You are Here: Home >< Maths

# M1 - Suvat

Announcements Posted on
Would YOU be put off a uni with a high crime rate? First 50 to have their say get a £5 Amazon voucher! 27-10-2016

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.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 7, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### University open days

Is it worth going? Find out here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams