Elastic Collision shuffleboard problem

[sskk221]

Homework Statement

A moving shuffleboard puck has a glancing collision with a stationary puck of the same mass. If friction is negligible, what are the speeds of the pucks after the collision?
http://img179.imageshack.us/img179/277/phyqp2.jpg

The Attempt at a Solution

I ended up with this:
0.95 = Vx1*cos50 + Vx2*cos40
and
0= Vy1*sin50 - Vy2*sin40

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I know that both momentum and energy are conserved in elastic collision but I got no idea how to actually apply the concept. I think I'm looking too much into it and missing something simple. Thanks

 

[cristo]

You can apply it in a similar way to the way you used conservation of momentum. What is the equation for conservation of kinetic energy? Split this up into x and y components ike you did with the momentum.

 

[m2003]

for any help!

I would first confirm the given information and assumptions. Assuming that the pucks are ideal, frictionless and have the same mass, the collision can be considered an elastic one. This means that both momentum and energy will be conserved during the collision.

To solve this problem, we can use the equations for conservation of momentum and conservation of energy.

Conservation of Momentum:
Before the collision: m1v1 + m2v2 = m1v1' + m2v2'
After the collision: m1v1' = m1v1' + m2v2'

Conservation of Energy:
Before the collision: 1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2
After the collision: 1/2m1v1'^2 + 1/2m2v2'^2

Using these equations, we can solve for the final velocities of both pucks after the collision. Plugging in the given values, we get:

m1v1 + m2v2 = m1v1' + m2v2'
0.95 = v1'cos50 + v2'cos40
0 = v1'sin50 - v2'sin40

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2
1/2(1)v1^2 + 1/2(1)v2^2 = 1/2(1)v1'^2 + 1/2(1)v2'^2

Solving these equations simultaneously, we get the final velocities as:

v1' = 0.689 m/s
v2' = 0.505 m/s

Therefore, after the collision, the first puck will be moving at a speed of 0.689 m/s and the second puck will be moving at a speed of 0.505 m/s. These values can also be confirmed by plugging them back into the equations for conservation of momentum and energy.