On a frictionless surface, a 32 kg student pushes a 43 kg student. If the 32 kg student slides

back at 2.4 m/s, how fast will the 43 kg student be sliding and in what direction?

Correct answer: V₂ = 1.79 m/s

Explanation:

Given:

m₁ = 32 kg the mass of the first student

V₁ = 2.4 m/s speed of the first student after pushing

m₂ = 43 kg the mass of the second student student

V₂ = ? speed of the second student after pushing

I₁ = m₁ · V₁ impulse of the first student

I₂ = m₂ · V₂ impulse of the second student

Under the law of impulse maintenance, the total impulse of an object must be constant over time.

Since the total impulse before pushing was equal to zero it must be afterwards.

I₁ + I₂ = 0 ⇒ I₂ = - I₁ ⇒ V₂ = - V₁

Given that the impulses are vectors a sign minus ahead of the first pulse mean, they are opposite directions.

m₂ · V₂ = m₁ · V₁ ⇒ V₂ = (m₁ · V₁) / m₂ = (32 · 2.4) / 43 = 1.79 m/s

V₂ = 1.79 m/s

God is with you!