Question

A ball of mass $M$ moving with a speed of $2 \,m / s$ hits another ball of mass $1 \,kg$ moving in the same direction with a speed of $1 \,m / s$. If the kinetic energy of center of mass is $\frac{4}{3}$ Joule, then the magnitude of $M$ is

• A. $1 \,kg$
• B. $0.25 \, kg$
• C. $0.50 \, kg$
• D. $2 \, kg$

Answer 1

Answer: (C)

According to given situation,

As there is no external force, velocity and hence $KE$ of centre of mass remains same before and after collision.
Now,
$v_{C M}=\frac{m_{1} v_{1}+m_{2} v_{2}}{m_{1}+m_{2}} $
$\Rightarrow v_{C M}=\frac{2 M+1}{M+1}$
Kinetic energy of centre of mass is given $\frac{4}{3} J$.
$\therefore \frac{4}{3}=\frac{1}{2}(M+1) \cdot v_{C M}^{2} $
$\Rightarrow \frac{4}{3}=\frac{1}{2}(M+1)\left(\frac{2 M+1}{M+1}\right)^{2} $
$\Rightarrow \frac{8}{3}=\frac{(2 M+1)^{2}}{M+1} $
$\Rightarrow 8(M+1)=3(2 M+1)^{2} $
$\Rightarrow 12 M^{2}+4 M-5=0$
$\Rightarrow 12 M^{2}+10 M-6 M-5=0$
$\Rightarrow 2 M(6 M+5)-1(6 M+5)=0 $
$\Rightarrow (2 M-1)(6 M+5)=0$
$\Rightarrow M=\frac{1}{2} $
or $ M=\frac{-5}{6} $
So, mass, $M=\frac{1}{2}=0.5 \,kg$