Physics impulse momentum problem

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Homework Help Overview

The problem involves a tennis ball's momentum and impulse during a serve and return scenario. Participants are tasked with calculating the return speed of the ball after it is hit back with a specified force over a given time interval.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of initial and final momentum, questioning the treatment of velocity direction and sign conventions. There is an exploration of how impulse relates to the change in momentum and the implications of the ball's direction change.

Discussion Status

Some participants have offered insights into the importance of considering the direction of the ball's velocity when calculating momentum. There is an ongoing examination of the assumptions made in the problem setup, particularly regarding the ball's speed and direction upon return.

Contextual Notes

Participants note confusion regarding the statement that the ball is still moving at 45.0 m/s when returned, prompting further discussion on the implications of this assumption on the calculations.

rzn972
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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?
 
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rzn972 said:

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.
 
Check the sign convention on your velocity.
 
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?
 

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