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12 July, 12:21

A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity of magnitude 3.00 m/s. (a) Calculate the velocity of the block (magnitude and direction) after a force of 5.00 N directed to the right has been applied for 4.00 s; (b) if instead a force of 7.00 N directed to the left is applied from t=0 to t = 4.00 s, what is the final velocity of the block?

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  1. 12 July, 12:27
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    a)

    We shall apply the concept of impulse.

    Impulse = force x time = change in momentum

    = 5 x 4 = 2 (V - 3), where V is final velocity of the object

    20 = 2V - 6

    V = 13 m / s

    b)

    Impulse applied = - 7 x 4 = - 28 kg m/s (negative as direction of force is opposite motion)

    If v be the final velocity

    2 x 3 - 28 = 2 v (initial momentum - change in momentum = final momentum)

    - 22 = 2v

    v = - 11 m / s

    object will move with 11 m / s in opposite direction.
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