Question A.7: A block of mass m = 2.00 kg travels in the positive x-direct...
A block of mass m = 2.00 kg travels in the positive x-direction at v_{i}=5.00 m / s , while a second block, of mass M = 4.00 kg and leading the first block, travels in the positive x-direction at 2.00 m/s. The surface is frictionless. What are the blocks’ velocities after collision, if that collision is perfectly elastic?
As can be seen in Chapter 6, a perfectly elastic collision involves equations for the momentum and energy. With algebra, the energy equation, which is quadratic in v, can be recast as a linear equation. The pair of equations are given by
\begin{array}{r}m v_{i}+M V_{i}=m v_{f}+M V_{f} (1)\\v_{i}-V_{i}=-\left(v_{f}-V_{f}\right) (2)\end{array}
Substitute the known quantities v_{i}=5.00 m / s \text { and } V_{i}=2.00 m / s into (1) and (2):
\begin{aligned}18 &=2 v_{f}+4 V_{f} (3)\\3 &=-v_{f}+V_{f} (4)\end{aligned}
Multiply (4) by 2 and add to (3):
\begin{aligned}18 &=2 v_{f}+4 V_{f} \\6 &=-2 v_{f}+2 V_{f} \\\hline \\24 &=6 V_{f} \rightarrow V_{f}=4.00 m / s\end{aligned}
Substituting the solution for V_{f} \text { back into (4) yields } v_{f}=1.00 m / s .