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Two cars (A and B) of equal mass have an elastic collision. Prior to the collision, car A is moving at 15 m/s in the +x-direction, and car B is moving at 10 m/s in the –x-direction. Assuming that both cars continue moving along the x-axis after the collision, what will be the velocity of car A after the collision?
  1. same as the original 15 m/s speed, opposite direction
  2. equal to car B’s velocity prior to the collision
  3. equal to the average of the two velocities, in its original direction
  4. equal to the average of the two velocities, in the opposite direction
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Final Answer

(b)

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 8, Problem 25 (Test Prep for AP® Courses) (2:27)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Conservation of momentum says that mass A times <i>vA</i> plus mass B times <i>vB</i> equals the total momentum after the collision. Since the masses are the same we can cancel all these m's and we get <i>vA</i> plus <i>vB</i> equals <i>vA prime</i> plus <i>vB prime</i>. Now we're going to look at conservation of kinetic energy which is what we have in an elastic collision and again we can cancel all the m's and cancel the one halves and we get this expression. Then we can move this term to the left side and move this one to the right side and we get the difference of squares which we can then factor out. So we get <i>vA</i> minus <i>vA prime</i> times <i>vA</i> plus <i>vA</i> plus <i>vA prime</i>, equals <i>vB prime</i> minus <i>vB</i>, times <i>vB prime</i> plus <i>vB</i>. The reason that's useful is because we can rearrange this conservation of momentum formula and create one side of the equation saying <i>vA</i> minus <i>vA prime</i> and the other side saying <i>vB prime</i> minus <i>vB</i> and these two are equal. So since they are equal these are common factors as it turns out so they can be canceled. So this conservation of energy formula reduces to <i>vA</i> plus <i>vA prime</i> equals <i>vB prime</i> plus <i>vB</i>. We can rearrange that to solve for <i>vA prime</i> which is ultimately what this question is asking us for, what will the velocity of car A be after the collision. Well, we don't have an answer here because we don't know what the velocity of car B will be after the collision, so we consider the conservation of momentum again and solve for <i>vB prime</i>. It is <i>vA</i> minus <i>vA prime</i> plus <i>vB</i>. We substitute that in for <i>vB prime</i> here and that's what is shown in this line. We get <i>vA prime</i> equals <i>vA</i> minus <i>vA prime</i>, plus <i>vB</i>, plus <i>vB</i> minus <i>vA</i>. The vA's cancel and we move this to the left side by adding it to both sides so we get two <i>vA prime</i> equals two <i>vB</i>. Divide both sides by two and you get that the velocity of cart A after the collision will equal the velocity of cart B before the collision. So the answer is B.