Momentum problem involving collision of two objects.

[angotta]

1. A tennis player swings her 1000 g racket with a speed of 5.00 . She hits a 60 g tennis ball that was approaching her at a speed of 13.0 . The ball rebounds at 37.0 .
Part Aow 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.
part B:If the tennis ball and racket are in contact for 15.0 , what is the average force that the racket exerts on the ball?


2. Homework Equations :
I would assume p_f=p_i and also p=mv where p is momentum



3. The Attempt at a Solution
I found the initial momentum of the racket to be 5Ns and the initial momentum of the ball to be -.78 Ns
Then I attempted to find the final velocity of the racket and was stuck...I tried to first use the equation for momentum with a mass of 1.06 kg b/c that is the mass when the objects are in a system. the velocity final should be 37 m/s due to the fact that the rebound velocity is that, but I don't know where to go or what to do.

 

[LowlyPion]

OK. Using your numbers you have 5N-s -.78N-s before the collision. After the collision you have the ball at 60*37 = 2.22N-s (Note your units are not N, but N-s)

That means you need to still account for 2N-s with the racket.

Its mass is 1 kg so it's going at 2N-s/1kg = 2m/s then right?

Your units on the time are not available for Part B. But you will want to consider the Δmv of the ball over the time interval to determine the average force.

F = m*a = mΔv/Δ t = Δmv/Δt

 

[angotta]

OK. Using your numbers you have 5N-s -.78N-s before the collision. After the collision you have the ball at 60*37 = 2.22N-s (Note your units are not N, but N-s)

That means you need to still account for 2N-s with the racket.

Its mass is 1 kg so it's going at 2N-s/1kg = 2m/s then right?

Your units on the time are not available for Part B. But you will want to consider the Δmv of the ball over the time interval to determine the average force.

F = m*a = mΔv/Δ t = Δmv/Δt

That helped a lot thank you

 

[talaroue]

I have almost the exact question...
A tennis player swings her 1120 g racket with a speed of 10.6 m/s. She hits a 50 g tennis ball that was approaching her at a speed of 20.5 m/s. The ball rebounds at 43.9 m/s
I found the Vf of the racket but for the second part it asks...
If the tennis ball and racket are in contact for 10.52 ms, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?


I thought i found the correct anwser by subtracting the velocities, then multiply by mass of the racket, then divide by the time. It is saying it is wrong... What am I doing wrong?

 

[LowlyPion]

I have almost the exact question...
A tennis player swings her 1120 g racket with a speed of 10.6 m/s. She hits a 50 g tennis ball that was approaching her at a speed of 20.5 m/s. The ball rebounds at 43.9 m/s
I found the Vf of the racket but for the second part it asks...
If the tennis ball and racket are in contact for 10.52 ms, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?

I thought i found the correct anwser by subtracting the velocities, then multiply by mass of the racket, then divide by the time. It is saying it is wrong... What am I doing wrong?

What they gave you was the change in velocity of the ball. Shouldn't you want to consider the Δp of the ball that involves the mass of the ball and not the racket?

Keep in mind too that the velocities are vectors and the velocity of the ball reverses.

 

[talaroue]

What they gave you was the change in velocity of the ball. Shouldn't you want to consider the Δp of the ball that involves the mass of the ball and not the racket?.

Pi=Pf so if the momentuem Pb(momentuem of the ball), and Pr(momentuem of racket). Since intitally they are coming at each other Pi= Pb+Pr...correct, Then Pf=Pb-Pr since then are going opposite directions?

velocity of the ball reverses[/B].

so are you saying instead of subtracting, I should add them because of the fact that one is going in the positive direction and then the negative direction?

 

[LowlyPion]

Pi=Pf so if the momentuem Pb(momentuem of the ball), and Pr(momentuem of racket). Since intitally they are coming at each other Pi= Pb+Pr...correct, Then Pf=Pb-Pr since then are going opposite directions?

so are you saying instead of subtracting, I should add them because of the fact that one is going in the positive direction and then the negative direction?

I'm less certain why you are interested in the momentum of the racket, since they are asking for the force on the ball aren't they?

But as to your question about the velocities, they are subtracted, but as you say they are in opposite directions so ...
... it does result in |v| + |v'|.

 

[talaroue]

o ok... when i said racket i meant ball i understand now. thanks