Question

A bomb of mass $10kg$ explodes into $2$ pieces of mass $4kg$ and $6kg$. The velocity of mass $4kg$ is $1.5m/s$, the K.E. of mass $6kg$ is ....…

• A. $3.84J$
• B. $9.6J$
• C. $3J$
• D. $2.5J$

Answer 1

The correct option is C

$3J$

Step 1: Given data:

1. Mass of body $m=10kg$
2. Mass of body $A$, $m_A=4kg$
3. Mass of body $B$, $m_B=6kg$
4. The final velocity of body $A$, $v_A=1.5m/s$

Formula Used: Law of conservation of momentum, $Totalinitialmomentum=Totalfinalmomentumofthesystem$

Step 2: 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 $B$,

$m_A{v}_A+m_B{v}_B=4\times 1.5+6\times v_B=6+6v_B$

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

$6(1+v_B)=0$ $\Rightarrow v_B=-1m/s$……$\left(i\right)$

The velocity of the bigger part is $1m/s$

Step 3: Find the kinetic energy of mass B using $\left(i\right)$

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

$\frac{1}{2}{m}_B{v_B}^2=\frac{1}{2}6{\mathrm{kg}\times \left(v_B\right)}^2{\mathrm{m}}^2{s}^{-2}=3{\left(1\right)}^2=3J$

Hence, the momentum of mass $B$ is $3J$

Option C is correct.