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[STPM] Physics

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Thread: [STPM] Physics

  1. #1
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    Default [STPM] Physics

    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?

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    College Student chiunlin's Avatar
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    Default Re: Physics STPM

    Quote Originally Posted by 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.

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    Default Re: Physics STPM

    Quote Originally Posted by chiunlin
    Quote Originally Posted by 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 ?

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    Default

    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

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    Default

    Quote Originally Posted by 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
    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??

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    Default

    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)

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    Thanx David.. i could barely understand it

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    Default

    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.

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    Default

    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...
    All I ask is a tall ship, and a star to steer her by - John Masefield, Sea Fever

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    Default

    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.......

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