Question

6) A person with mass 50.0 kg, standing still, throws an object with mass 12.0
kg. If the velocity of the person after throwing is +2.0 m/s, what is the
velocity at which he threw the object? Assume the surface to be frictionless.

Answer 1

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(50.0 kg) (0 m/s) + (12.0 kg) (0 m/s) = (50.0 kg) (2.0 m/s) + (12.0 kg) v

v = -8.33 m/s