Question

Two objects (A and B) of equal mass collide elastically. Mass A is initially moving 5.0 m/s in the +x-direction prior to the collision. Mass B is initially moving 3.0 m/s in the –x-direction prior to the collision. After the collision, mass A will be moving with a velocity of 3.0 m/s in the –x-direction. What will be the velocity of mass B after the collision?

- 3.0 m/s in the +x-direction
- 5.0 m/s in the +x-direction
- 3.0 m/s in the –x-direction
- 5.0 m/s in the –x-direction

Final Answer

(b)

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This is College Physics Answers with Shaun Dychko.
Conservation of momentum says that mass A times velocity A plus mass B times velocity B before the collision, this total has to be the same as the total momentum after the collision, mass A

*vA prime*plus*mB vB prime*. We're told that the masses are all the same so*mA*and*mB*are common factors and we can divide everything by that common factor. So we end up with*vA plus**vB equals**vA prime*plus*vB prime*. We can solve this for*vB prime*by subtracting*vA prime*from both sides and then switching the sides around. So the velocity of the object B after collision will be velocity A plus velocity B both before the collision, minus velocity A after collision. So before collision mass A has a positive 5.0 meters per second velocity. Mass B is moving in the negative x direction so we put in negative three meters per second and then minus the velocity of mass A after the collision. We're told that it's moving at a velocity of three in the negative x direction. So we substitute negative three meters per second in place of*vA prime*. So we have a minus three here and this is a positive three in the end after the minus sign in the equation and the negative of a number and so these make zero and we're left just with five. That's positive five and so mass B will be moving in the positive x direction with a velocity of five meters per second after the collision