Understanding Momentum Conservation in Particle Collisions

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Homework Help Overview

The discussion revolves around the conservation of momentum in a particle collision scenario. A particle of mass m collides with a stationary particle of mass 2m, resulting in the first particle coming to rest and the second particle splitting into two equal-mass pieces moving at angles with respect to the x-axis. Participants are exploring the implications of momentum conservation in this context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of momentum equation and its application to the collision scenario. There is a focus on the implications of the resulting speeds of the new particles and whether they violate momentum conservation. Some participants attempt to derive relationships between the speeds of the particles before and after the collision.

Discussion Status

There is ongoing exploration of the problem, with some participants expressing confusion about the results and questioning their understanding of momentum conservation. Others suggest breaking down the velocities into components to analyze the situation further. The discussion reflects a mix of interpretations and attempts to clarify the mechanics involved.

Contextual Notes

Participants are grappling with the implications of the collision setup, particularly regarding the initial conditions and the resulting motion of the particles. There is an acknowledgment of the need to consider both x and y components of the velocities in the analysis.

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Homework Statement



A particle of mass m is moving along the x-axis
with speed v when it collides with a particle of
mass 2m initially at rest. After the collision, the
first particle has come to rest, and the second
particle has split into two equal-mass pieces that
move at equal angles q > 0 with the x-axis.

What is true about the speed of the new particles?

Homework Equations



conservation of momentum:
mv = mv

The Attempt at a Solution



According to the answer, each new particle moves with a speed greater than v/2. This seems to violate conservation of momentum to me...but as most people get this question right, I'm probably missing something obvious...
 
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Ok, so conservation of momentum, m1v1 + m2v2 = m1v'1 + m2v'2. Let's call m1 the mass that hits and m2 the mass initially at rest. Since the target is initially at rest and after collision the mass that m1 is at rest, our equation reduces to: m1v1=m2v'2
v'2=(m1/m2)*v1, and since m2=2*m1. v'2 = (1/2)*v1
 
According to the answer, each new particle moves with a speed greater than v/2. This seems to violate conservation of momentum to me...but as most people get this question right, I'm probably missing something obvious...[/QUOTE]


Before collision total momentum is mv. (It is rigt)

After collision , 2m split 2 particle into equal mass m but moving at equal angles q > 0 with the x-axis.
Let say after collison velocity V'. You must divide 2 componets for each particle and consider for x components. Use law of conservation of momentum for x-axis , find V' then think!
 
got it. I'm dumb haha.
 
yeah you have to break in into components because the x components together add to the initial.. the y components cancel each other out. This is why v > 1/2v
 

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