A 90 kg man lying on a surface of negligible friction shoves a 69 g stone away from himself, giving it a speed of 4.0 m/s.

What speed does the man acquire as a result?

v = 3.06 x 10⁻³ m/s

Explanation:

given,

mass of man, M = 90 kg

mass of stone, m = 69 g = 0.069 Kg

speed of the stone, u = 4 m/s

speed of the man, v = ?

using conservation of momentum

initial velocity of the man and stone is equal to zero

(M + m) V = M v + m u

(M + m) x 0 = 90 x v + 0.069 x 4

90 v = 0.276

v = 0.00306 m/s

v = 3.06 x 10⁻³ m/s

speed of the man on the frictionless surface is equal to 3.06 x 10⁻³ m/s