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5 December, 12:33

A 3-kg skateboard is rolling down the sidewalk at 4 m/s when it collides with a 1-kg skateboard that was initially at rest. If the 1-kg skateboard moves off at 6 m/s after the collision, what is the new velocity for the 3-kg skateboard?

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Answers (2)
  1. 5 December, 14:43
    0
    2 m/s

    Explanation:

    Parameters given:

    Mass of first skateboard, m = 3 kg

    Initial speed of first skateboard, u = 4 m/s

    Mass of second skateboard, M = 1 kg

    Initial speed of second skateboard, U = 0 m/s

    Final speed of second skateboard, V = 6 m/s

    Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.

    Momentum is the product of mass and velocity. This implies that:

    m*u + M*U = m*v + M*V

    (3*4) + (1*0) = (3*v) + (1*6)

    12 + 0 = 3v + 6

    => 3v = 12 - 6

    3v = 6

    v = 6/3 = 2 m/s

    The final speed of the 3 kg skateboard is 2 m/s
  2. 5 December, 15:33
    0
    2m/s

    Explanation:

    Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision.

    Momentum = mass * velocity.

    Before collision:

    Momentum of 3kg body moving at 4m/s = 3*4

    = 12kgm/s

    Momentum of 1kg skateboard initially at rest = 1*0

    = 0kgm/s (the velocity 0m/s because the body is initially at rest)

    After collision:

    1-kg skateboard moves off at 6m/s. Its momentum after collision will be;

    1*6 = 6kgm/s

    For the 3kg body moving with an unknown velocity v, its final momentum will be;

    3*v = 3v

    Applying the law as stated above;

    12+0 = 6+3v

    12 = 6+3v

    12-6 = 3v

    6 = 3v

    v = 6/3

    v = 2m/s

    This means that the new velocity of the 3kg skateboard after collision is 2m/s
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