Understanding Momentum Conservation in Particle Collisions

[EventHorizon4]

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...

 

[wwshr87]

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

 

[inky]

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!

 

[EventHorizon4]

got it. I'm dumb haha.

 

[joshmdmd]

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