A 95 kg man lying on a surface of negligible friction shoves a 65 g stone away from himself, giving it a speed of 4.0 m/s. what speed does the man acquire as a result?

Total momentum of the system must stay constant, so M1*V1 = - M2*V2. Substituting in the values and rearranging, we find - V2 = M1*V1 / M2. 65 grams = 0.065 kg, so

0.065kg * 4.0m/s / 95kg = - 2.7*10^-3 m/s. The negative indicates he is travelling in the opposite direction as the stone. Speed is independent of direction, so the final answer is 2.7*10^-3 m/s.

We can solve this problem by conserving linear momentum.

We have then that:

P1 = P2

m1v1 + m2v2 = m1v1 ' + m2v2'

Substituting values and clearing:

95 (0) + 0.065 (0) = 95v + 0.065 (4)

0 = 96v + 0.26

96v = - 0.26

v = - 0.0027 m / s

The man moves in the opposite direction to the stone.

answer:

the speed that a man acquire as a result is v = - 0.0027 m / s