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11 January, 10:27

Two blocks slide without friction. Block 1 has a mass of 1.6 kg and a velocity of + 5.5 m/s. Block 2 has a mass of 2.4 kg and a velocity of + 2.5 m/s. After the collision, block 2 has a velocity of + 4.9 m/s. What is the velocity of block 1 after the collision?

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  1. 11 January, 10:56
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    The velocity of block 1 after the collision is 1.9 m/s.

    Explanation:

    Hi there!

    The momentum of the system composed by the two blocks is conserved (i. e. it remains constant) because no external force is acting on the blocks at the moment of the collision. Then, the momentum of the system before the collision is equal to the momentum after the collision. The momentum of the system is calculated adding the momenta of the two blocks.

    initial momentum of the system = final momentum of the system

    p1 + p2 = p1' + p2'

    m1 · v1 + m2 · v2 = m1 · v1' + m2 · v2'

    Where:

    p1 = initial momentum of block 1.

    p2 = initial momentum of block 2.

    p1' = final momentum of block 1.

    p2' = final momentum of block 2.

    m1 = mass of block 1.

    m2 = mass of block 2.

    v1 = initial velocity of block 1 (before the collision).

    v2 = initial velocity of block 2.

    v1' = final velocity of block 1.

    v2' = final velocity of block 2.

    Let's write the equation with the data we have:

    m1 · v1 + m2 · v2 = m1 · v1' + m2 · v2'

    1.6 kg · 5.5 m/s + 2.4 kg · 2.5 m/s = 1.6 kg · v1' + 2.4 kg · 4.9 m/s

    Solving for v1':

    14.8 kg · m/s = 1.6 kg · v1' + 11.76 kg · m/s

    (14.8 kg · m/s - 11.76 kg · m/s) / 1.6 kg = v1'

    v1' = 1.9 m/s

    The velocity of block 1 after the collision is 1.9 m/s.
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