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Say we have equal masses $M_1$ and $M_2$, traveling at equal but opposite velocities, $v$ and $-v$. Assume elastic collision. The forces are equal and they come to stop, then accelerate away from one another at the same initial speeds.

But what if now we do the same but one velocity is $V$, other is $-2v$. Is the force from the above example equal to the forces felt in this case? Wouldn't it have to be right, because the masses don't know how fast one is relative to the other, and the forces have to be equal by Newton's third law? Just now they act longer, causing the change in momentum, such that the balls switch velocities compared to before the collision.

Clarification edit: We know the forces felt by two cars colliding at equal and opposite speeds are essentially the same as hitting a wall. Like MythBusters.

What I'm asking, is the magnitude of the force between the masses during a collision with speeds $V$ and $-2v$ the same magnitude as they would be for a collision with $v$ and $-v$? I feel they would be.

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  • $\begingroup$ It's not exactly clear what you're asking. Could you maybe rephrase the question? $\endgroup$ Aug 21, 2020 at 1:48
  • $\begingroup$ Only when $V=0$ this will be true. $\endgroup$ Aug 21, 2020 at 4:11
  • $\begingroup$ I got confused the other day if we human will or can affect earth due to all our actions and N3L. My brains been thinking of examples that would make it such that we would need to include for force from earth, say as we fall or come to a stop. But I believe i am comparing apples to elephants, and earth is too massive with too many things going on to get a clear picture, and I am confusing myself. $\endgroup$ Aug 21, 2020 at 14:55
  • $\begingroup$ I guess that is my new questions. Do we care or have to take into account how earth may or may not be moving say, when we crash into a wall? and If not, why? $\endgroup$ Aug 21, 2020 at 14:56

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Have one of the masses a massless spring attached in the direction the collision takes place. So now when they collide, the spring compresses and then expands back. Momentum is conserved during this process.

While they're colliding, the spring compresses. The compression applies forces on both the masses, slowing both of them down.

  1. When the masses were moving with -v and +v : The total momentum is 0. When they collide, the both come to rest at some point (total momentum is still 0). Their total kinetic energy ($=mv^2$) is now stored as potential energy in the compression of the spring. This potential energy is a function of the compression of the spring. The force applied by the spring at any instant is also a function of the compression at that instant. So we can say that throughout the collision, the spring force (and hence the force at the masses) varies between the force corresponding to a potential energy of 0 (no compression) to a potential energy of $mv^2$ (maximum compression). Same goes for when the spring is elongating.

  2. When the masses are moving at $-2v$, $v$- The total momentum is $-mv$. During the collision, the spring force first slows both of them down until they're moving with the same velocity. This same velocity can be found by conserving momentum: $-2mv+mv=(m+m)v_{const}$, giving $v_{const}=\frac{-v}{2}$. So, in this case, both the masses are still moving at $\frac{-v}{2}$ when the spring compression is maximum. Their initial KE was $\frac{1}{2}m(2v)^2+\frac{1}{2}mv^2=\frac{5}{2}mv^2$, while thie KE at maximum compression is $\frac{1}{2} m (\frac{-v}{2})^2+ \frac{1}{2} m (\frac{-v}{2})^2=\frac{1}{4} m v^2$. The difference between initial and final is of $\frac{9}{4} mv^2$. Now this is stored as the PE of the spring at maximum compression.

So in this case, the force on the masses varies between 0 (no compression) to the force corresponding to a compression of PE=$\frac{9}{4}mv^2$ (at maximum compression). The force varies similarly in the opposite direction during elongation phase.

You see that the force varies throughout the collision phase, the but the range of forces applied is different in both cases because of different compressions of the spring.

When there is no spring attached, the electrostatic forces between the charges on the colliding surfaces act like springs. Those forces are also a function of the distance between charges, just like the spring force is a function of the elongation/compression.

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For collisions, analysing forces is not really useful, since the time of collision is too short to measure. It is better to understand collision using the concept of impulse, which is basically change in momentum. You can see it as the same force acting for a longer time, or a larger force acting for the same time, or anything in between. The net effect is the same change in momentum, which is measurable, and therefore relevant.

Note: In reality, the force will change continuously during the collision as the deformation increases and decreases. However, analysing this for the amount of time they are colliding is very difficult.

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I believe that the total force in the second collision would have a magnitude 1 1/2 times that of (or 50% greater than) the total force in the first collision. In order to double the total force, you would have to double the velocities of both masses. But since you're only doubling the velocity of one mass, you're only going halfway to doubling the total force.

Remember that episode of MythBusters? Two identical cars hitting each other, one travelling at V and the other at -V, is equivalent to one car hitting a wall while travelling at V. This is because the wall is fixed. It can't move, so when the collision happens it exerts a force on the car equal and opposite to the force which the car exerts on it. If you replace the wall with a stationary car in neutral gear, then the equivalence goes away. Please let me know if this makes any sense. :)

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    $\begingroup$ It helps but im not sure. Have to read up more $\endgroup$ Aug 21, 2020 at 3:12
  • $\begingroup$ Sure, do your thing. And give us an update if you figure it out. This stuff can be a little counterintuitive. :) $\endgroup$ Aug 21, 2020 at 3:22
  • $\begingroup$ @KevinCSpeltz In order to examine the force, you have to know the time profile of the interaction. If the masses are stiff (and elastic) the interaction is short and the max. force is high. Imagine steel balls bouncing on concrete. If the masses are softer (but still elastic) the interaction is longer and the max. force is lower (racquet ball?). The momentum transfer is force X time. $\endgroup$
    – Bill N
    Sep 22, 2020 at 20:07
  • $\begingroup$ @KevinCSpeltz Okay, I understand after reading and thinking more, you're looking at the same two masses, so their material properties aren't changing. $\endgroup$
    – Bill N
    Sep 22, 2020 at 20:27
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Even though the masses "don't know" how fast they are going they are carrying different momentum and it shouldn't be violated. Maybe you are asking about the exact moment when the masses interact and the "impulse function" can clarify your doubts. Please let me know if you find it helpful

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  • $\begingroup$ Tried to make it clearer above. Trying to figure out if magnitude of force between the two collisions would be the same despite the different velocities. Obviously there would be a different amount of time, therefore a greater impulse/change of momentum. But the force should be same between collisions $\endgroup$ Aug 21, 2020 at 2:06
  • $\begingroup$ The forces are different even though the mass is the same remember that F=dp/dt if you have a larger amount of momentun changing in the same interval of time you get a greater force $\endgroup$ Aug 21, 2020 at 3:50
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This only holds for $V=0$. In the first case $v_1=v$ and $v_2=-v$. In the second case, the velocities are $v_1=0$ and $v_2=-2v$. To go from the first case to the second you can perform a Galilean transformation (which leaves the physics unchanged), after which you end up in the frame of motion in which the velocity of the COM is zero. The ensuing situation will be the same as the first case. So, yes, the forces felt by both masses are the same.

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