Question

A bomb is $12kg$ and divides into two parts whose ratio of masses is $1:3$. If the kinetic energy of a smaller part is $216J$, then the momentum of a bigger part in $kg-m/sec$ will be

• A. $36$
• B. $72$
• C. $108$
• D. Data is incomplete

Answer 1

The correct option is A

$36$

Step 1: Given data:

1. Mass of body $m=12kg$
2. The ratio of masses of smaller part $A$ to bigger part $B$, $m_A/m_B=1/3$
3. The final kinetic energy of body $A$, $K{E}_A=216J$
4. The final velocity of body $B$, $v_B$

Step 2: Formula Used: Law of conservation of momentum, Total initial momentum $=$ Total final momentum of the system

Step 3: Calculate the ratio of velocities of masses $A$ and $B$:

Let the Final velocity of the first part $A$ $v_A$$m/s$ and the final velocity of body $B$, $v_B$$m/s$

The initial momentum of the system is zero.

The final momentum of the system is the sum of momentums of $A$ and $B$, as the mass of $B$ is three times of $A$

$m_A{v}_A+m_B{v}_B=m_A(v_A+3v_B)$

The initial momentum of the system = Final momentum of the system

$m_A(v_A+3v_B)=0$ $\Rightarrow v_B=-v_A/3$……$\left(i\right)$

The velocity of the bigger part is one-third the velocity of the smaller part.

Step 4: Find the velocity of mass A using $\left(i\right)$

As we know the kinetic energy of mass $A=216J$ (Given)

$$\begin{gathered} \frac{1}{2}{m}_A{v_A}^2=216 \\ \Rightarrow \frac{1}{2}3{\left(v_A\right)}^2=216 \\ \Rightarrow {v_A}^2=(2\times 216)/3=144 \\ \Rightarrow v_A=12m/s \end{gathered}$$

The velocity of mass $A$ is $v_A=12m/s$

Step 5: Find the momentum of mass $B$,

$m_B{v}_B=9(-v_A/3)=-3v_A$

The magnitude of the momentum of mass $B$ is $3v_A=3\times 12=36kgm/s$

Hence the momentum of mass $B$ is $36kgm/s$

Option A is correct.