Question

A mass of $1\text{kg}$ collides elastically with a stationary mass of $5\text{kg}$. After collision, the $1\text{kg}$ mass reverses its direction and moves with a speed of $2\text{m/s}$. Which of the following statement(s) is/are correct ?

• A. Total momentum of the system is $3\text{kg m/s}$
• B. Momentum of $5\text{kg}$ mass after collision is $4\text{kg m/s}$
• C. Kinetic energy of the centre of mass of system is $0.75\text{J}$
• D. Kinetic energy of the centre of mass of system is $1.75\text{J}$

Answer 1

Answer: (A), (C)

The correct options are
A Total momentum of the system is $3\text{kg m/s}$
C Kinetic energy of the centre of mass of system is $0.75\text{J}$


Considering the system of $1\text{kg}\&5\text{kg}$ mass, during collision $F_{ext}=0$. Therefore velocity of $\text{COM}$ will not change.
${\overrightarrow{v}}_{\text{com}}=\text{constant}$
$\Rightarrow {\overrightarrow{v}}_{\text{com}}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}$
Equating the velocity of centre of mass before and after collison,
$\Rightarrow \frac{(1\times u)+(5\times 0)}{1+5}=\frac{(-1\times 2)+(5\times v)}{1+5}$

$\Rightarrow u=5v-2....\left(1\right)$
For elastic collision, coefficient of restitution $\left(e\right)$ is equal to $1$.
$\Rightarrow e=\frac{\text{speed of separation}}{\text{speed of approach}}=1$
$\text{speed of separation}=v+2,\text{speed of approach}=u$
$\Rightarrow \frac{v+2}{u}=1$
$\Rightarrow v+2=u....\left(2\right)$
From Eq.$\left(1\right)$ and $\left(2\right)$,
$v+2=5v-2$
$\therefore v=1\text{m/s}$
Since $F_{\text{ext}}=0$, total momentum at any instant is constant i.e $P=\text{constant}$
$\Rightarrow P_i=P_f$
Hence total momentum of system is,
$P_f=-(2\times 1)+\left(5v\right)=-2+(5\times 1)$
$\Rightarrow P_i=P_f=3\text{kg m/s}$
Momentum of $5\text{kg}$ block after collision is, $P=5v=5\times 1=5\text{kg m/s}$
Velocity of centre of mass is constant and just after collison it is given as,
${\overrightarrow{v}}_{\text{com}}=\frac{(1\times -2)+(5\times 1)}{1+5}=\frac{1}{2}\text{m/s}$
Kinetic energy of centre of mass of system,
$K{E}_{\text{com}}=\frac{1}{2}{m}_{\text{com}}{v}_{\text{com}}^2$
$\therefore K{E}_{\text{com}}=\frac{1}{2}\times 6\times {\left(\frac{1}{2}\right)}^2=0.75\text{J}$
$\Rightarrow$Options $\left(a\right),\left(c\right)$ are correct.