Question:

A 0.5 kg ball is thrown horizontally with an initial velocity of 10 m/s. The ball strikes a wall and bounces back with a speed of 4 m/s. The collision lasts for 0.02 seconds. Calculate:

a) The impulse experienced by the ball during the collision. b) The change in momentum of the ball. c) The average force exerted on the ball by the wall during the collision.

Given: Mass of the ball, m = 0.5 kg Initial velocity, v₁ = 10 m/s Final velocity, v₂ = -4 m/s Collision duration, Δt = 0.02 s

Assume that the positive direction is towards the wall.

Answer:

a) The impulse experienced by the ball during the collision can be calculated using the equation:

Impulse (J) = Change in momentum (Δp)

Since the ball is moving horizontally, the momentum in the x-direction is given by:

p = m * v

The initial momentum (p₁) is given by:

p₁ = m * v₁

Substituting the given values, we have:

p₁ = (0.5 kg)(10 m/s) = 5 kg·m/s

The final momentum (p₂) is given by:

p₂ = m * v₂

Substituting the given values, we have:

p₂ = (0.5 kg)(-4 m/s) = -2 kg·m/s

Therefore, the change in momentum (Δp) is:

Δp = p₂ - p₁

Substituting the values, we have:

Δp = -2 kg·m/s - 5 kg·m/s

Δp = -7 kg·m/s

So, the impulse experienced by the ball during the collision is -7 kg·m/s.

b) The change in momentum of the ball is the same as the impulse experienced:

Change in momentum (Δp) = -7 kg·m/s

c) The average force exerted on the ball by the wall during the collision can be calculated using the equation:

Average force (F) = Impulse / Collision duration

Substituting the given values, we have:

F = (-7 kg·m/s) / (0.02 s)

F = -350 N

Therefore, the average force exerted on the ball by the wall during the collision is -350 N. The negative sign indicates that the force is in the opposite direction to the motion of the ball.