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Mass A (2.0 kg) has an initial velocity of 4 m/s in the +x-direction. Mass B (2.0 kg) has an initial velocity of 5 m/s in the –x-direction. If the two masses have an elastic collision, what will be the final velocities of the masses after the collision?
  1. Both will move 0.5 m/s in the –x-direction.
  2. Mass A will stop; mass B will move 9 m/s in the +x-direction.
  3. Mass B will stop; mass A will move 9 m/s in the –x-direction.
  4. Mass A will move 5 m/s in the –x-direction; mass B will move 4 m/s in the +x-direction.
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OpenStax College Physics for AP® Courses Solution, Chapter 8, Problem 41 (Test Prep for AP® Courses) (3:31)

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This is College Physics Answers with Shaun Dychko. Momentum conservation means that mass A times velocity A plus mass B times velocity B which is the total momentum before the collision, equals the total momentum after the collision. You'll notice that if you look at the amount of each mass they are two kilograms each. So the masses are the same. So that means that <i>mA</i> and <i>mB</i> are actually the same quantity so it's a common factor that can be canceled. So we have <i>vA</i> plus <i>vB</i> equals <i>vA prime</i> plus <i>vB prime</i>. We want to find what <i>vA prime</i> is or <i>vB prime</i> but I'm going to solve for <i>vA prime</i> first. We can't do it with just one equation so we need to turn our attention to another equation which we get from conservation of energy, conservation of kinetic energy since we know that this is an elastic collision. So we have one half <i>m vA</i> squared plus one half <i>m vB</i> squared, that's the total kinetic energy before collision, equals one half <i>m vA prime</i> squared plus one half <i>m vB prime</i> squared. The one half m's are common factors everywhere and so this becomes <i>vA</i> squared plus <i>vB</i> squared equals <i>vA prime</i> squared plus <i>vB prime</i> squared. Now this can be simplified quite a bit by moving this term to the left and moving this term to the right by subtracting them each from both sides. We have a difference of squares here that can be factored into <i>vA<i> minus <i>vA prime</i> times <i>vA</i> plus <i>vA prime</i>. On the right side we have <i>vB prime</i> minus <i>vB</i> times <i>vB prime </i> plus <i>vB</i>. Equation one can be rearranged to have <i>vA</i> minus <i> vA prime</i> on the left and <i>vB prime</i> minus <i>vB</i> on the right to show that these differences are equal. That means this is a common factor with that, so equation two can be simplified to <i>vA</i> plus <i>vA prime</i> equals <i>vB prime</i> plus <i>vB</i>. We'll solve that for <i>vA prime by subtracting vA from both sides.</i> So we have <i>vA prime</i> equals <i>vB prime plus vB minus vA.</i> Equation one can be changed yet again to solve for <i>vB prime</i>, the velocity of mass B after the collision, and say that it is <i>vA</i> plus <i>vB</i> minus <i>vA prime</i>. Then we'll substitute that in for <i>vB prime</i> here in equation two b. We get that <i>vA prime</i> is then <i>vA</i> plus <i>vB</i> minus <i>vA prime</i> which is our expression for <i>vB prime</i> and then plus <i>vB</i> minus <i>vA</i>. The vA's make zero because this is positive <i>vA</i> and that's negative <i>vA</i>. This can be added to both sides in which case you have two <i>vA prime</i> and then you have two vB's as well and so this says that the velocity of mass A after the collision will be equal to the velocity of mass B before the collision. We were told that mass B had a velocity of negative five meters per second and so the only option among A, B, C , and D here with mass A moving five meters per second in the negative direction is option D. So we don't even need to find the velocity of mass B because we've already narrowed it down to this option here.