Newtonian Mechanics – Does Newton’s Third Law Apply to the Kinetic Energy Formula?

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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.

Best Answer

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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