A 89 kg man lying on a surface of negligible friction shoves a 89 g stone away from himself, giving it a speed of 6.9 m/s. What speed does the man acquire as a result ?

Oct 25, 2015

$0.0069 m . {s}^{- 1}$

Explanation:

We can use the principle of conservation of momentum to solve this problem.

Total initial momentum = total final momentum, ${p}_{i} = {p}_{f}$

We will take the initial state to be before the man shoves the stone – so ${p}_{i} = 0$.
It therefore follows that the final momentum is also zero from the above: ${p}_{i} = {p}_{f} = 0$

Now consider the final state (after the man has shoved the stone). The total momentum is zero, so we can write an expression for that and then solve for the unknown velocity.
p_f=p_m – p_s=0
Where ${p}_{m}$ is the momentum of the man and ${p}_{s}$ is the momentum of the stone.

⇒ m_m×v_m - m_s×v_s = 0

⇒ v_m = (m_s×v_s)/m_m = (89×10^(-3)×6.9)/89 = 0.0069 m.s^(-1)

Note that the velocity of the man is equal to the ratio of the masses multiplied by the velocity of the stone. The mass of the man is 1,000 times larger than that of the stone, so it follows that the velocity of the man must be 1,000 times smaller than that of the stone.