. That act of passage through the plane, I believe is the 'dQ' that you are seeking to understand.

[okaywhy]

Here comes another trivial question.

1. I know that when electron drops from a higher energy level to a lower one, it emits electromagnetic waves. But, why is it that? How does the transmission creates photon?

Can a negatively charged ball does the same thing as it falls down?


2. How does friction releases heat? When we are rubbing something, aren't we breaking the "inter-lattice" bond? Breaking bonds requires energy, it should be endothermic, isn't it?


Sorry for being so 'zestful' about these trivial questions... my teachers and friends would scold me if I were to throw these questions to them... hehe

[bush]

Some of the questions that you ask are like, 'why does a body move when you exert a force on it?' or 'why does two like charges repel when brought close to each other?' or better, 'why is time part of spacetime?'



It has excess energy, hence EM waves are emitted =P.Well, Bohr made a postulation(2nd) that it was such and his postulation was a workable one, as proven in the absorption and emission spectra........................a photon is essentially a discrete bundle of energy which is either absorbed or radiated when a quantum jump is made. That transmission itself is a change in energy levels, when you have a difference in energy levels, you get excess energy, which can be taken to be a photon in this case.

Friction is a force. If friction is 'exerted' over a distance, work is done. So isn't it correct to say that you're doing work to produce heat? Just like you do work to raise a body to a certain height?
Endothermic.........I think you are jumbling inter- and intra-molecular bonds together.

[okaywhy]

Some of the questions that you ask are like, 'why does a body move when you exert a force on it?' or 'why does two like charges repel when brought close to each other?' or better, 'why is time part of spacetime?'
:P :oops: Haha, you're absolutely correct. :lol: :oops:

Anyway, thank you for taking time to answer my dumb questions. :wink: :D

[bush]

:P :oops: Haha, you're absolutely correct. :lol: :oops:

Anyway, thank you for taking time to answer my dumb questions. :wink: :D

I don't think they're dumb. Just bizarre.

[chen_doggie87]

question????
I would like to ask any one there who really know the tactics to score the physic paper 2?
Any important key word to use when answering the question?
my school teacher really many me confuse tat?
wat I answer is right...
but he dun wan to accept it coz din have the key words on his marking guide line. just some word..
if yes, pls guide me....
really sad.... :cry:

[ice111]

Hi everyone i am new here
Refering to the previous post , i need some mentor to guide to score for paper 2 for physic , since it did badly for it , i want to know how to answer those questions , the real format of it?
Can good samaritan helps me?
Hey , may I know where to get physics questionsssssssssssssssssss besides work books which are mainly available in the big bookshops?
Where to find STPM physics questions??????????????????
Thanks........

[littlepiglet]

as i know, from my experience, i answer my stpm physics question more on using formula. from the formula then i explain it. it a quite well thing to do because my teacher give me quite high mark in that, i dono whether is her way of marking is lenient or she can understand well what i write, but, i think you can try it. or ask your teacher before you try it ba...is better....and safer...

[lsw1990]

Hello, im new to this forum. going lower six this year.

i just started to study physics myself before school starts.

Im using Fajar Bakti's Physic volume 1 by Poh Liong Yong, S.Nagappan.

On page 7, (about deriving physical equation using dimensions)

it was stated that "suppose that T directly proportional to l^x . g^y "

and example 1.2, "Let v = λ^x . γ^y . p^z ..."

My question is.... where do x y and z came from??? why suddenly need to add x, y and z????? Any reasons????

Thanks in advance. Hope my post is not confusing.... :amuse

[withoutwax]

well.. i think u saw the POWERFUL words..i.e. 'let', 'suppose' asking u to ASSUME the equations are real..so x, y, z simply mean they are number..real number..

Hope my reply helps..

[lsw1990]

thanks for the response

hm....why must assume so? any reasons?

[withoutwax]

Well..they are all true mathematically (dimensionally)..but..some of them are theoretically wrong..you'll noe it when u proceed to further chapter..honestly..dun study 1st chapter..it confuses you..it's onli about maths..so no need to worry..

[lsw1990]

ohhh okay.... thanks

[starlemon]

i do think that Physics is sth abstract that is useless to study.
we just crapping our mind into the pilling of formulae just to face the exam.
Wat for we study Physics that is so inapplicable?(maybe there is some usage for architek or mathematician,or even engineer)
Can someone clarify that wat for we used to remember such a long formulae in physics? is it so applicable in our future lives?

[lsw1990]

sounds like... you hate physics!! LOL XD

[Caprio]

i do think that Physics is sth abstract that is useless to study.
we just crapping our mind into the pilling of formulae just to face the exam.
Wat for we study Physics that is so inapplicable?(maybe there is some usage for architek or mathematician,or even engineer)
Can someone clarify that wat for we used to remember such a long formulae in physics? is it so applicable in our future lives?

Haha... I see your frustration here. But, face it, this is what the conventional education is all about. Since you say Physics is so inapplicable, how about maths? Chapters like calculus, trigonometry and so on. It is more abstract...

However, let's look from the positive side. Though Physics may be inapplicable from your point of view, I would like to tell you Physics laws govern our whole universe. So it good for you to learn how to appreciate the beauty of Physics.:P

In addition, I would like to say that studying and solving Physics problems train us to develop logical and critical thinking. A lot of thinking needs to be done before you can reach the correct answer. Of course, what I mean here is that the Physics problems you attempt must be reached to a certain standard.

Enjoy Physics.

[youngyew]

i do think that Physics is sth abstract that is useless to study.
we just crapping our mind into the pilling of formulae just to face the exam.
Wat for we study Physics that is so inapplicable?(maybe there is some usage for architek or mathematician,or even engineer)
Can someone clarify that wat for we used to remember such a long formulae in physics? is it so applicable in our future lives?
The fact that you are typing this posts to share your dissatisfaction with so many people across the world... it's all physics (and its twin brother, maths) behind it. Your keyboard, your monitor, your processor, your internet connection etc etc.

The fact that our roofs do not collapse when we are grunting about physics... it's about physics.

Everywhere we go, we just can't run away from physics.

We can definitely argue that if you are going to study, say, business or finance in the future, then physics is inapplicable; but just like a scientists would should know about the world economy crisis, businessman and bankers should at least know how their BMW's engine produce the torque. In the modern era, it's almost a prerequisite for people to be multi-dimensional instead of knowing only their own field.

We touched a bit about "form 6 has too many useless hard things" in another thread (read conversation between myself and bush)
http://recom.org/forum/showthread.php?t=3593

[HimoYoto]

Hi..I'm new here too...is it physics need a lot of application of formula and concepts??my maths and add maths is good when in form 5..but my physic is like ok ok only...maybe is because of my laziness when i was in form 5..so i didn't score good result in spm...i'm still frustrating about which subj to choose when in form 6...bio or phy??really don't know...but my phy alw score better than bio...but many said that phy is tougher than bio in form 6 ,is that true??so scare that i can't cope with it..haizzz...Frustratedxx...

[uoykcuf]

Hi..I'm new here too...is it physics need a lot of application of formula and concepts??my maths and add maths is good when in form 5..but my physic is like ok ok only...maybe is because of my laziness when i was in form 5..so i didn't score good result in spm...i'm still frustrating about which subj to choose when in form 6...bio or phy??really don't know...but my phy alw score better than bio...but many said that phy is tougher than bio in form 6 ,is that true??so scare that i can't cope with it..haizzz...Frustratedxx...

firstly, yes it is. physics in form6 needs alot of formula. in whole syllabus i think got 200 formula in form6 if u assume it has 7-8 formula in each chapter.. i knw u r surprise with the numbers of formula u have to remember, but if u understand the concept in physics, actually 200 formula is nth.. lol.. so, how to memories these formula? jz do many question, and familar with the trend of the question will do.actually in stpm, the question is jz about like tat, wont be very very tricky o consuming ur brain juice o turn here turn there like olympiad question. jz apply the formula will do. stpm only test whether u knw the formula o not only.

secondly, bio o physics. i can tell u tat, in stpm, physics is more easily to score compared to bio. take my skul for example, the number of ppl score A in bio is less than physic student although the number of student taking bio is more than physics student. mebe mpm encourage student to take physic rather than bio since everyone want to become doctor o working in farmasy.. lol so they make the student easier to take A in physic. in bio, u need to write alot alot of things, memories alot of things, but in physics, u jz nid to memories 200 formula which is significantly less than wat bio need to memories. and the question, you only nid to write the formula and solve the question tat's all. however in physics, ur understanding have to b good. bio sumtimes u need not to be very understand and jz memories the thing and u can answer the question (this is wat i heard from my friend who takes bio, not my personal experience)

so which type of person u r? r u a person who hates to memories and prefer understand thing? o r u a person who like to memories, and not prefer to understand thing? so think about it and make ur own decision. if u r the first one, i encourage u to take physics, if not, then take bio. you oso nid to consider in future wat r u going to b. oh ya, in form6, if u 12 study physics, ur math have to b good.. tat's all i 12 say..

[HimoYoto]

firstly, yes it is. physics in form6 needs alot of formula. in whole syllabus i think got 200 formula in form6 if u assume it has 7-8 formula in each chapter.. i knw u r surprise with the numbers of formula u have to remember, but if u understand the concept in physics, actually 200 formula is nth.. lol.. so, how to memories these formula? jz do many question, and familar with the trend of the question will do.actually in stpm, the question is jz about like tat, wont be very very tricky o consuming ur brain juice o turn here turn there like olympiad question. jz apply the formula will do. stpm only test whether u knw the formula o not only.

secondly, bio o physics. i can tell u tat, in stpm, physics is more easily to score compared to bio. take my skul for example, the number of ppl score A in bio is less than physic student although the number of student taking bio is more than physics student. mebe mpm encourage student to take physic rather than bio since everyone want to become doctor o working in farmasy.. lol so they make the student easier to take A in physic. in bio, u need to write alot alot of things, memories alot of things, but in physics, u jz nid to memories 200 formula which is significantly less than wat bio need to memories. and the question, you only nid to write the formula and solve the question tat's all. however in physics, ur understanding have to b good. bio sumtimes u need not to be very understand and jz memories the thing and u can answer the question (this is wat i heard from my friend who takes bio, not my personal experience)

so which type of person u r? r u a person who hates to memories and prefer understand thing? o r u a person who like to memories, and not prefer to understand thing? so think about it and make ur own decision. if u r the first one, i encourage u to take physics, if not, then take bio. you oso nid to consider in future wat r u going to b. oh ya, in form6, if u 12 study physics, ur math have to b good.. tat's all i 12 say..

I think I'll probably choose physics...memorise the formula I think I can face it...Cause in form 5,the add maths formula I no need to refer back cause I memorise almost all the formula...hehe...and I like add maths cause is very fun .....physics..sometime after I read I'm not quite understand but after reading repeatedly then I'll understand...haha...biology...didn't like it very much..cause have to memorise many things like a robot...and must have the key words to score the marks....haizz...wondering why can't we explain in our own words...:huh....after understand the concept it is easy to memorise than you memorise it without understand it....:laugh

Anyway,thanks for your advice...it really does helps me a lot...:))

[Caprio]

Undeniably, Physics is easier to score than Biology, comparatively.

For those who are going to enter form6, choose wisely which stream to belong.

Besides choosing the stream bases on your interest, do think what profession you wish to take up. I have a friend who studies accounting after studying for physics for one and a half year in form 6. Though it has nothing and in fact he managed to score, why waste your time? It is better to go straight where your heart want to go.

[day_dreamer]

Hello, im new to this forum. going lower six this year.

i just started to study physics myself before school starts.

Im using Fajar Bakti's Physic volume 1 by Poh Liong Yong, S.Nagappan.

On page 7, (about deriving physical equation using dimensions)

it was stated that "suppose that T directly proportional to l^x . g^y "

and example 1.2, "Let v = λ^x . γ^y . p^z ..."

My question is.... where do x y and z came from??? why suddenly need to add x, y and z????? Any reasons????

Thanks in advance. Hope my post is not confusing.... :amuse

I believe the writer is trying to be general... x,y,z as assumed in Algebra , can be any real number ..

By doing so the writer makes the formula a general formula ..

*just my two cents .. correct me fellow maths champ

jz apply the formula will do. stpm only test whether u knw the formula o not only.

Ish ... don't follow this dude's advise .. he's too advance ( a National Physic Olympiad team member ) ..

STPM Physic test your basic maths skills .. your formulas .. your application of formulas .. your knowledge regarding the concept and definition of course ~

Good luck to all physic stream students..

[ccter]

Undeniably, Physics is easier to score than Biology, comparatively.

For those who are going to enter form6, choose wisely which stream to belong.

Besides choosing the stream bases on your interest, do think what profession you wish to take up. I have a friend who studies accounting after studying for physics for one and a half year in form 6. Though it has nothing and in fact he managed to score, why waste your time? It is better to go straight where your heart want to go.

wow, interesting... Is that friend a recom user too? what a coincidence.. I did physics in stpm as well, and am planning to take up accounting in uk if i manage to secure a scholarship this time.

The reason i entered form 6 was because I failed to secure JPA scholarship. in fact i was still waiting for the JPA appealing results during my orientation week. i was confused with what i wanted to do in the future initially, which is why I opted for physics. i thought it could give me more leeway compared to bio, as the field i could go into after that is more diverse. someone told me that physics students could do a pharmacy course in university even though they didn't learn bio in form 6. Besides, i never really liked bio. Hence, physics stream. anyhow studying physics does prepare me for accounting and finance because it helps make me more analytical. i never regretted my form 6 education even though it is a longer route.

Hello, im new to this forum. going lower six this year.

i just started to study physics myself before school starts.

Im using Fajar Bakti's Physic volume 1 by Poh Liong Yong, S.Nagappan.

On page 7, (about deriving physical equation using dimensions)

it was stated that "suppose that T directly proportional to l^x . g^y "

and example 1.2, "Let v = λ^x . γ^y . p^z ..."

My question is.... where do x y and z came from??? why suddenly need to add x, y and z????? Any reasons????

Thanks in advance. Hope my post is not confusing.... :amuse

I have used the Fajar Bakti Physics volume one before, and trust me, it isn't a very good reference, especially for your first time reading. The Pelangi version (blue colour, also by Poh Liong Yong), on the other hand, applies a more comprehensive approach, hence effectively cutting down your revision time considerably. Not that the Fajar Bakti version is not good, but from experience, it requires a first hand understanding of the topic before we can effectively absorb what it is trying to explain. The Fajar Bakti version is proven to be good for cross referencing with the Pelangi version, but not very good if you are rushing to revise for exams. Besides, it was the Pelangi version that helped me survive thru STPM Physics without having to attend any Physics tuition classes in Form 6.

Now, back to your question,

The velocity v of waves in a ripple tank depends on the wavelength λ, surface tension Y and density p of water. Deduce an equation to show the relationship between these quantities.

the solution:

let v=kλ^(x)Y^(y)^p (z) , where k,x,y,z are dimensionless constants.

*x,y and z are brought in because we didn't know how v depends on λ, Y and P initially, as in v could be directly or inversely (or even "square-rootly" :P, as shown in the worked answer) proportionatel to the λ, Y and P variables.

however, we already know that v's dimensions are LT^(-1), because it's unit is in metres per second. (note: L=length, T=time and M=mass)

similarly, the dimensions for wavelength, λ, is in L (metres); for tension, Y, (given) is in MT-2; and for density, p, is in ML-3 (kgm-3).

Hence what we do is to equate the dimensions on both sides of the eqn., simplify and then equating the indices(x,y,z) from both sides, and ultimately determining how v is dependent on the λ, Y and P variables. say for eg if x is -1/2, then v is dependant on λ^(-1/2). Hence the xyz is brought in to determine how v is dependant on the variables.

Hope this makes sense to you. :)

[cathy]

y many people said girls better take bio than phy??i prefer phy but i am so worry cos many people say stpm phy is very hard..i scare i cannot get good result...so can anyone tell me is phy really hard to study??bio is easier?

[jayden]

Don't let what people say influence you on your choice. There's no such thing about gender and the subjects that you study. Though I have to say that in form 6 in my school, the number of girls in bio class is more than the guys and vice versa.

Actually bio is all about memorizing facts and making sure that you can write everything down as precisely as possible in the exam.

Physics would focus more on counting and understanding. And you do get marks for 'jalan kerja' even though the final answer is wrong right?

Everyone says STPM is hard. All you need is hard work and perseverance. Follow your heart. Besides, being in a class with more guys could be fun :P

[day_dreamer]

Don't let what people say influence you on your choice. There's no such thing about gender and the subjects that you study. Though I have to say that in form 6 in my school, the number of girls in bio class is more than the guys and vice versa.

Actually bio is all about memorizing facts and making sure that you can write everything down as precisely as possible in the exam.

Physics would focus more on counting and understanding. And you do get marks for 'jalan kerja' even though the final answer is wrong right?

Everyone says STPM is hard. All you need is hard work and perseverance. Follow your heart. Besides, being in a class with more guys could be fun :P

Yes and no .. Physics is also about memorising 300+ formulas .. AND marks are also awarded if the answer is wrong, using the concept of "error carry forward" or "ecf" written on your answering sheets by your teacher.

[jayden]

Yes and no .. Physics is also about memorising 300+ formulas .. AND marks are also awarded if the answer is wrong, using the concept of "error carry forward" or "ecf" written on your answering sheets by your teacher.

Well, at least you get marks. Unlike SPM Kimia where wrong cancels right :cry

[newney]

Can anyone recommend reference books for Physics STPM?

[okaywhy]

Can anyone recommend reference books for Physics STPM?
I suppose you're still having your holiday. I would suggest you not to be too hasty on reading the STPM stuff first. Instead, try to go through some relatively light weighted physics books that concentrate more on conceptual understanding. One requires solid base before venturing into the realm of STPM.

For basic level of understanding you can read Homework Helpers: Physics by Greg Curran, a friendly introduction to all the basic ideas in physics including Mechanics, Thermodynamics, Waves and Light, Electromagnetism and Quantum physics.

For electromagnetism, you can read the Basic Electricity by Van Valkenburgh. It is quite a simple and friendly introductory book on electricity and it is used as text book for the US Navy.

Most importantly, take a brief look at the STPM past years questions to get an idea on what kind of questions are likely to be asked.

[Caprio]

Can anyone recommend reference books for Physics STPM?

You may try Pelangi or Longman.

[bluez_aspic]

Free fall your Malaysian textbooks into the ocean and get Fundamentals of Physics (Halliday, Resnick and Walker). Trust me on this one.

[youngyew]

But, will it help you pass your STPM exams?

Anyway, Malaysian STPM books aren't that bad.

[bluez_aspic]

But Halliday, Resnick and Walker is teh awesome! :D

Standard university text but there's heavy overlapping of material with the STPM syllabus - and there's a lot of application of principles to cute phenomena. Take Newtonian Mechanics for example - there's a fully worked out example why performing an aerial somersault more than four times is almost impossible; Judo techniques from the Physics POV; why cats are more likely to survive a fall from the 32th floor than the 5th floor etc.

After this book makes you fall in love with Physics, you can then slay the STPM exam questions =)

[youngyew]

But Halliday, Resnick and Walker is teh awesome! :D

Standard university text but there's heavy overlapping of material with the STPM syllabus - and there's a lot of application of principles to cute phenomena. Take Newtonian Mechanics for example - there's a fully worked out example why performing an aerial somersault more than four times is almost impossible; Judo techniques from the Physics POV; why cats are more likely to survive a fall from the 32th floor than the 5th floor etc.

After this book makes you fall in love with Physics, you can then slay the STPM exam questions =)
Okay you piqued my interest. Give me the answers. :P

Cat: No time to do their flips?
Aerial somersault: The angular momentum is lost to the air resistance?

[Gabrielle90]

is there anyone takes physics as 5th subject??? i mean in SPTM... how do u cope with it??? how about those experiments???

[bluez_aspic]

Okay you piqued my interest. Give me the answers. :P

Cat: No time to do their flips?
Aerial somersault: The angular momentum is lost to the air resistance?

re: aerial somersault - the book does this by modelling the situation and computing the angular speed for a quadruple somersault, which is about 3.23 revolutions per second. This angular speed is so fast that the aerialist wouldn't be able to see his surroundings clearly or fine-tune his rotation by adjusting his tuck. Even to make a 4.5-somersault flight, you'd have to increase the angular speed, which by the conservation of angular momentum, would require a tighter tuck to decrease rotational inertia. Quite physically unfeasible. The record is a four-somersault btw.

re: cat - The cat starts off plunging while being tucked in, then stretches out. This increases the drag coefficient. Assuming that this stretching doubles the area of the body (in the perpendicular direction which it is plunging), the speed can be reduced by 30%. The book goes on to quote an incident where a cat fell 32 floors and suffered only slight damage to its thorax and one tooth.

This reminds me of the 'why cats usually land on two feet' thing. The cat's spine is quite flexible, so if the first half of the cat's body rotates in a, say, clockwise direction, by the conservation of angular momentum again the second half rotates anti-clockwise - and spinal flexibility allows its body to be aligned in the correct manner.

How about we start throwing cats off the 32nd floor? :D The dog in my avatar promises he won't call RSPCA hehe

[day_dreamer]

is there anyone takes physics as 5th subject??? i mean in SPTM... how do u cope with it??? how about those experiments???

Choose between physic or bio as a primary sub. Make sure the primary sub stays strong.

If the primary sub is weakening , let go secondary sub temporary ..

When marks bounce back on primary sub, maintain at all cost ! Mean while .. fight back on secondary sub.

This is my tactic .. hope it works for you ~

[Xon]

Choose between physic or bio as a primary sub. Make sure the primary sub stays strong.

If the primary sub is weakening , let go secondary sub temporary ..

When marks bounce back on primary sub, maintain at all cost ! Mean while .. fight back on secondary sub.

This is my tactic .. hope it works for you ~

i dont get you. i choose physics. lower six now. any advice?

[GreenTea]

Can anyone here recommend some quality workbooks or reference books for physics STPM?

[Caprio]

Can anyone here recommend some quality workbooks or reference books for physics STPM?

I would recommend the Physics workbook by Federal Publications. I am not too sure about the author's name, just barely remember the surname is Lam.

This book contains a lot of exercises to reinforce the concept you learned. What's more that a step-by-step solution and explaination is available. If you cracked you brain and can't get the answer right, you may then refer to the solution manual provided at the back of the book.

Hope that helps.

[kpchen]

Free fall your Malaysian textbooks into the ocean and get Fundamentals of Physics (Halliday, Resnick and Walker). Trust me on this one.

Get this book. I've used it myself and I still love the book though what I'm studying now doesn't require much of Physics.

In addition get A Levels Physics by Nelson & Parker. It's an old book. Get the photocopied version at University Bookstore opposite Taylor's College provided it's still available.

[withoutwax]

Hey guys..help me out in this question..gosh..ve been scratching my head...

Suppose N electron can be placed in either of two configurations. In configuration 1, they are all placed on the circumference of a narrow ring of radius R and uniformly distributed so that the difference between adjacent electron is the same everywhere. In configuration 2, N-1 electrons are uniformly distributed on the ring and one electron is placed in the center of the ring. What is the smallest value of N for which the second configuration is less energetic than the first?

[sugan]

have ayone heard of this university bokk titled 'UNIVERSITY PHYSICS' by young and freedman.
anyone using it or heard of it.
any comment on it.

[withoutwax]

Heard of it, saw before (just few flips).
I think not bad la.