A collision occurs between two discs $A$, of mass $0.4kg$, and $B$, of mass $0.8Kg$, moving in the same direction with speeds $6$ $m/s$ and $2$ $m/s$ respectively. Given that the coefficient of restitution is $0.5$, calculate the energy change of each disc in the compression and restitution phases of the collision.
ANS: $4.98J$ loss, $2.84J$ gain; $1.42J$ loss, $1.96J$ gain
(Assuming the first 2 answers are for the compression phase and the last 2 for the restitution phase).
In the compression phase:
Let the common speed at max compression be $z$.
By conservation of momentum:
$$(0.4 * 6) + (0.8 * 2) = (0.4 + 0.8)z$$
so $$z = 3\frac{1}{3}$$
Also the approach speed is $(6 - 2) m/s$ = $4 m/s$ and so the separation speed is $(0.5 * 4)$ = $2 m/s$.
(the separation speed is the approach speed multiplied by the coefficient of restitution).
So after the collision, disc $A$ is moving at $xm/s$ and the velocity of disc $B$ is $(x+2)m/s$ (the separation speed is also the speed of disc $B$ minus the speed of disc $A$ after the collision).
By the conservation of momentum:
$$ (0.6*6) + (0.8*2) = 0.4x + 0.8(x+2)$$
which gives $x=2m/s$ so disc $A$ is moving at $2m/s$ and the velocity of disc $B$ is $4m/s$ after the collision.
Now for the kinetic energies:
Compression phase:
Disc $A$: $$\frac{1}{2}*0.4*6^2 - \frac{1}{2}*0.4*z^2 = 4.98J$$ which is a loss of $KE$.
Disc $B$: $$\frac{1}{2}*0.8*z^2 - \frac{1}{2}*0.8*2^2 = 2.84J$$ which is a gain of $KE$ as $\frac{1}{2}*0.8*z^2$ is the $KE$ at maximum compression and $\frac{1}{2}*0.8*2^2$ is the $KE$ before the collision.
Restitution phase:
Disc $A$: $$\frac{1}{2}*0.4*z^2 - \frac{1}{2}*0.4*2^2 = 1.42J$$ which is a loss of $KE$.
Disc $B$: $$\frac{1}{2}*0.8*4^2 - \frac{1}{2}*0.8*z^2 = 1.96J$$ which is a gain of $KE$.