Question

A 56 0-kg person holding two 0.750-kg bricks stands on a 3.00-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 18.0 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

At all times the total momentum must be conserved. Since the person was initially at rest, so the initial momentum is zero. Hence the final momentum after the person throws the bricks must also be zero.

$\small (m_{person}+m_{skateboard})\times v_{recoil}+ m_{bricks}\times v_{bricks}=0 \\\Rightarrow v_{recoil}= -\frac{m_{bricks}\times v_{bricks}}{m_{person}+m_{skateboard}}=-\frac{2\times 0.75 \times 18}{56+3}=-0.46 ms^{-1}$

So the recoil velocity is 0.46 m/s and the negative sign implies that the person moves in a direction opposite to that of the bricks.