Question

A 2.0-kg ball with an initial velocity of (4i + 3j) m/s collides with a wall and rebounds with a velocity of (–4i + 3j) m/s. What is the impulse exerted on the ball by the wall?

Answer 1

Answer:

-16i kgm/s

Explanation:

Impulse I = m(v - u) where m = mass of ball = 2.0 kg, u = initial velocity of ball = (4i + 3j) m/s  and v = final velocity of ball = (-4i + 3j) m/s.

So, the impulse is thus

I = m(v - u)

= 2.0 kg[(-4i + 3j) m/s - (4i + 3j) m/s]

= 2.0 kg[(-4i - 4i) + (3j - 3j) m/s]

= 2.0kg[-8i + 0j] m/s

= -16i kgm/s