Reflection of a falling object on an Inclined Plane

AI Thread Summary
A ball falling on a smooth inclined plane rebounds elastically after reaching a distance h. The discussion focuses on determining the angle of rebound and the distance from the impact point for the second rebound. Key insights include the conservation of momentum, where the normal component changes while the parallel component remains constant. The angle of incidence is equal to the angle of reflection, leading to the conclusion that the rebound angle is 180° - α. The conversation emphasizes the importance of resolving momentum components to understand the ball's trajectory after the collision.
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Homework Statement


(Irodov - Problems in General Physics - 1-31)
A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle α with the horizontal. Having fallen the distance h, the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time?

Homework Equations


Kinematic Equations, Conservation laws

The Attempt at a Solution


I can solve this question if I can get one piece of information - the angle with the inclined plane with which the ball recoils. Using geometry I found that the incoming ball makes an angle 90+α with the plane. However I can't find the recoil angle. I suppose application of the conservation laws is required, but they bear no result for me - any clues on how to proceed?
 
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When the ball reaches the surface, the normal force acts on it and makes it reflect. The normal force changes the normal component of the momentum, but the component parallel with the incline is conserved.
The situation is completely analogous to the reflection of light. Image a light ray falling vertically onto a mirror that makes an angle α with the horizontal. What do you know about the angle of reflection?

ehild
 
The angle of refraction is equal to the angle of incidence...so here the angle with the normal to the plane must be same before and after reflection. But what do you mean by component parallel to the incline? Shouldn't it be zero, as the ball is simply falling downwards?
 
The momentum is vertical, and the vertical direction makes an angle 90+α with the incline, as you found it out already. The momentum has a parallel component with respect to the plane which is unchanged during the collision.

ehild
 
Ok I get what you meant. Before the impact the ball has an angle α with the normal to the incline. If I set up a coordinate system with the y-axis as the normal to the incline and the x-axis as surface of the incline, I can write the momentum before the collision. How do I use this to explicitly show the angle after collision?
 
What are the x and y components of the momentum before the collision in terms of the angle α?
After collision, the x component stays the same, the y component changes sign. What will be the direction of the momentum after collision? Make a drawing, you will see at once.

ehild
 
I can see from the drawing through vector addition, but not from the algebra of it. Since the ball makes an angle 90+α with the plane, the angle with the normal is α. If I resolve the components along these axes the x-component would be mvsin(α) and the y would be -mvcos(α). If the angle after recoiling is θ, the components would be mvsin(θ) and mvcos(θ). If I equate these two using what you said I just get θ=α. So I must have resolved the momentum wrongly?
 
sinα=sinθ; -cosα=cosθ ----> θ=180°-α.

ehild
 

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Ok I get it - thanks for the help!
 
  • #10
Irodov ... i miss my school days :(
 
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