How Is Force Calculated When Momentum Change Appears Zero?

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Homework Help Overview

The problem involves a 100 g snooker ball striking a cushion at an angle of 45° and bouncing back with the same speed and angle. The objective is to determine the average force exerted by the cushion on the ball during the contact time of 0.05 s, despite the apparent zero change in momentum.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of momentum before and after the collision, noting that the initial and final momenta appear equal. Questions arise about the implications of zero change in momentum for calculating force.

Discussion Status

Some participants suggest systematic approaches to break down the problem, while others point out potential mistakes in the subtraction of momentum components. There is acknowledgment of errors, but basic principles are affirmed without reaching a consensus on the final calculation.

Contextual Notes

Participants are navigating the implications of impulse and momentum in the context of a collision, questioning the assumptions made regarding the direction and magnitude of forces involved.

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Homework Statement



A 100 g snooker ball with speed 50 cm/s strikes a cushion at an angle of 45° and bounces back off the same speed and angle. If the ball is in contact with the cushion for 0.05 s, what is the average force exerted by the cushion on the ball?

The Attempt at a Solution



I get final momentum=initial momentum because the angle, mass and the speed are the same before and after.

[tex]p_i=(0.1)(0.5 cos 45) = p_f[/tex]

The change in momentum is 0.

So, the impulse is equal to [tex]I=\Delta \vec{p}[/tex]. To find the force I think we can use the formula

[tex]\vec{I}=\sum \vec{F} \Delta t[/tex]

But how can I find force when the change in momentum is zero? Is there something wrong with my strategy or did I make some mistakes?
 
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You need to approach these questions systematically
ie.
Vi=50cm/s* cos45i+50cm/s* sin45j
vf=-50cm/s*cos45+50cm/s* sin45j
j=pf-pi
=0.5cos45i*2*100i+0.5cos45*2*100j as cos 45=sin 45
=0.0707i+0.0707j
J=0.1=Ft
F=0.05/0.1
F=0.5N

Did you understand the other momentum problem from before??
 
pat666 said:
You need to approach these questions systematically
ie.
Vi=50cm/s* cos45i+50cm/s* sin45j
vf=-50cm/s*cos45+50cm/s* sin45j
j=pf-pi
=0.5cos45i*2*100i+0.5cos45*2*100j as cos 45=sin 45
=0.0707i+0.0707j
J=0.1=Ft
F=0.05/0.1
F=0.5N




Did you understand the other momentum problem from before??
I think you made a mistake when subtracting the initial and final momenta
 
whys that what answer did you get? the plus is there as its a double negative
 
pat666 said:
whys that what answer did you get? the plus is there as its a double negative
you made a mistake when subtracting the j-components for the impulse. If you notice, it's equal for both the final and initial momenta so the change in the j-direction is zero.
 
yep your right,,,, . that's my bad sorry but the basic principals are still right. agree??
 
pat666 said:
yep your right,,,, . that's my bad sorry but the basic principals are still right. agree??

Yes, that's right.
 

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