Finding the radius of the sphere

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kaspis245
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Homework Statement


In a spherical bowl a small ball is jumping around. The ball hits the bowl in two spots and moves in two different trajectories in times T1, T2. Find the radius of the bowl.

sphere_image.png


Homework Equations


Newton's laws of dynamics.

The Attempt at a Solution


The only thing I could figure out is how the horizontal velocities of different trajectories are related:
[tex]s=v_{x1}⋅T_1=v_{x2}⋅T_2 ⇒ \frac{v_{x1}}{v_{x2}} = \frac{T_2}{T_1}[/tex]
But I can't find a way to get any expression with the radius. Please help.
 

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What do you know about the incoming and outgoing angle for a perfectly elastic collision?

Unrelated to that: You will probably need the trajectories in both parts.
 
mfb said:
What do you know about the incoming and outgoing angle for a perfectly elastic collision?
Momentum is conserved. Then the velocities of both trajectories are equal. I still can't see how the radius is related to any of this.
 
kaspis245 said:
Momentum is conserved. Then the velocities of both trajectories are equal. I still can't see how the radius is related to any of this.
The ball interacts with the bowl which is much bigger than the ball. You can consider the bowl of infinite mass. The velocities are not the same, but energy might be conserved. What would it mean for the speeds?
The ball performs two different projectile motions with the same range.
upload_2017-12-28_14-28-36.png
 

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kaspis245 said:
$$
s=v_{x1}⋅T_1=v_{x2}⋅T_2 ⇒ \frac{v_{x1}}{v_{x2}} = \frac{T_2}{T_1}$$
Are you saying the two different trajectories have the same path length ?
 
Ok, so the velocities are not equal. I can’t really apply the momentum conservation law since the direction of motion changes after the impact, and I don’t have any angles.

As for the energy I can only say that [tex]v_1<v_2[/tex] since the path [tex]T_1[/tex] has more potential energy (jumps higher).

BvU said:
Are you saying the two different trajectories have the same path length ?
The trajectories have the same horizontal distance, which I denoted by s.
 
kaspis245 said:
Ok, so the velocities are not equal. I can’t really apply the momentum conservation law since the direction of motion changes after the impact, and I don’t have any angles.

As for the energy I can only say that [tex]v_1<v_2[/tex] since the path [tex]T_1[/tex] has more potential energy (jumps higher).
What do you mean on v1 and v2?
The problem said
In a spherical bowl a small ball is jumping around.
that means the ball repeats its motion, so the collisions can be considered elastic.
The ball has both horizontal and vertical components of velocity. It jumps higher if the vertical component is bigger.
T1 and T2 are the times between two collisions with the bowl.