M1 - Suvat

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

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

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.

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.

$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ and $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and $\displaystyle \frac{\sqrt{a}}{a} = \frac{a^{1/2}}{a} = a^{1/2 -1} = a^{-1/2} = \frac{1}{\sqrt{a}}$

So: $\displaystyle \frac{2\sqrt{gh}}{g\sqrt{2}} = \frac{2\sqrt{h}\sqrt{g}}{g\sqrt{2}} = \frac{2\sqrt{h}}{\sqrt{2}\sqrt{g}} = \frac{2\sqrt{h}}{\sqrt{2g}} = 2\sqrt{\frac{h}{2g}}$ .

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?

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.