# Maximum height after collision in a bowl

1 vote
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Consider two masses $m, M$ that fall from the top of a hemispherical bowl (from rest) and slide down without any friction.

We want to find how high each of the masses will go after the collision assuming the collision is elastic.

This is how far I can go. It seems to me there are too many unknowns and I am missing some key simplification.

edited Oct 8, 2018
Instead of using the conservation of energy you can use the Law of Restitution, which says that for an elastic collision the relative velocity of approach equals the relative velocity of separation. You still have to use the conservation of momentum also.

You are making the solution harder than it needs to be.  There are 3 steps :

1. Calculate the velocities at collision, which occurs at the bottom of the circle and is $\sqrt{2gh}$ for each .

2. Calculate the velocities after collision. Think of it as a linear collision.

3. The final heights are again given by $v^2=2gh$  but using the velocities after collision.
How do I know the collision occurs at the bottom of the bowl? Is there a mathematical/physical proof that that is true? I can see that they would hit each other at the bottom if they had equal masses.
For the same reason that all bodies accelerate at the same rate $g$ due to gravity.

You can think of the hemispherical bowl as an inclined plane with a variable angle of inclination $\theta$. The acceleration down the incline is $g\sin\theta$ which is independent of mass. Both blocks start from rest at the same height and accelerate at the same rate, so they have the same speed at every height. They reach the bottom at the same time and with the same speed.

If the two blocks have the same mass and the collision is elastic (KE conserved) then they will both return to the height from which they started, because of conservation of energy.

If the masses are different but the collision is still elastic, the smaller mass will rise higher than it started and the larger mass will rise lower than it started. When it reaches the top of the hemisphere the smaller mass will keep going straight up. This is actually quite important because if the hemisphere was instead a sphere then the smaller block would continue moving in a circle. Unless it has enough KE to "loop-the-loop" then it wil lose contact with the sphere at some point and become a projectile with a horizontal component of velocity.  As a result it would not climb as high as when it moves straight up, because it is not stationary at its highest point.

If the two blocks were attached to a smooth circular track then after the collision the following relation holds, unless the smaller block does a "loop-the-loop". (It would be very interesting to calculate (i) what minimum ratio of $\mu=M/m$ is required for the smaller mass to "loop-the-loop", and (ii) how many collisions there are for each value of $\mu$ - there will be a collision each time the smaller block reverses direction and "loops-the-loop" .)

Using conservation of energy (ie gravitational PE), and assuming the blocks are stationary at their maximum heights, the maximum heights after the collision will be related by $$mgh+Mgh=mgh'+Mgh''$$  I was tempted to suggest that the COM will remain in the same position horizontally throughout, but this is not true. The COM starts on the right of the centre line (assuming $M \gt m$) but it is on the centre line when the blocks meet at the lowest point. So the COM does move, because there are external forces acting horizontally on the two blocks - namely the normal contact forces from the hemisphere. (If the hemisphere were resting on a frictionless surface, the COM of the hemisphere and blocks would stay in the same position horizontally, because there are no external forces acting horizontally on this system.)

Another nuance which might affect the result is that the blocks as drawn would actually rotate - ie change their orientation in space. This *might* affect the speed with which they slide down the hemisphere, and also the position at which they collide. It would not affect the above relation, because all kinetic energy (rotational as well as translational) is zero when the blocks come to rest.
Outstanding explanation.
I shall leave you to obtain the solution to the problem. In a few days I will post a solution.

1 vote

The two blocks reach the bottom of the bowl at the same time with the same speed $u=U=\sqrt{2gh}$, as discussed in the comments. These are the initial speeds in the collision.

Momentum and kinetic energy are both conserved during the elastic collision. Instead of applying conservation of KE it is simpler to apply the Law of Restitution with a coefficient $e=1$. This implies that the relative velocity of separation equals the relative velocity of approach.

If the final speeds are $v, V$ for small and large blocks respectively, taking the +ve direction as left for the smaller block and right for the larger, then conservation of linear momentum and the law of restitution give $$(M-m)u=mv-MV$$ $$2u=v+V$$ From these we get $$v=\frac{3M-m}{M+m}u$$ $$V=\frac{3m-M}{m+M}u$$ Note that if $M \gt 3m$ then the larger block does not rebound ($V\lt 0$); it continues moving in its initial direction, chasing the smaller block. The smaller block always rebounds because $3M \gt m$ if $M \gt m$ hence $v \gt 0$ always.

The heights reached after the collision by the smaller and larger blocks are respectively $$h'=\frac{v^2}{2g}=(\frac{3M-m}{m+M})^2h$$ $$H'=\frac{V^2}{2g}=(\frac{3m-M}{M+m})^2h$$

answered Oct 9, 2018 by (26,660 points)
selected Oct 9, 2018
1 vote

This is my solution. It is brute force algebra so not as elegant as the solution above.

There is also energy conservation for each body after the collision

Linear momentum conservation means that

and because the collision happens at the bottom of the bowl $v{My}'=v{my}'=0$.

Multiplying this equation by itself gives

and substituting in the equations for energy conservation for each mass after the collision we get the heights they reach after the collision

answered Oct 9, 2018 by (270 points)
edited Oct 10, 2018