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So, this problem should be easy, and I feel pretty silly even asking this question given my physics knowledge:

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, $\theta$, with the x -axis. Which statemente correctly describes the speeds of the two pieces?

Apparently, the answer is:

Each piece moves with speed greater than $v/2$

... Meaning that the momentum of each piece of mass $m$ is greater than $mv/2$—so their total momentum is greater than $mv$. How can this be possible? Apparently no one here has a problem with it, and I certainly understand the mathematics used to justify it, but it seems to fly in the face of one of the most basic physical assumptions. Is the assumption here that there's some other force that propelled the two pieces, as if you were splitting a magnet or something?

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    $\begingroup$ ::doing my best Steve Ballmer impersonation:: Vectors! Vectors! Vectors! $\endgroup$ – dmckee Aug 9 '16 at 18:43
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For the two dimensional collision detailed in the problem statement, momentum must be conserved in both the x direction, and in the y direction. For the x direction, the total x-momentum after the collision must equal the total x-momentum before the collision. Due to this, each particle after the collision must move with speed v/2 in the x direction.

Momentum must also be conserved in the y direction. For the particles after the collision, one particle moves up with a velocity the depends on the angle under which the collision took place, and the other particle moves down with this same velocity, ensuring that there is zero total y-momentum before and after the collision.

The velocity of each particle after the collision is the vector sum of the x and y components of each particle's velocity. This means that the final velocity of each particle after the collision must be greater than v/2.

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  • $\begingroup$ Oh. You pointed out what I was totally neglecting—the fact that the y-momentums cancel one another out (that is, $mv_y$ and $-mv_y$). That went completely over my head and I am a bad person. ty $\endgroup$ – AmagicalFishy Aug 9 '16 at 18:19
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The velocity of each secondary particle is something like:

$\vec{u}=\frac{v}{2}\hat{i}+f(\theta)\hat{j}$

Each particle has equal but opposite speed in the y-axis, and the momentum in the x-axis is divided equally between the two particles (so that each particle moves at half the speed of the first)

You will notice that since there is a y-component to the velocity (and the x-component has magnitude $\frac{v}{2}$, the total speed is greater than $\frac{v}{2}$

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