Question
Cart A is moving with a velocity of +10 m/s toward cart B, which is moving with a velocity of +4 m/s. Both carts have equal mass and are moving on a frictionless surface. The two carts have an inelastic collision and stick together after the collision. Calculate the velocity of the center of mass of the system before and after the collision. If there were friction present in this problem, how would this external force affect the center-of-mass velocity both before and after the collision?
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Final Answer

The velocity of the center of mass is the same before and after collision. It is 7 m/s7 \textrm{ m/s}.
If friction were present, the center of mass velocity would gradually decrease before and after collision.

Solution video

OpenStax College Physics for AP® Courses, Chapter 8, Problem 20 (Test Prep for AP® Courses)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Two carts with equal mass are both moving in the same direction— one is moving at 10 meters per second and the other is moving at 4— and this cart A, which is moving faster, is catching up to cart B and when they eventually collide, there's an inelastic collision, which is to say they stick together after they collide. So calculate the velocity of the center of mass of the system before and after the collision. Alright! So the center of mass is in the combined object after the collision so after carts A and B are stuck together, the center of mass is right in the middle because they both have equal masses and we just have to figure out what is v prime, the velocity that they have after sticking together. So the total momentum before collision is the same as the total momentum after collision and that is mass times v A plus the same mass times v B equals the combined mass of them both, which is 2 times the mass of one of them times v prime and the masses do not get subscripts A or B here because since they are the same, there's no need for the subscript. So we can divide both sides by 2m to solve for v prime and we get v prime then is v A plus v B over 2. So that's 10 meters per second plus 4 meters per second divided by 2 which is 7 meters per second so that's the velocity of the center of mass after collision. And the velocity of the center of mass before collision well, the total momentum is the total mass times the velocity of the center of mass and so we can solve for v cm by dividing both sides by 2m here—this side and this as well— and we know what the total momentum is after collision, it's 2m times v prime I guess I am doing this division by 2m in the next step here by the way sorry so here's where we divide by 2m. So the velocity of the center of mass before the collision, it's actually the same as the velocity of the center of mass after collision and that will always be true— it's another way of saying that momentum is conserved— and we have 7 meters per second. So if friction were present, the center of mass velocity would gradually reduce before the collision and after the collision and momentum would not be conserved in that perspective but it would be conserved in a broader perspective including the Earth but, you know, the Earth is not gonna have a measurable change in velocity so we can't practically do that. Okay... and there we go!