Change in momentum problems usually involve figure skaters, bullets, and other ridiculous scenarios.
But don’t be fooled by the flash and dance – there’s a method to the madness.
And in this video I go through an example of a conceptual change in momentum problem from a Physics 1 exam, step-by-step.
Change In Momentum Problem Statement
A ball of mass = 10g slides on a frictionless table and strikes a barrier in two different cases as shown in the figure. In A, it bounces right back to retrace its path; in B, it bounces diagonally (path shown in dotted line). The initial speed of the ball is the same in both cases and no energy is lost in either collision. Which of the following is true?
(A) There is not enough information to determine which case has a greater change in momentum
(B) The change in the ball’s momentum is zero for both cases
(C) The change in the ball’s momentum is greater for case A than for B
(D) The change in the ball’s momentum is the same in both cases (but not zero)
(E) The change in the ball’s momentum is greater for case B than for A
0:40 – Breaking down the problem statement and final answer: It says the correct answer is C, the change in the ball’s momentum is greater for case A than for case B. Right off the bat we notice that in case A the ball is bouncing directly back in the horizontal direction, whereas in case B it’s taking a ricochet off the wall. Why is that?
2:22 – How conservation of energy is used to determine the initial and final velocity: We know that energy is conserved, what does that mean? If we remember, that means that the energy in equals the energy out. In this case this would just be kinetic energy. Because the mass stays the same, this implies that the initial velocity and final velocity of the ball in both cases are equal.
4:13 – Analyzing the change in momentum equation and how it’s used to determine the difference between A and B: If we remember, the change in momentum would be the final momentum minus the initial momentum. Starting with case A, you actually only have velocity and momentum in the horizontal direction. This means that you end up with a change in momentum of 2*m*v because the velocity completely changes direction. Case B is a little bit more complicated because it’s not only in the horizontal direction, it also has velocity in the vertical direction. Here we have to use trig, and end up with a change in momentum of 2*m*v*cos(theta), which is less than the change in case A because the vertical component of the velocity involves no change in direction.
That’s how I would break down a conceptual momentum problem like this.
It’s tempting with these types of problems to want to analyze it based on your intuition, but it always helps to go back through, look at the principles and assumptions involved, and actually put together the equations for yourself even if it’s just symbolically.
By putting together the equations you understand the phenomenon better and you’re able to make a determination based on the numbers and not only based on your gut.