Question

A 57 g tennis ball is traveling at 45 m/s to the right, when it hits a racket. The ball reversed direction and travels at 33 m/s. The ball is in contact with the racket for 0.0085 s.

What is the magnitude of the force that was exerted on the ball?

A. 1.8N
B. 12N
C. 81N
D. 520N

Answer 1

Answer:D 520N

Step-by-step explanation:

Answer 2

D. 520 N

Step-by-step explanation:

According to the impulse theorem, the change in momentum of the ball is equal to the impulse exerted on the ball, mathematically:

`\Delta p = I = F \Delta t`

where

`\Delta p` is th change in momentum

F is the average force exerted on the ball

`\Delta t` is the time of contact

The change in momentum of the ball is

`\Delta p = m(v-u)`

where

m = 57 g = 0.057 kg is the mass of the ball

u = 45 m/s is the initial velocity

v = -33 m/s is the final velocity (in the opposite direction)

Substituting,

`\Delta p = (0.057)(-33-45)=4.45 kg m/s`

The time of the collision is

`\Delta t = 0.0085 s`

Therefore, the average force is

`F=(\Delta p)/(\Delta t)=(-4.45)/(0.0085)=-523 N`

So, the magnitude is approximately 520 N.