Velocity from an elastic collision

AI Thread Summary
In an elastic collision between two equal mass bodies, one moving and the other stationary, the final velocities can be determined using the conservation of momentum and kinetic energy. The initial momentum of the moving body is equal to the final momentum of the stationary body after the collision, resulting in the moving body coming to rest and the stationary body moving with the initial velocity. The key equation derived is that the relative velocity reverses, leading to the conclusion that the velocity of body X becomes 0 and the velocity of body Y becomes equal to the initial velocity v of body X. This understanding is reinforced by examining the transfer of momentum and the concept of reduced mass in the collision. The principles of conservation applied here clarify the outcomes of elastic collisions effectively.
Millie Baker
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Homework Statement


A body X moving with a velocity v makes an elastic collision with a stationary body Y of equal mass on a smooth horizontal surface. Which statement gives the velocities of the two bodies after the collision? (multiple choice question)

The Attempt at a Solution


The answer according to the mark scheme is
velocity of X = 0
velocity of Y = v
So I can't understand how this is worked out. But this as far as I got:

MUx+MUy = MVx + MVy
Ux +Uy = Vx +Vy
Ux = Vx + Vy

How do you know that Vx is 0 and Vy is V (or Ux)?
 
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You've used the conservation of momentum which is true for all collisions. Since the question specifies that it's an elastic collision, you know that kinetic energy will be conserved. Try including that in your equations.
 
triso said:
You've used the conservation of momentum which is true for all collisions. Since the question specifies that it's an elastic collision, you know that kinetic energy will be conserved. Try including that in your equations.
Thank you, I understand now!
 
When you have conservation of both energy and momentum the two equations combined produce this neat result: the relative velocity reverses. I.e. v1f-v2f=v2i-v1i. See "Newton's Experimental Law" for a generalised version.
 
This is another instance in which we can illuminate the problem by examining the equal and opposite transfer of momentum in an elastic collision.

Δp = 2μΔv where μ is the reduced mass [ m1 * m2 / (m1 + m2) ] of the colliding objects and Δv is their relative velocity. So in this case we determine:

Δp = 2 m^2 v / (2m) = mv. Thus momentum of moving body will be mv - Δp = 0 and momentum of stationary body will be 0 + Δp = mv.
 
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