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Impulse and average force 
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#1
Mar2309, 05:30 PM

P: 45

1. The problem statement, all variables and given/known data
A 2.18 kg steel ball strikes a massive wall at 12.6 m/s at an angle of a = 55.1o with the plane of the wall. It bounces off with the same speed and angle (as seen in the figure below). If the ball is in contact with the wall for 0.152 s, what is the average force exerted on the ball by the wall? 2. Relevant equations mv/change in t = average force v=sq. root 2gh 3. The attempt at a solution I've tried finding the impulse and then dividing it by the .152 sec. but I keep getting the wrong answer. Could someone please walk me through the equations and steps? Thanks! 


#2
Mar2309, 05:44 PM

HW Helper
P: 5,343




#3
Mar2309, 06:29 PM

P: 45

momentum= mass(velocity)...2.18*12.6=27.5



#4
Mar2309, 06:36 PM

HW Helper
P: 5,343

Impulse and average force
Think x,y components. 


#5
Mar2309, 06:40 PM

P: 133

Resolve the initial and final velocity vectors to find out which component undergoes a change.



#6
Mar2309, 06:49 PM

P: 45

v1 prime would be v1
v2 prime would be 0 how does that fit in to the problem? 


#7
Mar2309, 07:10 PM

HW Helper
P: 5,343

Now not all the momentum changes here. For instance the momentum  to the wall is unchanged after contacting the wall. But the ⊥ component has a change. Figure that change as I think you will be needing it. 


#8
Mar2309, 07:13 PM

P: 45

I don't understand... Is that not just the negative of the velocity?



#9
Mar2309, 07:39 PM

HW Helper
P: 5,343

It did not. 


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