Scientists have created a new type of lightweight foam and are performing experiments to investigate the properties of the foam.
- Mayya Alocci
- Apr 17
- 3 min read
PHYSICS C: MECHANICS 2023 Test
Scientists have created a new type of lightweight foam and are perfomring experiments to investigate the properties of the foam. The mass of Cart A is a 1000 kg and the mass of Cart B is 2000kg. A piece of foam with negligible mass is attached to the front of Cart A as shown. Cart A moves with a constant speed toward Cart B, which is initially at rest. At time t=0s, the foam connected to cart A makes contact with Cart B. The foam remains in contant for 0.5 seconds, after which the carts separate and both carts move with constant velocities.
To answer question 1.(a) i, we must go back to the definition of what displacement is. From AP Calc, we know that the displacement is the integral of velocity with respect to time. Graphically speaking, the integral of a function is the area of the region under the curve of the function. Therefore, we can calculate estimate the displacement of Cart A by estimating the area of the region under the velocity curve in the given graph.
For 1.(a) ii, we must conceptually know that the total momentum of the whole system must be conserved since there is no net external forces acting on the carts. We can easily set up the equation by setting
We are given with values of masses to be m1=1000kg and m2=2000kg and v1i=5m/s and v2i=0m/s, if we plug in v1f = 1m/s, we can easily solve that v2f = 2m/s.
In order to correctly graph the velocity of cart 2 as a function of time during the collision, the key is to pick some time points where the velocity of cart 1 is easy to read off from the given graph. Once we read off those values of velocities of cart 1, we can just reapply the conservation of momentum equation to evaluate a set of velocities of cart 2 that corresponds to the velocities of cart 1. For example, if we look at the velocity curve at t=0.1s, v1f=4.5m/s. Using this information to solve for v2f, we get v2f= 0.25m/s. Repeat this process many times, we can get a set of velocities of cart 2 in red dots as shown below. The curve of velocity of cart 2 during the collision should just be all these red dots connected together as shown below with the red line.
To solve this problem, we must recall Newton’s 2nd law that Fnet=ma. The maximum net force will therefore have the maximum acceleration according to this formula. To find the maximum acceleration, we have to first take a derivative of v(t) to find a(t) and then differentiate a(t) to set it equal to zero to find the critical points of acceleration function during the given time interval. At the end, make sure to plug in both the end points of the time interval as well as all the critical points within the time interval into the acceleration function to evaluate for the maximum acceleration.
We clearly see that the value of t=0.25s will make the derivative of acceleration function equal to zero and that is within the given time interval. Finally, let’s evaluate the acceleration function at t=0s,0.25s,0.5s.
Since the question only asks for the greatest magnitude of the net force, a=-12 should be the answer and Fnet= 1000*(-12) = -12000N.
To graph the function of force acting on A as a function of time, we must multiply the a(t) function we obtained previously by the mass of A=1000kg. We would get
We see that this is a parabola so the graph of Fnet is shown in the red line in the above graph (since we only care the magnitude of Fnet, the graph is the reflection across the x-axis of the Fnet function we obtained).
In a nutshell, since the change of momentum will be the same, regardless of time of contact and F=ΔP/ΔT The force is larger for the scenario with shorter contact time. So F1<F2
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