Question

Particle 1 moving with velocity $v=10m/s$ experienced a head-on collision with a stationary particle 2 of the same mass. As a result of the collision, the kinetic energy of the system decreased by $\eta =1.0\%$. Find the magnitude of the velocity of particle 1 after the collision in m/s.

Answer 1

Since, the collision is head on, the particle 1 will continue moving along the same line as before the collision, but there will be a change in the magnitude of its velocity vector. Let it starts moving with velocity $v_1$ and particle 2 with $v_2$ after collision, then from the conservation of momentum
$mu=m{v}_1+m{v}_2$ or, $u=v_1+v_2$.....(1)
And from the condition, given,
$\eta =\frac{\frac{1}{2}m{u}^2-(\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2)}{\frac{1}{2}m{u}^2}=1-\frac{v_1^2+v_2^2}{u^2}$
or, $v_1^2+v_2^2=(1-\eta )u^2$.... (2)
From (1) and (2),
$v_1^2+{(u-v_1)}^2=(1-\eta )u^2$
or, $v_1^2+u^2-2u{v}_1+v_1^2=(1-\eta )u^2$
or, $2v_1^2-2v_{1}u+\eta u^2=0$
So, $v_1=2u\pm \frac{\sqrt{4u^2-8\eta u^2}}{4}$
$\frac{1}{2}[u\pm \sqrt{u^2-2\eta u^2}]=\frac{1}{2}u(1\pm \sqrt{1-2\eta })$
Positive sign gives the velocity of the 2nd particle which lies ahead. The negative sign is correct for $v_1$.
So, $v_1=\frac{1}{2}u(1-\sqrt{1-2\eta })=5m/s$ will continue moving in the same direction.
Note that $v_1=0$ if $\eta =0$ as it must.