# Thread: Conservation of momentum in perfectly elastic collisions

1. In perfectly elastic collisions, we know that when an object with mass m1 and speed u collides with another object with mass m2 which is stationary(where m1 is much larger than m2), m1 will continue moving at u while m2 will move with a speed of 2u.
We can see that the total momentum before collision in the system is m1*u+ m2(0)=m1*u
However, the total momentum after collision is m1*u + m2(2u)
They are totally inequivalent and there's no external participated in the collision!!
How does it obey the conservation of momentum???

2.

3. Originally Posted by litterboy
In perfectly elastic collisions, we know that when an object with mass m1 and speed u collides with another object with mass m2 which is stationary(where m1 is much larger than m2), m1 will continue moving at u while m2 will move with a speed of 2u.
In general, that's not true. The final speed depends on the masses of both particles. If the particles have about the same mass then both will move at speeds less than u. If m1 >> m2 then what you said is approximately correct.

Note: By definition, a collision is elastic if kinetic energy is conserved.

If you use conservation of momentum and kinetic energy then you can find the speeds of the particles after the collision.

4. Originally Posted by litterboy
In perfectly elastic collisions, we know that when an object with mass m1 and speed u collides with another object with mass m2 which is stationary(where m1 is much larger than m2), m1 will continue moving at u while m2 will move with a speed of 2u.
We can see that the total momentum before collision in the system is m1*u+ m2(0)=m1*u
However, the total momentum after collision is m1*u + m2(2u)
They are totally inequivalent and there's no external participated in the collision!!
How does it obey the conservation of momentum???
If m1 is much larger than m2, then m2(2u) is negligible compared to m1u. Call it zero, and momentum is conserved. Remember, this is just an approximation.

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