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# Restitution Watch

1. See attachment for question. I need help with part three, where I thought that sqrt(0.05) was the answer. The answer is 0.5, by the way.
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2. (Original post by bbrain)
See attachment for question. I need help with part three, where I thought that sqrt(0.05) was the answer. The answer is 0.5, by the way.
By restitution, since the ground isn't moving:

"velocity after impact" = -e ("velocity before impact")

Post some working if it's not coming out.
3. The speed of B immediately before impact = 2 ms^-1
The speed of B immediately after impact = sqrt(2^2-2*g*0.05) = sqrt(3)
Then e is not 0.5

4. (Original post by bbrain)
The speed of B immediately after impact = sqrt(2^2-2*g*0.05) = sqrt(3)
That's not correct, but I don't know what you're doing to get there.
5. So what should I do?
6. Knowing it reaches a height of 0.05m, you can either use conservation of energy, or .

Since you're not telling me what you did use, I can't tell why it's incorrect.
7. Got it. Since v=2gh=2*10*0.05 =1, and the speed before impact =2, 1/2=0.5
8. (Original post by bbrain)
Got it. Since v=2gh=2*10*0.05 =1, and the speed before impact =2, 1/2=0.5

Hence and hence .

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