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20 May, 03:23

3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 20 m/s. The two carts stick together and move as a single object after the collision. Determine the post-collision speed of the two carts.

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  1. 20 May, 03:26
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    The speed of the two carts after the collision is 10 m/s.

    Explanation:

    Hi there!

    The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

    initial momentum = mA · vA1 + mB · vB1

    final momentum = (mA + mB) · vAB2

    Where:

    mA = mass of cart A = 0.500 kg

    vA1 = velocity of cart A before the collision = 100 m/s

    mB = mass of cart B = 1.50 kg.

    vB1 = velocity of cart B before the collision = - 20 m/s

    vAB2 = velocity of the carts that move as a single object = unknown.

    (notice that we have considered leftward as negative direction)

    Since the momentum of system remains constant:

    initial momentum = final momentum

    mA · vA1 + mB · vB1 = (mA + mB) · vAB2

    Solving for vAB2:

    (mA · vA1 + mB · vB1) / (mA + mB) = vAB2

    (0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

    vAB2 = 10 m/s

    The speed of the two carts after the collision is 10 m/s.
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