Physics impulse momentum problem

[rzn972]

Homework Statement

A 60.0 gram tennis ball is served at 45.0 m/s and is returned with an average force of 100 N over 50 ms. Find the speed with which the ball is returned. Assume the ball is still moving at 45.0 m/s when it is returned.

Homework Equations

Impulse = Average Force * time

The Attempt at a Solution

Pi= mv= .06*45 = 2.7
Pf= mv= .06*v
Impulse = Average Force * time= 100 * .05= 5
dp= pf-pi
5= .06v-2.7
v= 128.3
The answer should be 38.3. What did I do wrong?

 

[SammyS]

Homework Statement

A 60.0 gram tennis ball is served at 45.0 m/s and is returned with an average force of 100 N over 50 ms. Find the speed with which the ball is returned. Assume the ball is still moving at 45.0 m/s when it is returned.

Homework Equations

Impulse = Average Force * time

The Attempt at a Solution

Pi= mv= .06*45 = 2.7
Pf= mv= .06*v
Impulse = Average Force * time= 100 * .05= 5
dp= pf-pi
5= .06v-2.7
v= 128.3
The answer should be 38.3. What did I do wrong?

First of all: The statement in the above quote, which I highlighted is a bit puzzling. (Oh! I see what's meant. The ball is still moving at 45.0 m/s when it gets to the returner's racket.)What you did wrong was to ignore that the velocity ( and thus the momentum ) of the ball changes direction, thus it changes sign. This assumes that the ball is returned exactly 180° from the direction it was received.

 

[paisiello2]

Check the sign convention on your velocity.

 

[rzn972]

Okay, so abs (change in impulse) = force *times.
So I change the velocity final to -v.
abs (-.06v -2.7) = 5
v= -128.3
v= 38.3
I use the positive 38.3 because I already put in the negative sign?
And yeah "Assume the ball is still moving at 45.0 m/s when it is returned." confused me too... The ball returns at the same speed that it leaves?