# Collision of point mass and sphere in particular fashion

BvU
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I'm convinced. This makes the exercise physically despicable and the accompanying picture (purposely?) misleading. It's a mathematical figment of imagination -- and yet I'm still curious to know the trick hidden deep inside there. Almost enough to join 'Brilliant' to find out ... @BvU people have solved this problem in fact only 5 have but none have posted solution

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@BvU are you still trying that question i am getting nowhere i decided to give up on that sorry
meanwhile i have another question
why must a ball drop when drop it from a certain height as it is not going to get to lower energy level then why the motion of falling
also i am stuck on another similar but slightly easier problem should i post it here or separate thread?

BvU
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post it here or separate thread
Separate thread

haruspex
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As already deduced, the two masses are the same and they have equal and opposite vertical velocity components immediately after first impact.

When two equal masses collide elastically, the velocity components normal to the plane of contact simply swap over. The point mass therefore comes to rest in that direction, moving off tangentially to the sphere. Since that is 60 degrees above the horizontal, the line of centres must be 30 degrees below it.

It follows from the above that the sphere acquires a greater horizontal velocity than the point mass retains. It is therefore impossible for the point mass to overtake the sphere in the manner of the diagram.

Perhaps treating it as two collisions, point mass with sphere then sphere with ground, is wrong.

As already deduced, the two masses are the same and they have equal and opposite vertical velocity components immediately after first impact.

When two equal masses collide elastically, the velocity components normal to the plane of contact simply swap over. The point mass therefore comes to rest in that direction, moving off tangentially to the sphere. Since that is 60 degrees above the horizontal, the line of centres must be 30 degrees below it.

It follows from the above that the sphere acquires a greater horizontal velocity than the point mass retains. It is therefore impossible for the point mass to overtake the sphere in the manner of the diagram.

Perhaps treating it as two collisions, point mass with sphere then sphere with ground, is wrong.

if that is not the case then why does the sphere bounce ?

haruspex
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if that is not the case then why does the sphere bounce ?
It will bounce, for sure, but there really is no justification for treating it as two impacts in series, other than to make it simpler.
In reality, it would depend on the relative stiffness of the bodies, including the ground. If the point mass and the sphere are highly rigid (think, spring with very high k), but the ground is a softer spring, then it might approach such a serialisation. More likely, the ground would be the most rigid of the three.

A more accurate view would consider the two impacts occurring in parallel, but that gets quite hard to analyse. In principle, one of them would transit from compression phase to expansion phase at different times. A reasonable model would be to take the point mass and the ground as completely rigid and the sphere as the source of all elasticity.

One thing that makes it defy intuition somewhat is the assumption of no friction between the surfaces. We need to do that or we get tangled up with rotational energy too.