A 96 kg man lying on a surface of negligible friction shoves a 75 g stone away from himself, giving it a speed of 7.2 m/s. What speed does the man acquire as a result?

7.21m/s

Explanation:

Using the law of conservation of momentum expressed as;

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the mass of the man and the stone respectively

u1 and u2 are their respective initial velocities

v is their final common velocity

Given

m1 = 96kg

m2 = 75g = 0.075kg

u1 = ?

u2 = 0m/s (initial speed of the stone)

v = 7.2m/s

Substitute into the firmula and get u1

96u1 + 0.075(0) = (96 + 0.075)(7.2)

96u1 = 96.075(7.2)

u1 = 691.74/96

u1 = 7.21m/s

Hence the man acquired a speed of 7.21m/s