# Collision of ball with inclined plane problem

Hi friends,
I have an issue in solving a Collision of ball with inclined plane.

The problem is as:

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1379765_1432382863655486_1628611844_n.jpg

Attempt:

https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-ash3/q77/s720x720/1381857_1432380430322396_543886441_n.jpg

So friends when I place vale of tanθ, the answer arises e = 1 But according to question the answer of the problem is tan2θ, which is Option (A) . Please try to help me in this.
I will appreciate the help.

haruspex
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Yes, you will get tan θ if e=1, but e may be < 1. You have not made full use of the the information that the rebound is horizontal. Think about a different direction.

Yes, you will get tan θ if e=1, but e may be < 1. You have not made full use of the the information that the rebound is horizontal. Think about a different direction.

My friend haruspex,

I thought about the horizontal direction also. But the directions in which I am concentrating are along the line of impact and perpendicular to line of imapact.

If I conserve momentum along the line of impact,

mv cosθ = -mv' sinθ
that gives, v'/v = cotθ.
Putting this vale in equation (1), it gives, e = -1

Conserving momentum along the direction perpendicular to line of impact,

v sinθ = v' cosθ
it gives, v'/v = tanθ.

eq. (1) gives now the correct answer, e = tan2θ

But this is wrong way.
I can not conserve momentum along perpendicular to line of impact. ƩFnet ≠ 0 along this.

By the way tanks for reverting back.

Last edited:
haruspex
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2020 Award
My friend haruspex,

I am thought about the horizontal direction also. But the directions in which I am concentrating are along the line of impact and perpendicular to line of imapact.
As long as you use two directions it should be fine. You only used the normal to the plane, so try thinking about parallel to the plane.
If I conserve momentum along the line of impact,
Momentum is clearly not conserved in that line. On impact, what direction is the force?

As long as you use two directions it should be fine. You only used the normal to the plane, so try thinking about parallel to the plane.

Momentum is clearly not conserved in that line. On impact, what direction is the force?

Along parallel the force could be like mg cosθ !!!

haruspex
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Gold Member
2020 Award
Along parallel the force could be like mg cosθ !!!
No, gravity doesn't matter during the impact. I may have misled you by asking about force instead of impulse. The impact is taken as occurring in a very short time, and impulse is force*time. This means gravity supplies very little impulse during impact, so ignore it. Just think about the impulse from the plane on the particle. What direction is that in?

No, gravity doesn't matter during the impact. I may have misled you by asking about force instead of impulse. The impact is taken as occurring in a very short time, and impulse is force*time. This means gravity supplies very little impulse during impact, so ignore it. Just think about the impulse from the plane on the particle. What direction is that in?

Perpendicular to the Plane on the ball and on the floor separately. i.e. Impulse will be the change in momentum of the ball the direction perpendicular to the floor plane.

haruspex
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Gold Member
2020 Award
Perpendicular to the Plane on the ball and on the floor separately. i.e. Impulse will be the change in momentum of the ball the direction perpendicular to the floor plane.
Ok, so what can you say about momentum of the ball parallel to the plane?

• 1 person
Ok, so what can you say about momentum of the ball parallel to the plane?

For the ball-

I think , the force on the ball by the plane is only on the perpendicular direction hence in this direction momentum will change. But

In the direction parallel to the plane there is no force, hence momentum of the ball will not change in that direction. Hence

v sinθ = v' cosθ

i.e. v'/v = tanθ
and from eqn (1),

e = tan2θ