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20 May, 17:30

Block 1, of mass m1 = 2.70 kg, moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 53.0 kg, which was initially at rest. The blocks stick together after the collision. A: Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically. B: Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically. C: what is the change deltaK = Kfinal - K initial in the two block systems kinetic energy due to the collision? Express your answer numerically in joules.

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  1. 20 May, 18:06
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    a) Block 1 = 72.9kgm/s

    Block 2 = 0kgm/s

    b) vf = 1.31m/s

    c) ∆KE = 936.36Joules

    Explanation:

    a) Momentum = mass * velocity

    For block 1:

    Momentum = 2.7*27

    = 72.9kgm/s

    For block 2:

    Momentum = 53 (0) (body is initially at rest)

    = 0kgm/s

    b) Using the law of conservation of momentum

    m1u1+m2u2 = (m1+m2) v

    m1 and m2 are the masses of the block

    u1 and u2 are their initial velocity

    v is the common velocity

    Given m1 = 2.7kg, u1 = 27m/s, m2 = 53kg, u2 = 0m/s (body at rest)

    2.7 (27) + 53 (0) = (2.7+53) v

    72.9 = 55.7v

    V = 72.9/55.7

    Vf = 1.31m/s

    c) kinetic energy = 1/2mv²

    Kinetic energy of block 1 = 1/2*2.7 (27) ²

    = 984.15Joules

    Kinetic energy of block 2 before collision = 0kgm/s

    Total KE before collision = 984.15Joules

    Kinetic energy after collision = 1/2 (2.7+53) 1.31²

    = 1/2*55.7*1.31²

    = 47.79Joules

    ∆KE = 984.15-47.79

    ∆KE = 936.36Joules
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