A particle is projected up an inclined plane of base angle $\beta$ with the horizontal with an initial velocity $V$. The particle collides elastically with the incline and rebounds vertically. If the particle reaches back the point of projection after time
$\large T = \dfrac {aV}{g \sqrt{1+b\sin^2\beta}}$
then enter the value of $a + b$
Note: $g$ denotes the acceleration due to gravity.
Well first of all I am confused over the word "vertical". Vertical as in making an angle of $90^\circ$ with the plane or the horizontal?? Confused over this, I proceeded in two cases :-
CASE I From the plane
So when a particle is projected up an inclined plane, the time taken by it to complete it projectile motion is :
$T$ = $\frac{2u\sin(\alpha - \beta)}{g\cos\beta}$ ...... (i)
Where $\alpha$ is the angle the projectile makes with the horizontal(the smaller angle) and $\beta$ is the angle of the inclined plane.
When it is projected down an inclined plane :-
$T$ = $\frac{2u\sin(\alpha + \beta)}{g\cos\beta}$ ............. (ii)
Here, $\alpha$ and $\beta$ have the same meaning, angle from the horizontal and angle of the inclined plane respectively.
Now, from what is given in the question, we know that the range of the projectile will be the same in both cases. And we have to find the sum of the times to go up and down the plane.
When it is being projected down the plane, $\alpha + \beta$ is 90. Now since the projectile reaches back the point of projection, the ranges up and down the inclined plane should be equal.
The formulas for these are :
Up the plane : $\frac{u^2}{g\cos\beta}[\sin(2\alpha - \beta) - \sin\beta]$ ......... (iii)
Down the plane : $\frac{u^2}{g\cos\beta}[\sin(2\alpha + \beta) + \sin\beta]$ ............... (iv)
(Alpha and beta have the same meanings)
so (iii) = (iv) and $\alpha + \beta$ is 90. Using this, we get :-
$\sin(2\alpha - \beta) = 3\sin\beta$ ................... (v)
Now the total time would be (i) + (ii), which involves the variable \alpha. So I need to eliminate the variable $\alpha$. I could do that using (v), but that would involve writing $\cos2\alpha$ and $\sin2\alpha$ ONLY in terms of $\sin\alpha$ and then using that value of $\sin\alpha$ in order to eliminate $\alpha$ from (i) + (ii). However, I dont think the solution to be that long/untidy and am like 99% sure that there must be a shorter/neater way.
CASE II From the horizontal
I pretty much do the same steps as in case I. In this case I get from equating the ranges : -
$\sin(2\alpha - \beta) = \sin\beta$ which gives me $\alpha = \beta$. This not only makes the time up the plane to be 0, but also makes the time down the plane to be $\frac{2u}{g}$, which is the same for a freely falling body/1D. Also, in this case I think the particle will fall back to it's starting position, thus never getting back to its original position. So I am like 80% sure that this is the wrong case, but I havent just got the gist of projectile motion along an inclined plane, so I think that I may be wrong.
Also, the word "elastic" means that no energy is lost right, so the particle rebounds with the same velocity with which it struck. But I think I vaguely remember reading somewhere that it's also got to to do something about making equal angles with the normal or something like that. My best guess is that this might be used somewhere and thus providing us with an easier relation between $\alpha$ and $\beta$. Either this, or the worst case scenario that I have COMPLETELY misunderstood the problem. PLS correct me if I am wrong anywhere in this explanation. I tried my level best to keep in compliance with the homework policy. Pls be kind enough to point out any mistakes I might have made n this regard.
Thank you