Momentum, Collisions and Explosions Legacy Problem #19 Guided Solution

An effective problem solver reads a problem carefully and develops a mental picture of what's going on, identifying the known and the unknown quantities, and then uses conceptual reasoning skills and physics principles to reason from the known information to the unknown information. Here are pictures of a cart which is moving along the lamp table. We're given the mass of the cart, we're given the speed of the cart, and as it moves along, a brick is sort of dropped or set up on the cart, and we know the speed at which the brick and the cart move after the collision. What we're asked to determine is the mass of the dropped brick. There's two basic approaches to this problem based on the similar principles. The first approach I'm going to talk through is the approach that's based on the principle that the momentum change of one object is equal and opposite to the momentum change of the other object. Here in this problem, we know the pre-collision speed of the cart and the post-collision speed of the cart, and we also know the mass of the cart. That's sufficient information to calculate the momentum change of the cart. If I were to do that, I would take the mass of the cart and multiply it by the velocity change of the cart. The velocity change of the cart is simply the final value for velocity minus the initial value of velocity. After the collision, it was moving 15.7, and before the collision, it was moving 28.2 centimeters per second. So, if I subtract the initial from the final value, I would get negative 12.5 centimeters per second. If I multiply that velocity change by the mass of the cart, I would get a momentum change in the cart of negative 32.5 kilograms times a centimeter per second. That's the momentum lost by the cart, and the principle is that the momentum change of one object is equal and opposite to the momentum change of the other object. So the brick gains momentum, 32.5 kilograms per centimeter per second units of momentum. So if we say that that momentum change of the brick is equal to the mass of the brick times the velocity change of the brick, we should be able to figure out the mass of the brick. Because the velocity change of the brick was from zero centimeters per second to 15.7 centimeters per second, and the momentum change is 32.5 kilograms times a centimeter per second. So I say 32.5 is equal to the mass of the brick times 15.7 centimeters per second. I divide through by 15.7, and I get the mass of the brick is 2.07 kilograms, or about 2.1 kilograms. The second approach to solving this problem is based on a similar principle, but it's expressed differently. The principle is that the pre-collision momentum of both objects is equal to the post-collision momentum of both objects. That is, if you sum up the total momentum before the collision, it should be equal to the sum of the total momentum after the collision. Here for our cart and our brick, before the collision, it's only the cart that has momentum. We can find its momentum if we go 2.6 times the 28.2 centimeters per second. We find out what that is, and we set it equal to the total momentum of cart and brick after the collision. So we would go 2.6 times 28.2 is equal to the post-collision momentum, which would be expressed as 2.6 for the cart times 15.7 for the cart, plus the momentum of the brick, which would be mass of brick times 15.7. I'll say that again. Before collision momentum equals after collision momentum. So you write for before collision momentum, 2.6 times 28.2, and you set that equal to the after collision momentum, which would be 2.6 times 15.7 plus mass of brick times 15.7. This is a one equation with one unknown, and you should be able to solve for the unknown. You should get the same mass of 2.1 kilograms.