M3 Projectile help Watch

ubisoft
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Attachment 381743

Can someone help me with 7b?

7a is 2Usinθ/gcosα

aii: Component is -Usinθ

7b: Ive shown that the y-component after the rebound is eusinθ. But how do you work out the speed? Using Pythagoras with the x-comp is too complicated. Is the rebound angle θ again, and thus the speed just eu?
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ghostwalker
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(Original post by ubisoft)
...
I know it's been a few days and you've already had replies on another thread, but for the benefit of others....

Covering the whole question.

The easiest way to do this question is to let our x,y axes be along the plane and perpendicular to the plane, with the origin at O.

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Since acceleration is constant, suvat applies.

Using s=ut+\frac{1}{2}at^2 perpendicular to the plane. When the projectile hits the slope again, we have s=0.

So,0=ut+\frac{1}{2}at^2=t(u+\frac{1  }{2}at)

This has two solutions t=0 being the time at the point of projection, and 0=u+\frac{1}{2}at for when the projectile hits the plane again.

From which we get: t=-\frac{2u}{a}...(1) as the time from projection to hitting the plane. Note: time of flight is proportional to the initial velocity in the y-direction for fixed a.

We also note that when the projectile hits the plane again, using v^2=u^2+2as with s=0, we have v^2=u^2, so v=+/- u. And since the projectile has reversed its direction we have v=-u...(2)

So, regarding the actual question:
(a)(i).

Acceleration perpendicular to the plane is -g\cos\alpha, and initial velocity is U\sin\theta

By (1), time of flight is \frac{2U\sin\theta}{g\cos\alpha}

(a)(ii)

By (2) when the projectile strikes the hill the velocity has the same magnitude perpendicular to the hill, as when it was projected.

(b)

Perpendicular component of velocity on striking hill is -U\sin\theta, and hence perpendicular component of velocity when rebounding at A is eU\sin\theta

By (1) time of flight is proportional to perpendicular velocity and so ratio of times of flight is  U\sin\theta:eU\sin\theta which simplifies to 1:e
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ubisoft
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Thanks for that. So the rebound angle with the plane is theta again?
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ghostwalker
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(Original post by ubisoft)
Thanks for that. So the rebound angle with the plane is theta again?
Do you mean the angle the trajectory makes with the plane after the first impact? If so, no, it's not theta.

Try working out the components of the velocity parallel and perpendicular to the plane - though it may be clear to you without calculation, if you consider what the variables are for the y-component, compared with the x-component.
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ubisoft
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(Original post by ghostwalker)
Do you mean the angle the trajectory makes with the plane after the first impact? If so, no, it's not theta.

Try working out the components of the velocity parallel and perpendicular to the plane - though it may be clear to you without calculation, if you consider what the variables are for the y-component, compared with the x-component.
yeah like this https://www.physicsforums.com/attach...fig-jpg.55133/

If the y-comp after the bounce is eusin(theta), then shouldn't the actual velocity be just eu (resolving components)?

I get everything in your answer apart from the very last line
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ghostwalker
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(Original post by ubisoft)
yeah like this https://www.physicsforums.com/attach...fig-jpg.55133/

If the y-comp after the bounce is eusin(theta), then shouldn't the actual velocity be just eu (resolving components)?

I get everything in your answer apart from the very last line
The component of velocity parallel to the plane is uneffected by the bounce and its formula does not include "e" as one of its variables. You could work it out... v=u+at for OA parallel to the plane. Whereas the perpendicular component is going to vary depending on "e".
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