Puck 1 (0.5 kg) travels with velocity 40 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After the collisio

Question

Puck 1 (0.5 kg) travels with velocity 40 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After the collision, puck 1 moves with a velocity of 0 m/s. Assume that no external forces are present and therefore the momentum for the system of pucks is conserved. What is the final velocity (in m/s) of puck 2 after the collision

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Tài Đức 2 weeks 2021-07-22T03:34:13+00:00 2 Answers 0 views 0

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    0
    2021-07-22T03:35:14+00:00

    Answer:

    The final velocity of puck 2 after the collision is 20 m/s .

    Explanation:

    Let us assume , right direction is positive and left direction is negative .

    Given :

    Mass of puck 1 and 2 of 0.5 kg and 1 kg respectively .

    Initial velocity of puck  1 is 40 m/s .

    Initial velocity of puck  2 is 0 m/s .

    Final velocity of puck  1 is 0 m/s .

    Let , us assume final velocity of puck 2 is v_2 .

    Therefore , by conservation of momentum :

    0.5\times 40+1\times 0=0.5\times 0+1\times v_2\\v_2=20\ m/s

    Therefore , the final velocity (in m/s) of puck 2 after the collision is 20 m/s .

    0
    2021-07-22T03:35:26+00:00

    Answer:

    20 m/s

    Explanation:

    mass of puck 1, m1 = 0.5 kg

    initial velocity, u1 = 40 m/s

    mass of puck 2, m2 = 1 kg

    initial velocity of puck 2, u2 = 0 m/s

    final velocity of puck 1, v1 = 0 m/s

    Let the final velocity of puck 2 is v2.

    As there is not external force acting on the system, the momentum of the system is conserved.

    m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

    0.5 x 40 + 1 x 0 = 0.5 x 0 + 1 x v2

    20 = v2

    v2 = 20 m/s

    Thus, the velocity of puck 2 after the collision is 20 m/s.

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