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5 January, 17:10

Blocks A (mass 3.00 kg) and B (mass 9.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let + x be the direction of the initial motion of block A. Find the velocity of block B when the energy stored in the spring bumpers is maximum.

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  1. 5 January, 17:24
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    The velocity of block B when the energy stored in the spring bumpers is max = 2.5m/s

    Explanation:

    An head on collision is aka elastic collision and in elastic collision momentum and energy is conserved that is the total momentum before collision = total momentum after collision

    Given

    Mass of block A mA = 3kg, initial Velocity of block A vA1 = 5m/s, Final velocity = vA2

    Mass of block B mB = 9kg, Initial velocity of block B vB1 = 0m/s, Final velocity = vB2

    Therefore we have the equation

    1) mAvA1 + mBvB1 = mAvA2 + mBvB2

    Also in elastic collision between 2 objects, the relative velocities before and after the collision have the same magnitude but opposite direction

    vA1 - vB1 = vB2 - vA2

    vA2 = vB2 - vA1 when vB1 = 0

    Substitute into the equation we have

    (3*5) + (9*0) = 3 * (vB2-vA1) + (9*vB2)

    15 = 3vB2 - 3vA1 + 9vB2

    15=3vB2 - (3*5) + 9vB2

    15+15 = 12vB2

    vB2 = 2.5m/s
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