What is the velocity of the wedge after collision?

[stealthezio]

1. In the figure shown , a ball of mass m collides perpendicularly on a smooth stationary wedge of mass M , kept on a smooth horizontal plane. If the coefficient of restitution is e , then determine the velocity of the wedge after collision.

https://postimage.org/][/PLAIN]

Given
mass of ball = m
mass of wedge = M
coefficient of restitution = e
velocity of wedge after collision = ?
answer to problem =
(1+e)mv sinθ / M+m sin^2 θ2.
mv = mv1 + Mv2
v1 = velocity of m after collision
v2 = velocity of M after collision
e = relative velocity after collision / relative velocity before collision

The Attempt at a Solution

So I tried conservation of momentum along the common normal , ie at the line of impact but since the wedge can only move in x direction I am stumped and cannot get an answer after solving. Tried center of mass approach and still arriving at weird solution. Any advice will be appreciated. Thanks.[/B]

 

[Mastermind01]

When can we conserve momentum?

 

[stealthezio]

momentum is always conserved that i know of.

 

[Mastermind01]

momentum is always conserved that i know of.

Nope, look up Netwon's second law, momentum is conserved only in the absence of external forces. Now in which direction are there no external forces?

 

[stealthezio]

perpendicular to the line of impact?
or along the slide of wedge

 

[Mastermind01]

perpendicular to the line of impact?
or along the slide of wedge

Ok, let me rephrase it - which is the external force on the system (ball + wedge)?

 

[stealthezio]

Normal force from the ground?
So we conserve momentum in x direction

 

[Mastermind01]

Normal force from the ground?

Yes that is one. What are the others?

 

[stealthezio]

No other external forces.
Rest of the forces are from within the system.

 

[Mastermind01]

No other external forces.
Rest of the forces are from within the system.

What about gravity, is it not external?

 

[stealthezio]

yes sorry it is.

 

[Mastermind01]

yes sorry it is.

It's ok. So the Normal force balances the wedge's weight. But the ball's weight is not balanced by anyone, so there is a net force in the direction of the ball's weight. So in which direction is there no external force.?

 

[stealthezio]

x direction

 

[Mastermind01]

x direction

Correct, so momentum can be conserved in the x-direction. So conserve momentum along x-axis, use the equation for e and I think you're done.

 

[stealthezio]

Ok so I did this
mv sinθ = Mv1 + mv2x ------------1
mv2x = velocity of m in x direction
e = (v1 + v2) / v
ev = v1 + v2
v2 = ev - v1
Now velocity in x direction for m will be v2 sinθ
v2 sinθ = (ev - v1) sinθ -------2
substituting 2 in equation 1
mv sinθ = Mv1 + m sinθ (ev-v1)
v1 = mv sinθ (1-e) / (M-m sinθ)

This is not the answer , what am I doing wrong?

 

[Mastermind01]

Well your e-equation in wrong. The e-equation is valid only for velocities along the line of impact.

 

[stealthezio]

Is this correct?
ev = v1 sinθ + v2
then substitute as
v2 sinθ = (ev-v1 sinθ) sinθ

 

[Mastermind01]

I think so, does that get you your answer?

 

[stealthezio]

Well almost ,
It seems as I have to take
ev = v1 sinθ - v2
As by assigning each velocity a direction . Silly mistake by me.
Thank you for the help.