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Coefficient of restitution

10c

I've said that assuming u>0 then if direction were changed then the speed of P

\dfrac{u}{2}(5-3e)<0 \text { which leads to } e >\dfrac{5}{3}

which can't be true as

0 \leq e \leq 1

.

Is this fine?

Thanks :smile:

Sent from my SM-G925F using Tapatalk

(edited 8 years ago)

Reply 1

Duke Glacia

Original post by Kvothe the arcane

Part c

I've said that assuming u>0 then if direction were changed then the speed of P

\dfrac{1}{2}(5-3e)\dfrac{5}{3}

which can't be true as

0 \leq e \leq 1

.

Is this fine?

Thanks :smile:

Sent from my SM-G925F using Tapatalk

Looks fine. Can u post the question ?

Reply 2

Kvothe the Arcane

Original post by Duke Glacia

Looks fine. Can u post the question ?

Thanks

. Pleaze see edit.

Reply 3

Duke Glacia

Original post by Kvothe the arcane

Thanks

. Pleaze see edit.

Yup perfect.

Lovely handwriting btw

Reply 4

Ayman!

Original post by Kvothe the arcane

I've said that assuming u>0 then if direction were changed then the speed of P

\dfrac{u}{2}(5-3e)<0 \text { which leads to } e >\dfrac{5}{3}

which can't be true as

0 \leq e \leq 1

I'm not sure if yours sufficiently shows it or not, but this is how I'd do it:

\displaystyle 0 \leq \mathrm{e} \leq 1 \Rightarrow u \leq \mathbf{v}_{\text{p}}\leq \frac{5u}{2} \Rightarrow \mathbf{v}_{\text{p}}> 0

Reply 5

Zacken

Original post by Kvothe the arcane

...

It's a bit... weird.

It'd be better to say that since

0 \leq e \leq 1

, then

\frac{u}{2} (5-3e) > 0

. i.e: a direct proof is easily found here so using an indirect one (contradiction) feels off.

Reply 6

Kvothe the Arcane

Original post by Duke Glacia

Yup perfect.

Lovely handwriting btw

Thanks. It's okay but at least it's legible :h:

Original post by Ayman!

I'm not sure if yours sufficiently shows it or not, but this is how I'd do it:

\displaystyle 0 \leq \mathrm{e} \leq 1 \Rightarrow u \leq \mathbf{v}_{\text{p}}\leq \frac{5u}{2} \Rightarrow \mathbf{v}_{\text{p}}> 0

That makes sense :yep:

. Thanks.

Original post by Zacken

It's a bit... weird.

It'd be better to say that since

0 \leq e \leq 1

, then

\frac{u}{2} (5-3e) > 0

. i.e: a direct proof is easily found here so using an indirect one (contradiction) feels off.

I see. Fair enough. That's what made sense to me at the time so I'm glad I posted :smile:

. Thanks.

Reply 7

Zacken

Original post by Kvothe the arcane

Thanks.

Cheers.

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Coefficient of restitution

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