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   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!!!

some think that theories become laws, but theories and laws have separate and distinct roles in the scientific method. a law is a description of an observed phenomenon in the natural world that hold true every time it is tested. it doesn't explain why something is true; it just states that it is true.

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting