Question

Two masses, $2m$ and $m$ are moving in opposite directions, with velocities $2v$ and $v$ respectively. After collision, mass $2m$ comes to rest and mass $m$ moves with $3v$. What are the kinetic energies of the two masses, $2m$ and $m$ respectively after collision and the coefficient of restitution$\left(e\right)$ ?

• A. $0,\frac{9}{2}m{v}^2,e=1/3$
• B. $0,\frac{9}{2}m{v}^2,e=1$
• C. $0,\frac{9}{2}m{v}^2,e=0$
• D. None of the above

Answer 1

The correct option is B $0,\frac{9}{2}m{v}^2,e=1$

According to conservation of momentum principle,

$2m\left(2v\right)+m(-v)=2m\left(0\right)+m\left(v_2\right)$

$\therefore v_2=3v>0$

This means after collision, mass $m$ starts moving in the initial direction $($before collision$)$ as of mass $2m$.

Now,

For $2m,K{E}_f=\frac{1}{2}\left(2m\right)(0{)}^2=0$

For $m,K{E}_f=\frac{1}{2}\left(m\right)(3v{)}^2=\frac{9m{v}^2}{2}$

And coefficient of restitution is,

$e=\frac{v_2-v_1}{u_1-u_2}=\frac{3v-0}{2v-(-v)}=1$

Hence, option $\left(B\right)$ is the correct answer.