Question

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the horizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.066 s.

1)

What is the magnitude of the initial momentum of the racquet ball?

kg-m/s

2)

What is the magnitude of the change in momentum of the racquet ball?

kg-m/s

3)

What is the magnitude of the average force the wall exerts on the racquet ball?

Answer 1

Step-by-step explanation:

Given that,

Mass of the, m = 0.256 kg

Initial peed of the ball, v = 11.8 m/s

The ball is in contact with the wall for t = 0.066 s

(1) The magnitude of the initial momentum of the racquet ball is given by :

p = mv

`p=0.256\ kg* 11.8\ m/s\\\\p=3.02\ kg-m/s`

(2) It is projected at angle of 29 degrees. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal.

The change in momentum of the racquet ball is given by :

`\Delta p = m(v\cos \theta - (-v\cos \theta))\\\\\Delta p = m(v\cos \theta - (-v\cos \theta))\\\\\Delta p = 2mv\cos \theta\\\\\Delta p = 2* 0.256 * 11.8* \cos (29)\\\\\Delta p=5.28\ kg-m/s`

(3) Impulse is equal to the change in momentum.

`\Delta p=Ft\\\\F=(\Delta p)/(t)\\\\F=(5.28)/(0.066 )\\\\F=80\ N`

Hence, this is the required solution.