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Simple Elastic Collision

  • Thread starter v3r
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  • #1
v3r
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Homework Statement


Particle 1 of mass m moving with speed v in the +x direction has an elastic
collision with particle 2 of mass 3m which was originally at rest. After the collision,
particle 2 is moving in the +x direction. What is its speed?

Homework Equations


deltap = 0
pf1 + pf2 = pi1 + pi2

The Attempt at a Solution


Wouldn't it just be
pf2 = pi1
3mvf2 = mv1
vf2 = v1/3

How come the answer is v1/2 instead? Help please..
 

Answers and Replies

  • #2
Doc Al
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The Attempt at a Solution


Wouldn't it just be
pf2 = pi1
You can't assume that the first particle stops and all its momentum goes to the second particle.

Hint: What does elastic mean?
 
  • #3
v3r
5
0
Elastic means no change in internal energy
DeltaE = 0
DeltaK + DeltaEint = 0
DeltaK = 0
Kf1 - Ki1 + Kf2 = 0

0.5mvf1^2 - 0.5mvi1^2 + 0.5(3m)vf2^2 = 0
0.5(3m)vf2^2 = 0.5mvi1^2 - 0.5mvf1^2

There's two unknowns, vf2 and vf1
Is this setup wrong?
 
  • #4
Doc Al
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0.5mvf1^2 - 0.5mvi1^2 + 0.5(3m)vf2^2 = 0
0.5(3m)vf2^2 = 0.5mvi1^2 - 0.5mvf1^2

There's two unknowns, vf2 and vf1
Is this setup wrong?
No, it's fine. You just need another equation--conservation of momentum. Then you'll have two equations and two unknowns.

In any collision, momentum is conserved; in an elastic collision, kinetic energy is also conserved.
 

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