Question

A 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle as seen in the figure.
If the ball is in contact with the wall for 0.207 s, what is the average force exerted on the ball by the wall?

Answer 1

The magnitude of the average force exerted on the ball by the wall is 225.469 N.

What is impulse?

The change in momentum is equal to the product of impact force applied while colliding and time for that impact.

Impulse F. t = m (Vf -Vi)

where, Vf is the final velocity and Vi is the initial velocity.

Given is a 3.42 kg steel ball strikes a massive wall at 14.3 m/s at an angle of α = 61.5° with respect to the plane of the wall. It bounces off with the same speed and angle. The ball is in contact with the wall for 0.207 s

Substitute the values into the expression, we get

Impulse = 2 x mvcosθ

Impulse= 2 x 3.42 x 14.3 x cos 61.5°

Impulse = 46.672 kg.m/s

The impact force can be written as

F.t = I

Put the given values, we have

F = 46.672 /  0.207

F = 225.469 N

Thus, the magnitude of force exerted by the wall is 225.469 N

Learn more about impulse.

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