A 49.0 kg person holding two 0.850 kg bricks stands on a 2.10 kg skateboard.?

Question

A 49.0 kg person holding two 0.850 kg bricks stands on a 2.10 kg skateboard. Initially, the skateboard and the person are at rest. The person now throws the two bricks at the same time so that their speed relative to the person is 21.8 m/s. What is the recoil speed of the person and the skateboard relative to the ground, assuming the skateboard moves without friction?

Answer 1

The key to this problem is conservation of momentum. The system at rest has zero momentum. Since no external forces act on the system, the final momentum must also be zero. If we assume that the two bricks are thrown in the same direction, then the person and skateboard must have a momentum with magnitude equal to that of the bricks' momentum, but in the opposite direction.

Let us say that the bricks are thrown in the "backwards" (negative) direction. This will cause the person and skateboard to move "forwards" (positive). So if the person and skateboard have a final velocity of v, the bricks have a velocity of v - 21.8 (which should be less than zero), so that they have a relative speed of 21.8 m/s. The final momentum of the system is therefore (49.0 kg + 2.10 kg)*v + (2*0.850 kg)(v - 21.8), which we know is equal to zero. Simplifying just a little bit (and leaving off the units), you simply need to solve the equation 51.1v + (1.70)(v - 21.8) = 0, a simple algebra problem.