# Use of kinetic energy formula yielding additional solutions

I was answering a simple momentum question:

Three smooth particles $$P$$, $$Q$$ and $$R$$, having masses $$2m$$, $$m$$ and $$2m$$ respectively, are placed at rest on a smooth horizontal plane such that $$P$$, $$Q$$ and $$R$$ lie in a straight line and in that order. The particle $$P$$ is projected, with speed $$U$$, in the direction $$PQ$$.

Given that $$P$$ and $$Q$$ coalesce on the first impact and proceed to collide perfectly elastically with $$R$$, what are the speeds of the combined particle and $$R$$ after this second collision.

The second collision will be a collision between a combined $$P$$ and $$Q$$ of mass $$3m$$ travelling at speed $$\frac{2}{3}U$$ towards a stationary $$R$$ of mass $$2m$$.

If we define the velocities of the combined particle and $$R$$ after the collision as $$v_{1}$$ and $$v_{2}$$ respecitvely we can form the equation: $$2U = 3v_{1} + 2V_{2}$$ Using conservation of momentum. Then, as the question states the collision is perfectly elastic we can use $$e = \frac{\textrm{speed of rebound}}{\textrm{speed of approach}}$$ letting $$e = 1$$ to obtain: $$\frac{2}{3}U = v_{2} - v_{1}$$ Combining these two equations gives $$v_{1} = \frac{2}{15}U$$ and $$v_{2} = \frac{4}{5}U$$

However if I were to have used the kinetic energy formula instead of the coefficient of restitution forming the equation: $$\frac{1}{2}(3m)(\frac{2}{3}u)^{2} = \frac{1}{2}(3m)(v_{1})^{2} + \frac{1}{2}(2m)(v_{2})^{2}$$ I would have obtained an additional solution where $$v_{1} = \frac{2}{3}U$$ and $$v_{2} = 0$$

So my question is simply what is the relevance of these two solutions? Are they correct for what the question is asking? And if so why is the use of coefficient of restitution insufficient in that it doesn't yield these additional solutions?

• Look carefully at your additional solution. What is the physical meaning of those velocities? Does the result now make sense, given that the only conditions you have imposed are that the total momentum and energy are the same as before?
– user856
Sep 8, 2019 at 14:00

$$v_{1} = \frac{2}{3}U,\>\>\>v_{2} = 0$$