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Here is a classic problem:

A 60kg boy and a 40kg girl are sitting in front of each other on skateboards on a frictionless horizontal surface. The boy pushes the girl and the girl gains a velocity of 0.3m/s during the 0.5s the boys hand was in contact with her. What is the boy's resulting velocity?

Considering Newton's Third Law, the best way is to find the total force applied and rearrange the formula to find the boy's speed.

Newton's Law Method:

$F = m*a$

$a=\frac{0.3m/s}{0.5s}$

$F = 40kg * 0.6m/s^2$

$F = 24N$

Therefore the boy's speed will be:

$\frac{24N}{60kg} = a$

$a = 0.4m/s^2$

$v = 0.4m/s^2 * 0.5s$

$v = 0.2m/s$

Kinetic Energy Method:

$E_{k}=\frac{1}{2}*m*v^2$ the girl's kinetic energy

$E_{k}=\frac{1}{2}*40kg*(0.3m/s)^2$

$E_{k}=1.8J$

Since every action has an equal and opposite reaction the boy should have the same amount of kinetic energy.

$E_{k}=\frac{1}{2}*60kg*(0.2m/s)^2$

$E_{k}=1.2J$

As you can see, using Newton's Third Law, the kinetic energy was not the same. So what's going on here? I thought that in this case they should both have the same amount of kinetic energy due to Newton's Law. I must be missing something since there is more energy going in one direction than the other after meeting. My book used the first method as the answer and, well, it feels wrong.

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    $\begingroup$ Why do newton's laws mean they have to have the same energy? $\endgroup$ – Brian Moths Dec 10 '13 at 3:57
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    $\begingroup$ Newton's law say nothing about kinetic energy. Never have. For that matter the "pushing off" experiment can not even be modeled as an elastic collision it is emphatically inelasitc, though a little odd in that it is sees a gain in total kinetic energy. $\endgroup$ – dmckee Dec 10 '13 at 4:04
  • $\begingroup$ @dmckee What happens then if that girl with 1.8J of kinetic energy headed to the right runs into a different boy on a skateboard of mass 60kg. If she transfers all of her kinetic energy (law of conservation of energy) to him then he will be moving at 0.245m/s [right] which is faster than the first boy currently moving 0.2m/s [left]. Is this correct? $\endgroup$ – Klik Dec 10 '13 at 4:56
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    $\begingroup$ Energy conservation is a curvy babe, I know, but be careful: just because you insist on giving this second boy a full 1.8J of love and affection doesn't mean our little heroine doesn't have some energy left over for herself afterwards. You may think this collision is elastic, but it ain't: kinetic energy goes up again. The girl rebounds. You gotta conserve momentum. $\endgroup$ – Geoffrey Dec 10 '13 at 5:21
  • $\begingroup$ @Klik You must conserve momentum in every collision. If you are presuming that the girl is stopped after that second collision then you can only conserve momentum. $\endgroup$ – dmckee Dec 10 '13 at 5:26
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In both of your solutions, you attempted to use Newton's 3rd law: $$\vec{F}_{1\rightarrow2}=-\vec{F}_{2\rightarrow1}.\tag{Newton's 3rd law}$$ You did this correctly in your first method ("Newton's law method") but incorrectly in your second method ("Kinetic energy method").

In your first method, you explicitly set the magnitude of the forces equal to each other. Is is the way to use Newton's 3rd law.

In your second method, you assumed that "equal and opposite reaction" applies to other quantities other than force. It does not. It only applies to forces. Not kinetic energy. In addition, it seems you want to assign a direction for kinetic energy. It does not have one.

I also want to make the comment that the phrase

for every action there is an equal and opposite reaction

in Newtonian mechanics is extremely easy to misinterpret. I think that is what happened in your second method. I always prefer the labeled equation above.

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  • $\begingroup$ What happens then if that girl with 1.8J of kinetic energy headed to the right runs into a different boy on a skateboard of mass 60kg. If she transfers all of her kinetic energy to him then he will be moving at 0.245m/s [right] which is faster than the first boy currently moving 0.2m/s [left]. Is this correct? $\endgroup$ – Klik Dec 10 '13 at 4:54
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Equal and opposite forces do not imply equal energies.

Conservation of momentum is what is important here. You'll notice that the two children end up with equal and opposite momenta when all is said and done. But energy is not being conserved because this is an inelastic collision (i.e. work is being done).

Moreover, energy is a scalar, so don't get confused that there should be some sort of "equal and opposite energy." Kinetic Energy is always positive and has no direction. The boy is heavier, so he moves more slowly than the girl to conserve momentum. The girl will have a higher energy because she is moving more quickly and energy goes as the square of velocity (while momentum is linear in velocity).

Momentum is always conserved when only conservative forces are at play. Be not seduced by the painted succubus of energy conservation. Well, at least wait until you learn some Lagrangian mechanics.

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