Elastic Collision of Two Particles: Solving for Final Velocity

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Homework Help Overview

The problem involves an elastic collision between two particles, where one particle is initially at rest. The original poster seeks to determine the final speed of the second particle after the collision, given the masses and initial conditions of both particles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and kinetic energy in elastic collisions. The original poster attempts to relate the final velocity of the second particle to the initial velocity of the first particle but questions the validity of their reasoning. Others suggest that the assumption of the first particle stopping is incorrect and prompt a reconsideration of the definitions and implications of elastic collisions.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the need for additional equations to solve for the unknowns, specifically emphasizing the conservation of momentum alongside kinetic energy.

Contextual Notes

Participants are navigating the complexities of elastic collisions, including the implications of conservation laws and the relationships between the variables involved. There is an acknowledgment of the need for more information to resolve the two unknowns in the equations presented.

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..
 
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v3r said:

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?
 
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?
 
v3r said:
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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