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# M1 - Suvat watch

1. For the time between the first and second bounce I used suvat

s = 0 u = sqrt(gh) found from first part a = -9.81 t = t

using S = ut + 0.5at^2

t = 2 * sqrt(g/h)

How do I do the last part because I don't know how to get u as there is no 2nd height given in the question.

Thanks
2. (Original post by khanpatel321)

For the time between the first and second bounce I used suvat

s = 0 u = sqrt(gh) found from first part a = -9.81 t = t

using S = ut + 0.5at^2

t = 2 * sqrt(g/h)

How do I do the last part because I don't know how to get u as there is no third height given in the question.

Thanks
Use the coefficient of restitution: the velocity of the ball after the second bounce will be given by coefficient of restitution * velocity before the second bounce. You've found the coefficient of restitution in an earlier part, no?
3. (Original post by khanpatel321)

For the time between the first and second bounce I used suvat

s = 0 u = sqrt(gh) found from first part a = -9.81 t = t

using S = ut + 0.5at^2

t = 2 * sqrt(g/h)

How do I do the last part because I don't know how to get u as there is no 2nd height given in the question.

Thanks
A ball falling from a height 1/2h after rebounding from the floor, is no different to a ball falling from a height 1/2h after being dropped at that height.Hence the third height will be 1/4h.
4. (Original post by Zacken)
Use the coefficient of restitution: the velocity of the ball after the second bounce will be given by coefficient of restitution * velocity before the second bounce. You've found the coefficient of restitution in an earlier part, no?
Ok so is the coefficient of restitution for a given surface constant?

I get the the answer 2*sqrt(gh) / g*sqrt2 this is equivalent to the answer in the mark scheme 2*sqrt(h/2g) how do you simplify my answer to the one given? I can't do it.
5. (Original post by khanpatel321)
Ok so is the coefficient of restitution for a given surface constant?

I get the the answer 2*sqrt(gh) / g*sqrt2 this is equivalent to the answer in the mark scheme 2*sqrt(h/2g) how do you simplify my answer to the one given? I can't do it.
and and

So: .
6. (Original post by khanpatel321)
Ok so is the coefficient of restitution for a given surface constant?

I get the the answer 2*sqrt(gh) / g*sqrt2 this is equivalent to the answer in the mark scheme 2*sqrt(h/2g) how do you simplify my answer to the one given? I can't do it.
Use g=sqrt(g^2) and that sqrt(a)/sqrt(b)=sqrt(a/b) and then use cancelling.
7. (Original post by Zacken)
and and

So: .
(Original post by Dalek1099)
Use g=sqrt(g^2) and that sqrt(a)/sqrt(b)=sqrt(a/b) and then use cancelling.
Ohh ok nice thanks.

One last thing I can use the coefficient of restitution found in the earlier part because i'm using the same ball right? If I changed the balls mass I would have to work out e again for that particular ball right?
8. (Original post by khanpatel321)
Ohh ok nice thanks.

One last thing I can use the coefficient of restitution found in the earlier part because i'm using the same ball right? If I changed the balls mass I would have to work out e again for that particular ball right?
Nopes, coefficient of restitution doesn't depend on mass. If you're using the same ball and same floor then the same coefficient of restitution applies.
9. (Original post by Zacken)
Nopes, coefficient of restitution doesn't depend on mass. If you're using the same ball and same floor then the same coefficient of restitution applies.
oh, okay thank you.
10. (Original post by khanpatel321)
oh, okay thank you.
No problem.

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