Question

A 3.0-kg ball with an initial velocity of (4.0i + 3.0j) m/s collides with a wall and rebounds with a

Answer 1

The impulse exerted on the ball by the wall is 24i kgm/s.

Impulse exerted by the ball on the wall

The impulse exerted by the ball on the wall is the change in the linear momentum of the ball.

J = ΔP

ΔP = Pf - Pi

P = mv

where;

• m is mass of the ball
• v is the velocity

ΔP = 3(4.0i + 3.0j) - 3(-4.0i + 3.0j)

ΔP = (12i + 9j) - (-12i - 9j)

ΔP = 24i kgm/s

Thus, the impulse exerted on the ball by the wall is 24i kgm/s.

Learn more about impulse here: brainly.com/question/904448

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