# Question relating to energy transfer in between two particles

How would I go about calculating the maximum kinetic energy transferred from one particle when it forms an elastic collision with another particle. For example, if I had two billiards one with mass m and the other mass $M$, where mass $m<M$, how could I find the maximum kinetic energy the ball of mass m transferred to the ball of mass M upon impact? For simplicity, assume there on a horizontal, frictionless surface.

Here's a general overview of how to approach this:

Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And finally, we can use conservation of kinetic energy (as that is the definition of 'elastic' collision).

We are trying to find the maximum amount of kinetic energy of the ball of mass $m$ which we can transfer to $M$. I am assuming that what we are allowed to vary during this 'maximization' is how we shoot $m$ at the ball $M$. Since there is no limit to how much kinetic energy we can transfer to $M$ (just shoot $m$ faster), I assume by maximum kinetic energy "of" ball $m$, you mean the maximum percentage of is initial kinetic energy.

The most kinetic energy we can transfer is all of it. So a reasonable approach will be to guess that is the correct maximum and see if we can always find a way to shoot $m$ at $M$ such that all its kinetic energy transfers to $M$.

We can further simplify by seeing if this is possible in general for head-on collisions, so we can describe everything with one dimensional mechanics to simplify.

Our inputs are the masses $m$,$M$, and the initial velocity of $M$ which is $V_i$. We have as a controllable variable the initial velocity of $m$ which is $v_i$. With these we should be able to find the final velocities of $m$, $M$, which are $v_f$,$V_f$ respectively.

Conservation of energy: $$\frac{1}{2}mv_i^2 + \frac{1}{2}MV_i^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2$$

Conservation of momentum: $$mv_i + MV_i = mv_f + MV_f$$

And our constraint, we are hoping to find a solution with 100% of the kinetic energy of $m$ transferred to $M$, so $$v_f = 0$$

For 1D, there is no angular momentum, and we already have enough equations to solve for $v_i$ in terms of $m$,$M$,$V_i$ so this approach looks doable. It is just algebra after this.

I'm curious as to the solution now, so let's see what happens:

$$\frac{1}{2}mv_i^2 + \frac{1}{2}MV_i^2 = 0 + \frac{1}{2}MV_f^2 \quad \Rightarrow \quad V_f^2 = \frac{m}{M}v_i^2 + V_i^2$$

$$mv_i + MV_i = 0 + MV_f \quad \Rightarrow \quad V_f^2 = \left(\frac{m}{M}v_i + V_i\right)^2$$

combining:

$$\frac{m}{M}v_i^2 + V_i^2 = \left(\frac{m}{M}v_i + V_i\right)^2 = \left(\frac{m}{M}\right)^2 v_i^2 + 2\frac{m}{M}v_i V_i + V_i^2$$

$$\left(\frac{m}{M}- \frac{m^2}{M^2}\right) v_i^2 - 2\frac{m}{M}v_i V_i = 0$$

So either $v_i=0$ (which is if $m$ had no kinetic energy to give in the first place), or

$$\left(\frac{m}{M}- \frac{m^2}{M^2}\right) v_i - 2\frac{m}{M} V_i = 0 \quad (\text{eq } 1)$$

Which, except for $m=M$ gives

$$v_i = \frac{2\frac{m}{M}}{\frac{m}{M}- \frac{m^2}{M^2}} V_i$$

$$v_i = \frac{2}{1 - \frac{m}{M}} V_i$$

So when $m \ne M$, we can always arrange a collision such that $m$ transfers all its kinetic energy to $M$.

Now, in the special case when $m=M$, The result $(\text{eq } 1)$ becomes: $$0 - 2\frac{m}{M} V_i = 0$$ And so for the head-on setup with $m=M$ we can only transfer all the kinetic energy of $m$ to $M$ if $M$ is starting at rest ($V_i=0$). And if $M$ is starting at rest and $m$ ends at rest, we can see directly from the momentum conservation that $V_f=v_i$. So while this case is special and restricts the starting velocity of $M$ for all the energy to transfer, there is no restriction on the velocity of $m$.

So does this mean we can't always find a way to maximize the transfer of the kinetic energy? Well, it just turns out that for $m=M$, considering only the head-on case was insufficient. We need to look in two dimensions. Since $KE = \frac{1}{2}MV^2 = \frac{1}{2}M\left(V_x^2 + V_y^2\right)$, and the momentum in the $x$ and $y$ direction are individually conserved, we can always just shoot $m$ such that it hits $M$ perpendicular to $V_i$. Due to this setup, the equations in the direction of $m$ are like in the one dimensional case. With $M$ originally having no momentum in the direction $m$ is travelling, this is exactly like the one dimensional case. This gives $m$ coming to rest, and thus transferring 100% of its kinetic energy to $M$.

So the answer is that regardless of the current motion of $M$ it is always possible to shoot $m$ at $M$ such that it transfers all its kinetic energy to $M$ and thus the maximum amount is 100%.

When $m\ne M$ and $V_i=0$, the above solution gives $v_i=0$. While all the kinetic energy of $m$ is given to $M$ with this solution, it is the very boring trivial solution of transferring zero kinetic energy. So let's modify the situation I analyzed by making the very reasonable requirement that $m$ needs to start with some amount of kinetic energy. From above, this shows we cannot transfer 100% of the kinetic energy now (except for the $m=M$ case) and furthermore that this is the only case where we need to consider this modification, as the solution above shows $v_i$ will be non-zero if $V_i$ is non-zero).

$$\frac{1}{2}mv_i^2 + 0 = \frac{1}{2}mv_f^2 + \frac{1}{2}MV_f^2 \quad \Rightarrow \quad V_f^2 = \frac{m}{M}(v_i^2 - v_f^2)$$ $$mv_i + 0 = mv_f + MV_f \quad \Rightarrow \quad V_f = \frac{m}{M}(v_i - v_f)$$

combining, and using our new requirement $v_i \ne 0$ so that I can divide by it:

$$\frac{m^2}{M^2}(v_i - v_f)^2 = \frac{m}{M}(v_i^2 - v_f^2)$$ $$\frac{m}{M}\left(1 - \frac{v_f}{v_i}\right)^2 = 1 - \frac{v_f^2}{v_i^2}$$

Defining $A=\frac{v_f}{v_i}$, note that the fraction of kinetic energy transferred is

$$\frac{KE_i-KE_f}{KE_i} = \frac{\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2}{\frac{1}{2}mv_i^2}=1 - \frac{v_f^2}{v_i^2}=1 - A^2$$

For convenience, let's also define $r=\frac{m}{M}$. Now continuing on ahead $$r(1 - A)^2 = (1 - A^2)$$ Note that $A=1$ appears to be a solution! Technically it is, but this is just the boring solution of the balls not colliding at all (a trivial solution to the conservation equations).

$$r(1 -2A + A^2) - 1 + A^2 = 0$$ $$(1+r)A^2 -2rA + (r-1) = 0$$ $$A = \frac{2r \pm \sqrt{(2r)^2 - 4(1+r)(r-1)}}{2(1+r)} = \frac{2r \pm \sqrt{4r^2 - 4(r^2 -1)}}{2(1+r)} = \frac{2r \pm \sqrt{4}}{2(1+r)} = \frac{r \pm 1}{1+r}$$

Since we don't want the $A=1$ case, our final solution is:

$$\frac{KE_i-KE_f}{KE_i} = 1 - A^2 = 1 - \left(\frac{r-1}{r+1}\right)^2$$

This ratio is independent of our choice of $v_i$, so there is nothing to maximize and thus we are done. And as a check, we see that for $r=1$, 100% of the kinetic energy is transferred as expected from previous results.

• Let's say that $M$ and $m$ do not equal each other and that $V_i=0$. Then would'nt you get $(\frac{m}{M}-\frac{m^2}{M^2})v^2=0$? Since $v$ cannot be equal to zero, you would $(\frac{m}{M}-\frac{m^2}{M^2})=0$ and therefore $M=m$. But that would contradict what I just said, so what happened? – Oscar Flores Oct 1 '14 at 3:29
• You can see from the solution that for $m\ne M$ and $V_i=0$ that $v_i = 2 V_i / (1 - M/m) = 0$. So in this case, $v_i=0$ which is the boring trivial solution (all of the kinetic energy of $m$ is given to $M$, but this amount is boringly zero). I didn't notice that there was still this boring case in there (good catch!). If we make (the very reasonable) requirement that $v_i$ has to have some kinetic energy to start with, then this doesn't solve that problem. I'll add something tonight or tomorrow when I get some time. – CuriousKev Oct 1 '14 at 3:48

To calculate for a situation like this, consider the Law of Conservation of Momentum:

Pi = Pf

In the case of the billiards:

KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity)

KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity)

Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal.

KEi = KEf

Hope this helps.