Journey W. answered • 06/14/22
Rice Physics and Applied Math Grad- Perfect SAT Math, Perfect PSAT
The main idea behind this problem is the conservation of energy. Let's take our system to be the grocery cart and the spring- this just means that when we're accounting for total energy in the system, we'll take the individual energies of these objects and add them together. Any transfer of energy between the two objects won't affect the total energy of our system- only work applied by some other object outside the system would be able to change the system's total energy.
The interaction described in this problem is solely between the grocery cart and the spring, so we expect the total energy of our system to be conserved. All of tne energy lost by the shopping cart will be transferred to the spring.
Right before the grocery cart hits the spring, it has a certain kinetic energy, because it's moving. The spring is not compressed and is not moving, so it contributes nothing to the total energy of our system. At this point, the total energy of our system is just the kinetic energy of the shopping cart, which can be found purely from the formula KE = (1/2)*m*v2. Plugging in, the kinetic energy of the shopping cart (which is the total energy of the system at this point) is (1/2)*(12kg)*(v2) = 6v2. We don't know the shopping cart's speed yet- that's what we're trying to find ultimately- so we leave our expression for the initial total energy in this form for now.
During the collision, the shopping cart will compress the spring until it stops moving, at which point the spring will have attained its maximum compression of 80cm. At this second, final state, the grocery cart is stationary, and as a direct result it has no kinetic energy. The spring will be compressed, so it will have gained elastic potential energy at this point. The stationary grocery cart contributes nothing to the total energy, so this elastic potential energy of the spring will be the total energy of the system at this second state. Plugging into the Elastic Potential Energy formula, EPE = (1/2)*k*x2, we get that the elastic potential energy of the spring at max compression (which is the total energy of the system at this second state) is (1/2)*(220N/m)*(0.8m)2 = 70.4J.
Bringing it all together, we now have two expressions for the total energy of our system, one from each state of the system. Total energy at the first state (right before collision) is 6v2, Total energy at the second state (spring at max compression) is 70.4J. We said before that the total energy of this system should be conserved during this process, as all of the interactions involved in this process are entirely between the cart and the spring- entirely internal to our system. Therefore, we conclude that these quantities have to be the same. We state this mathematically by writing 6v2 = 70.4, then solve for v, the speed, by dividing both sides by 6 and finally taking the square root of both sides. We get a final answer of v = 3.425 m/s.
