A point mass of 1kg collides elastically with a stationary point mass of  5kg. After their collision, the 1kg  mass reverses its direction and moves with a speed of  2m/s. Which of the following statement(s) is (are) correct for the system of these two masses?
a) Total momentum of the system is  3kg m/s
b) Momentum of 5kg mass after collision is  4kg m/s
c) Kinetic energy of the centre of mass $\left(\frac{1}{2}\left(\text{total\hspace{0.17em}mass}\right)v_{cm}^2\right)$ is  0.75 J
d) Total kinetic energy of the system is 4J 
 

Given point
Mass of  m=1kg
Mass of stationary point m=5kg  
Speed  v=2m/s
By conservation of linear momentum

$u=5v-\mathrm{2...}\left(i\right)$

2. As we know, the coefficient of restitution is given by $e=\frac{v_2-v_1}{u_1-u_2}=1$  for elastic collision. Using this we get,  $u=v+\mathrm{2....}\left(ii\right)$
Using Equation (i) and (ii), we have  $v=1\text{\hspace{0.17em}m}/\text{s}$ and  $u=3\text{\hspace{0.17em}m}/\text{s}$

Initial momentum of the system,  $m_1{u}_1=\left(1\right)3=3kgm/s$

$$\begin{gathered} v_2=\frac{2m_1{u}_1}{m_1+m_2}=\frac{2\left(1\right)3}{1+5}=1 \\ \Rightarrow KE of 5kg =\frac{1}{2}{m}_2{v_2}^2=\frac{1}{2}\left(5\right){\left(1\right)}^2=2.5J \end{gathered}$$

$V_{com}=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}=\frac{1\left(3\right)+5\left(0\right)}{1+5}=\frac{3}{6}=\frac{1}{2}m/s$
Kinetic energy of the centre of mass  $=\frac{1}{2}{m}_{system}{v}_{CM}^2=\frac{1}{2}\left(1+5\right){\left(\frac{1}{2}\right)}^2=0.75\text{\hspace{0.17em}J}$

KE of system=$\frac{1}{2}{m}_1{v_1}^2+\frac{1}{2}{m}_2{v_2}^2=\frac{1}{2}\left(1\right){\left(3\right)}^2+\frac{1}{2}\left(5\right){\left(0\right)}^2=4.5J$
Therefore, the correct answer is (B).