Submitted by Meryjoy99 on Fri, 04/07/2017 - 11:34
Forums:
In part ii), my expression for w is the required but without the factor of cos(alpha) squared multiplying m in the denominator... where does it come from?? Perhaps I reached a similar expressions why a wrong method...
I think you'd need to post
I think you'd need to post your working so we can identify where anything went wrong.
I seem to have done exactly
I seem to have done exactly the same thing.
For part (i)
Using conservation of momentum I obtained $mu = Mv - mw$ where v and w are the speeds of cone and particle respectively after the collision
Using restitution I obtained $e = \frac{v+w}{u} $
Rearranging and solving for v gets $v = \frac{u(e+m)}{M+m}$ then multiply by $cos({\alpha})$ to get the horizontal velocity
On part (ii) I subbed $usin(\alpha)$ in for u as the restitution equation and conservation of momentum only apply parallel to the direction of motion which leaves me with the same expression as the answer but without the $cos^2(\alpha)$ in the denominator
At a quick glance, for the
At a quick glance, for the first part wouldn't conservation of momentum give $Mv - mu\cos \alpha = mu\cos \alpha$ and restitution $v\cos \alpha + w = eu$? So your final expression for $v$ isn't quite correct.
I think I tried to make
I think I tried to make everything parallel to the slope rather than working in the horizontal plane for the conservation of momentum and multiplied through by $cos(\alpha)$ later so the numerator looked right. But I didn't do the same for the restitution which is where the additional $cos(\alpha)$ comes from to get squared in the denominator.
After subbing using the new equations I get $$v=\frac{mu(e + 1)cos(\alpha)}{M + mcos^2(\alpha)}$$
For part (ii) $usin(\alpha)$ gets the velocity perpendicular to the cone's surface then use $usin(\alpha)cos(\alpha)$ for the horizontal momentum so: $$Mw - mvcos(\alpha)sin(\alpha) = mucos(\alpha)(sin(\alpha)$$
Then coefficient of restitution is: $$eusin(\alpha) = v + wcos(\alpha)$$ which doesn't get me the required value for w when substituting in.
Yep, you've fixed (i). In (ii
Yep, you've fixed (i). In (ii) should you really have a $v$ in the second term of the LHS of your horizontal momentum? Looks like it should be $Mw - mu\cos \alpha \sin \alpha mu\cos \alpha \sin \alpha$ and I agree with your restitution, so it should work out now.
The LaTeX messed up, I meant
The LaTeX messed up, I meant to say that it should be $Mw - mv_n\cos \alpha = mu\cos \alpha \sin \alpha$ where $v_n$ is the velocity of the particle after the collision, normal to the surface.
So I reformed my momentum
So I reformed my momentum equation as $Mw-mvcos(\alpha) = mucos(\alpha)sin(\alpha)$ which when used with the restitution gets the required answer but I think that was more guesswork than anything else. I think the issue is more a lack of understanding so if I attempt to explain my previous reasoning could you try and see where I've gone wrong?
Using w as the horizontal velocity of the cone after the collision, and v as the velocity of the particle normal to the surface - since the parallel velocity remains unchanged.
From conservation of momentum horizontally: $usin(\alpha)$ is the velocity of the particle perpendicular to the cone's surface before the collision so $usin(\alpha)cos(\alpha)$ is the horizontal velocity of the particle and $ucos^2(\alpha)$ the vertical
So momentum before collision is $musin(\alpha)cos(\alpha) + mucos^2(\alpha)$
After the collision w is in the positive direction so $Mw$ as well as the $mucos^2(\alpha)$
In the negative direction, you get $v$ which since it is normal to the surface goes to $vcos(\alpha)$
So the full equation becomes
$Mw-mvcos(\alpha)+mucos^2(\alpha) = mucos(\alpha)sin(\alpha)+mucos^2(\alpha)$
Looking at the restitution LHS is clearly $eusin(\alpha)$
Then RHS since the velocities are in opposite directions $v + wcos(\alpha)$ accounting for w being in the horizontal plane
So $eusin(\alpha) = v + wcos(\alpha)$
Amusingly having written all this I worked out the issue, since that reaches the desired equations, not defining my variables clearly and not drawing enough diagrams.
On a related note, what method do you use to work out which of $sin(\theta)$ or $cos(\theta)$ etc is the right angle to use in some question? Here I struggled slightly to see if it was $wcos(\alpha)$ or $wsin(\alpha)$ - I try and visualise rotations to a different plane but that doesn't work for all situations.
With enough practice, it
With enough practice, it becomes immediately obvious which directions correspond to $\sin$ and which to $\cos$. One alternative is to think about when $\theta = 0$ or $\theta = \frac{\pi}{2}$, should your force/velocity in that particular direction be of full magnitude or 0? That'll tell you what trig function corresponds to it, since $\sin 0 = 0$ and $\cos 0 = 1$, similarly for $\frac{\pi}{2}$.
Does that help? (good work with the answer btw)
Ah yes, thank you very much,
Ah yes, thank you very much, that makes life a lot easier. Also, thanks with all the help getting me to the answer
No problem!
No problem!
Just when you think it's over
Just when you think it's over...
I was playing around with angles to check my understanding and when I stated that the component of u parallel to the surface of the cone was $ucos(\alpha)$ so the part of that in the horizontal direction was $ucos^2(\alpha)$, that was wrong wasn't it? That part should have been $ucos(\alpha)sin(\alpha)$ right?