Question

Mass A is initially moving with a velocity of 2 m/s in the +x-direction. Mass B is initially moving with a velocity of 6 m/s in the –x-direction. The two objects have equal masses. After they collide, mass A moves with a speed of 4 m/s in the –x-direction. What is the final velocity of mass B after the collision?

- 6 m/s in the +x-direction
- 4 m/s in the +x-direction
- zero
- 4 m/s in the –x-direction

Final Answer

(c)

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This is College Physics Answers with Shaun Dychko.
The conservation of momentum means that the total momentum initially is the same as the total momentum after collision. So we have

*m vA*plus*m vB*total momentum before the collision, equals*m vA prime*plus*m vB prime*, total momentum after the collision. I did not write a subscript on the letter m because we're told that the objects have equal masses and so it's a common factor as it turns out and so it cancels everywhere after you divide both sides by*m*. Also we can move this*vA prime*term to the left by subtracting*vA prime*from both sides. We get*vB prime*, the velocity of mass B after the collision, is velocity of mass A before the collision, plus the velocity of mass B before the collision, minus velocity of mass A after the collision. So that's two meters per second plus negative six meters per second because we're told mass B is moving in the negative x direction, and then minus negative four meters per second because mass A is also moving in the negative x direction after the collision. So that's two plus negative six is negative four, and then minus negative four makes zero. So the answer is C.