Richard, you're right. The key is using the same reference system for the velocities. Choose an observer on teh ground for the reference system. And using Conservation of Momentum as there are no external forces (we assume no friction) acting on the system of the boy and the skateboard. The force the boy exerts on the skateboard is equal and opposite to the force acting on the boy from the skateboard, so the net force on the system from the internal forces is zero.
The equation is (54)(5) = (50)(2.5) + 4x. x is the final velocity of the skateboard.
The boy's velocity is +2.5 m/s (forward direction) relative to the ground. He is still jumping backwards at a velocity of 5 m/s relative to the skateboard. But, since the skateboard is moving to the right initially at 5 m/s, his relative velocity to the ground is 2.5 m/s forward.
The question was not clear about the reference frames, so I added some words to clarify it.
Thanks,
John
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