Question

A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the right. If the ball is in contact with the wall for 0.25 s, what is the constant force exerted on the ball by the wall?

Answer 1

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = $2.2\times 7.4=16.28$

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = $2.2\times -6.2=-13.64$

Thus change in moment is: $16.28-(-13.64)=29.92$

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

$\frac{29.92}{0.25}=119.68$

Hence, the constant force exerted on the ball by the wall is 119.68 N.