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Impulse Mechanics 1 Jan 2005 Watch

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    How do I work out part d?
    A stone S is sliding on ice. The stone is moving along a straight horizontal line ABC,
    where AB = 24 m and AC = 30 m. The stone is subject to a constant resistance to motion
    of magnitude 0.3 N. At A the speed of S is 20 m s–1, and at B the speed of S is 16 m s–1.
    Calculate
    (a) the deceleration of S,
    (2)
    (b) the speed of S at C.
    (3)
    (c) Show that the mass of S is 0.1 kg.
    At C, the stone S hits a vertical wall, rebounds from the wall and then slides back along
    the line CA. The magnitude of the impulse of the wall on S is 2.4 Ns and the stone
    continues to move against a constant resistance of 0.3 N.
    (d) Calculate the time between the instant that S rebounds from the wall and the instant
    that S comes to rest
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    Impulse is equivalent to change of momentum. The stone has been subjected to an impulse (change of momentum) in the opposite direction to its motion before impact. So what you have to do is work out the stone's momentum immediately before impact, apply the change of momentum to find its momentum immediately after impact, and finally work out the stone's new speed from its new momentum.
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    (Original post by old_engineer)
    Impulse is equivalent to change of momentum. The stone has been subjected to an impulse (change of momentum) in the opposite direction to its motion before impact. So what you have to do is work out the stone's momentum immediately before impact, apply the change of momentum to find its momentum immediately after impact, and finally work out the stone's new speed from its new momentum.
    What way would the impulse be?
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    (Original post by geography1294)
    What way would the impulse be?
    The stone has rebounded from the wall, so if you are treating the stone's original direction of travel as positive, the impulse imparted to the stone by the wall is negative.
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    (Original post by old_engineer)
    The stone has rebounded from the wall, so if you are treating the stone's original direction of travel as positive, the impulse imparted to the stone by the wall is negative.
    So would v be negative and u positive?
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    (Original post by geography1294)
    So would v be negative and u positive?
    Yes, if you are using u and v to denote velocity before and after impact respectively.

    Once you have worked out v in this way, you are free to make the new direction of travel positive for the remaining calculation steps if you wish.
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    (Original post by old_engineer)
    Yes, if you are using u and v to denote velocity before and after impact respectively.

    Once you have worked out v in this way, you are free to make the new direction of travel positive for the remaining calculation steps if you wish.
    Could you explain why the impulse acts in the direction it does?
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    An impulse acting in the same direction as an object is already travelling will speed it up in the same direction (i.e. add to its momentum).

    An impulse acting in the opposite direction to that in which an object is already travelling will either slow it down (i.e. reduce its momentum), or if the impulse is big enough it will reverse the object's direction of motion (also a reduction in momentum, remembering that momentum is a vector quantity).
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    (Original post by old_engineer)
    Yes, if you are using u and v to denote velocity before and after impact respectively.

    Once you have worked out v in this way, you are free to make the new direction of travel positive for the remaining calculation steps if you wish.
    I dont understand. Wouldnt then the impulse of 2.4 be negative as well then. what way are you taking the positive direction
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    (Original post by geography1294)
    I dont understand. Wouldnt then the impulse of 2.4 be negative as well then. what way are you taking the positive direction
    I fear we are at cross purposes here. I'll try to be clear.

    Momentum immediately before impact = mu, where m is the given mass and u is your answer from part (b).

    Momentum immediately after impact is (mu - 2.4). The impulse is taken as negative here because we know that it has caused the stone to reverse in direction.

    The velocity after impact, v, is given by starting with mv = (mu - 2.4), then rearranging for v.
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    (Original post by old_engineer)
    I fear we are at cross purposes here. I'll try to be clear.

    Momentum immediately before impact = mu, where m is the given mass and u is your answer from part (b).

    Momentum immediately after impact is (mu - 2.4). The impulse is taken as negative here because we know that it has caused the stone to reverse in direction.

    The velocity after impact, v, is given by starting with mv = (mu - 2.4), then rearranging for v.
    I drew a diagram more clearer and now I understand. Thanks
 
 
 
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