Question

A bomb of mass $3mkg$ explodes into two pieces of mass $mkg$ and $2mkg$. If the velocity of $mkg$ mass is $16m/s$. The total kinetic energy released in the explosion in $J$ is

• A. $192mJ$
• B. $96mJ$
• C. $384mJ$
• D. $768mJ$

Answer 1

The correct option is A

$192mJ$

Step 1: Given data:

Mass of bomb, $m=3mkg$

Mass of one part A, $m_A=mkg$

Mass of another part B, $m_B=2mkg$

The final velocity of body A, $v_A$$=16m/s$

Step 2: Formula Used:

Law of conservation of momentum, Total initial momentum $=$ Total final momentum of the system

Step 3: Calculate the velocity of mass 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,

$m_A{v}_A+m_B{v}_B=m(1*16+2*v_B)$

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

$16+2*v_B=0$ $\Rightarrow v_B=-8m/s$……$\left(i\right)$

The velocity of the bigger part is $8m/s$

Step 4: Find the kinetic energy of mass B using $\left(i\right)$

As we know the kinetic energy of mass A = $=\frac{1}{2}{m}_A{v_A}^2=\frac{1}{2}*m*{16}^2=128mJ$

Kinetic energy of mass B= $\frac{1}{2}{m}_B{v_B}^2=\frac{1}{2}2m{\left(8\right)}^2=64mJ$

Step 5: Sum the kinetic energies of masses A and B to find the kinetic energy of the system:

Total Kinetic energy = $128m+64mJ=192mJ$

Hence the total kinetic energy released in the explosion is $192mJ$

Hence, Option A is correct.