2026 AP Physics 2 Free Response #4

Added: 2026-05-10

[00:00:00]All right, FRQ number four for 2026 version J AP Physics 2. So, a small sphere not known positive charge Q0 Q has initial speed V0 in the positive X direction. So, it's moving this way.

[00:00:12]Cool.

[00:00:13]And kinetic energy KS when it when the sphere is at point S. So, point here it's moving to the right and it has kinetic energy KS. Okay, cool. Got it.

[00:00:22]The sphere moves to point T.

[00:00:24]Um the force from the electric field is the only force exerted on the sphere and the sphere moves between S and T. The sphere has a final speed VT KT when it's at point T. So, we're going to go through an electric field. Electric fields, by the way, go from um they always point from higher to lower potential. So, the electric field is going to point this way.

[00:00:41]Or um through this path and it's going to be perpendicular to the electric potential. So, it's going to be just like this. So, it's going to be moving in a straight line. So, just going to go horizontal.

[00:00:51]Um indicate whether it is Now, it's a positive charge. Well, so in this case, the electric field is pointed to the left. So, it's going to push it to the left. So, that means the force is going to be to the left cuz it's a positive charge. Positive charge experience force direction directed in the direction of the electric field, right? That's the definition of electric field is the direction that a positive charge would experience a force. So, the force is to the left. That's the opposite direction of the velocity. So, the acceleration is opposite direction of velocity. It's going to slow down. Slowing down means less kinetic energy. So, let's uh let's just do that. So, I um the electric field points left because um the higher potential is on the right.

[00:01:37]is to the right.

[00:01:41]The charge the positive charge experiences a left force left directed force.

[00:01:54]Which implies the acceleration to the left.

[00:02:00]Move to the left.

[00:02:02]Which implies acceleration is to the left, which implies it's slowing down.

[00:02:10]Which implies it has less kinetic energy. So, that is a forces analysis.

[00:02:14]Okay? It's my favorite way to think about it. Other way is through potential. You can look at as the Q times the change in potential is going to be the change in the kinetic energy, like the work being done there. And you can make an argument here.

[00:02:29]I never like this cuz the sign of this is always a little bit tricky to sort of like discern. Although, this is probably the way I'm going to do the analysis just cuz work is the change in energy.

[00:02:37]Like, fundamentally, we're doing work is change in energy over here, right? It's just the work is done by the electric field.

[00:02:45]Right? And so, um you're just um that's that's kind of all that's happening there when we're doing this calculation.

[00:02:51]Q times the change in potential. So, you can make an argument, higher potential, so the change in potential is technically positive, but it's like there's like a negative sign in here.

[00:03:01]This is why I don't really worry about it. I prefer like using an analysis as more in forces, but that's a personal preference. You can make an argument that you're going from lower to higher potential, so you're going to gain um electric potential energy, which means you're going to lose kinetic energy by conservation of energy. You can make that argument. Going so, the argument would be you're going to higher potential.

[00:03:22]Going to higher potential means that you're going to gain electric potential energy for a positive charge, which means you're going to lose kinetic energy for this charge electric field system.

[00:03:32]Okay, uh derive an equation in terms of this. So, we're just going to use this Q delta V. So, it's going to be Q and the delta uh the change in potential is going to be uh -3 V0 to 2 V0. So, that's going to be 5 V0.

[00:03:49]Oh, we should write this. Oh, yeah, Q delta V is the negative change in the kinetic energy.

[00:03:56]So, this is going to be here is going to be negative change in kinetic energy.

[00:04:00]So, that'll be initial kinetic energy minus the final kinetic energy.

[00:04:04]Right? And that's going to be KS minus KT is going to be 5 QV0. So, is going to be um KS minus 5 QV0.

[00:04:18]Like that. Just rearrange that like that.

[00:04:21]Okay, a new small sphere of the same mass as the original sphere but with a negative charge has the same initial speed as the original sphere when it's at point S.

[00:04:28]The force of the electric field is the only force exerted on the new sphere.

[00:04:31]The new sphere's kinetic energy K new when the new sphere is at point T.

[00:04:34]Same mass but a negative charge.

[00:04:37]Um well, K new is going to be greater.

[00:04:40]Is going to be greater. Why is it going to be greater? Cuz from a force's point of view you're going to be accelerating it, right? Cuz the negative charge is going to experience a force to the right.

[00:04:49]Um from the equation we did in here is because the um Oh.

[00:04:55]By referencing your derivation in part B. That's kind of interesting. Well, the Q will be negative so then it'll be a make this a positive charge. So, um the the K new Well, so the KS So, if you look at the KS minus 5 QV0 when Q becomes negative less than zero then KS minus 5 QV0 is going to be greater than KS.

[00:05:23]Um which is greater than the KT. Right? So, basically because this becomes negative, this whole thing becomes positive and thus you are going to be increasing the kinetic energy um compared to that. So, that's and that would be greater than the original one before. Okay.