Question

A (1 kg) soccer ball is heading toward a wall with a speed of 20 meters per second. After hitting the wall, the ball bounces back with a speed of 25 meters per second. The ball was in contact with the wall for 0.003 seconds. What is the average force the wall exerted on the ball?

Option 1: 5000 N
Option 2: 4166.67 N
Option 3: 3333.33 N
Option 4: 2500 N

Answer 1

The average force exerted by the wall on the ball is 15,000 N (in the opposite direction of the initial velocity).

Step-by-step explanation:

To find the average force exerted by the wall on the ball, we can use the impulse-momentum principle.

The impulse is the change in momentum of the ball during the collision. The momentum of an object is given by its mass multiplied by its velocity.

To calculate the impulse, we subtract the final momentum from the initial momentum. The average force can then be found by dividing the impulse by the time of contact.

Given:

1. Mass of the ball (m): 1 kg
2. Initial velocity of the ball (vi): 20 m/s
3. Final velocity of the ball (vf): -25 m/s (negative sign indicates a change in direction)
4. Time of contact (t): 0.003 s

Using these values, we can calculate the impulse:

Impulse = (Final momentum - Initial momentum) = (m * vf) - (m * vi)

Impulse = (1 kg * -25 m/s) - (1 kg * 20 m/s) = -25 kg·m/s - 20 kg·m/s = -45 kg·m/s

To find the average force, we divide the impulse by the time of contact:

Average force = Impulse / t = (-45 kg·m/s) / (0.003 s) = -15,000 N

Since force is a vector quantity, the negative sign indicates that the force exerted by the wall is in the opposite direction of the initial velocity of the ball.

Therefore, the average force the wall exerted on the ball is 15,000 N (in the opposite direction of the initial velocity).