# According to Newton's Third Law why is there a difference in movements between collisions of different mass (please read description)?

When an object of some mass hits and rolls towards another stationary object of the same mass, the ball that was rolling stops. However, if we use the same ball from that scenario, and roll it at the same speed toward a ball of more mass, then the ball goes backwards. How is this? Shouldn't the force from both of these situations be the same since the rolling ball is moving at the same velocity plus it is the same mass, ergo it should impart the same force at collision no matter what. Newton's third law says the reaction force should be the same for both of these, so shouldn't the reaction in both of these scenarios be the same because it is the same force? I understand that there is a difference in mass between the two balls, but the reaction force should be the same in these two scenarios so what happens to the rolling ball in the first scenario after the collision with the other ball should happen in the second scenario. But this does not happen, and it violates the conservation of momentum. However, what in Newton's Third Law explains how there is a difference in movement between these two scenarios?

• It's not exactly clear what your situation is, does the second mass conserve momentum by moving? Otherwise why would the first mass stop. If the second mass of equal size is stationary and can't move your initial premise isn't necessarily true. Commented May 5 at 18:59
• "hits and rolls towards" -- isn't that backwards? Doesn't it first roll and then hit? Commented May 6 at 15:06

Shouldn't the force from both of these situations be the same since the rolling ball is moving at the same velocity plus it is the same mass

No. Consider three cars, all with the same mass and all traveling at the same speed. One collides with a stationary insect, one collides with a stationary car, and one collides with a stationary train. The forces are different despite the fact that the moving cars all have the same initial mass and velocity. The mass of the object they are hitting matters too.

Newton's third law says the reaction force should be the same for both of these

No. Newton’s 3rd law says that the force of the insect on the car is equal and opposite the force of the car on the insect. It does not say that the force of the insect on the car is equal and opposite the force of the train on the car.

the reaction force should be the same in these two scenarios

No, it is not. See above.

this does not happen, and it violates the conservation of momentum

Conservation of momentum is not violated in any of the collisions. You are incorrectly comparing two different collisions. Two different collisions can have different outcomes while each still individually conserves momentum.

what in Newton's Third Law explains how there is a difference in movement between these two scenarios?

The forces are different in the different scenarios. The force from colliding with an insect is much lower than the force from colliding with a train.

• I was thinking, wouldn't net acceleration play a part too? If an object of mass M hits another stationary object of the same mass M, then the stationary object moves at the same speed as the initial moving object. This means that the acceleration it would apply backward is equal to the acceleration forward, so it cancels. But if the stationary object is larger, the acceleration is smaller, and it only cancels out a little bit of the acceleration backward. Am I correct on this or not?
– John
Commented May 5 at 20:13
• @John you could probably make something like that rule work out, but I think it is best to stick with the standard rules. So Newton’s laws when you can figure out the forces or conservation of momentum in collisions
– Dale
Commented May 5 at 20:48

The 3rd law means momentum is conserved. The change in momentum of an object (1) due to a force $$F_{2, 1}$$ from object 2 applied for a time $$\Delta t$$ is:

$$\Delta p_1 = F_{2, 1}\Delta t$$

The 3rd law says:

$$F_{2, 1} \rightarrow F_{1, 2} = -F_{2, 1}$$

where the arrow means "there exists" and the equation means "equal and opposite".

Thus:

$$\Delta p_2 = F_{1, 2}\Delta t = -\Delta p_1$$

and the total change in momentum is:

$$\Delta p_1 + \Delta p_2 = 0$$

i.e., conservation.

For your problem, I would (if I had graphing software rn), write the conservation of momentum as:

$$p' + k' = p$$

where the primed values are after the collision and $$p$$ ($$k$$) are for the moving (target) ball.

Then, define cartesian coordinates:

$$x = p'/p$$ $$y = k'/p$$

so that momentum conservation is:

$$x + y = 1$$

which is graphed as the line: $$y = 1 - x$$.

Notable cases have final states ($$S$$) as follows:

1. No collision: $$S_0 = (1, 0)$$

2. Billards style: $$S_1 = (0, 1)$$

3. Infinte mass target: $$S_{\infty} = (-1, 2)$$

To solve the problem, you need to consider energy conservation:

$$\frac{p'^2}{2m'^*} + \frac{k'^2}{2M'^*} = \frac{p^2}{2m^*}$$

Where $$m'^*$$ ($$M^*$$) are the effective masses of the balls, that is: $$m^* > m$$ because rolling includes rotational kinetic energy, and the $$'$$ indicates we've included a coefficient of restitution factor.

Which I am going to drop for now.

We can rewrite that equation as:

$${x^2} + \frac m M {y^2} = 1$$

where the extreme cases highlight the bound:

$$|x| \le 1$$

I dropped the * superscripts because if the balls have the same shape, the moment-of-inertia scale factors cancel.

At this point, if I had a graphing program I plot surfaces of constant energy:

$$y = \alpha\sqrt{1 - x^2}$$

with the parameter:

$$\alpha \equiv \sqrt{\frac M m}$$

noting that there is a constraint:

$$y \ge 0$$

as the target ball cannot go against the incoming ball (unless you magnetize them (and other changes), which is a major upgrade to the problem)

Those semi-conic sections will intersect the momentum line and show you a unique solution.