OpenStax College Physics for AP® Courses, Chapter 8, Problem 33 (Test Prep for AP® Courses)

This is College Physics Answers with Shaun Dychko. Since no external forces are acting on the system we know for sure that momentum is conserved. So that means we can write down this first statement that mA vA plus mB vB the total momentum before the collision, equals mA vA prime plus mB vB prime, the total momentum after the collision. Now, we can factor out the m because we're told the objects are of equal mass, so mA and mB are the actually the same factor and we can cancel it everywhere. So we get vA plus vB equals vA prime plus vB prime. Let's solve this for vB prime to see what this unknown velocity is, and we'll subtract vA prime from both sides. We get vB prime equals vA plus vB minus vA prime. So vA is the 12 meters per second initial velocity of object A and vB we're told is initially at rest and so it's zero. Then we're told after the collision object A which is vA prime, its velocity will be zero. So this works out to 12 meters per second. So the object B after the collision has a velocity of 12. Now, then we can turn our attention to the conservation of energy formula to see if it is true that the total kinetic energy before the collision is the same as the total kinetic energy after the collision. Now, the vB term here is zero because object B has no velocity initially and this term also is zero because object A has no velocity initially. We can multiply everything by two and divide everything by this common mass, there is no need to put a subscript there because mass A and B are the same. So what we're left with is vA squared equals vB prime squared. Now if this is true, then kinetic energy is also conserved. It is true because vA is 12 meters per second and 12 squared equals vB prime squared which also is 12. So we can say A, both momentum and kinetic energy are conserved.