# impulse and average force

by Knfoster
Tags: average force, impulse
 P: 46 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!
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P: 5,346
 Quote by Knfoster 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!
What is the Δ in momentum?
 P: 46 momentum= mass(velocity)...2.18*12.6=27.5
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P: 5,346

## impulse and average force

 Quote by Knfoster momentum= mass(velocity)...2.18*12.6=27.5
But what was the Δ in momentum.

Think x,y components.
 P: 133 Resolve the initial and final velocity vectors to find out which component undergoes a change.
 P: 46 v1 prime would be -v1 v2 prime would be 0 how does that fit in to the problem?
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P: 5,346
 Quote by Knfoster v1 prime would be -v1 v2 prime would be 0 how does that fit in to the problem?
Impulse requires knowing the Δ in momentum.

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.
 P: 46 I don't understand... Is that not just the negative of the velocity?
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P: 5,346
 Quote by Knfoster I don't understand... Is that not just the negative of the velocity?
Negative of the original velocity means that it returned in the direction it came.

It did not.

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