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A tennis player swings her 1000g racket with a speed of 10 m/s. She hits a 60g tennis ball that was approaching her at a speed of 20 m/s . The ball rebounds at 40m/s.

a. How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.

b. If the tennis ball and racket are in contact for 10 m/s, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?

Short Answer

Expert verified

a. The speed of racket after impact is 6.4m/s

b. The gravitational force is smaller than the force due to impulse.

Step by step solution

01

Part (a). Step 1. Given information

Mass of racket, m=1000g

Speed of racket, 10m/s

Mass of tennis ball60g60g

Speed of tennis ball, 20m/s

Final speed of tennis ball,40m/s

02

Step 2. Analyze

The collision in this issue is an elastic collision in which the laws of momentum conservation and energy conservation are applied. Mechanical energy is preserved in a perfectly elastic collision. The conservation of momentum law, which is expressed as an equation in the form

pโ†’i=pโ†’f

The momentum is given by, p=mv

Considering the parameters,

m1v1i+m2v2i=m1v1f+m2v2fm1v1f=m1v1i+m2v2i-m2v2fv1f=m1v1i+m2v2i-v2fm1

03

Step 3. Evaluate

On substitution of values,

v1,f=1gร—10m/s+0.06gร—(-20m/s-40m/s)1gv1,f=6.4m/s

04

Part (b) Step 1. Analyze

When force act for a small interval of time, impulse acts on the area,

so the equation becomes,

pf,x=pi,x+Jx

So evaluating the value of impulse,

Jx=0.06ร—40-0.06ร—(-20)=3.6kg.m/s

05

Step 2. Evaluating forces value

As we know,

Jx=Fโ–ณtF=3.6kgร—m/s10-2s=360N

Now the gravitational force of the ball,

Fg=mg=60gร—9.8=0.59N

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