Question

A bomb of mass $3m\text{kg}$ explodes into two pieces of mass $m\text{kg}$ and $2m\text{kg}$ . If the velocity of $m\text{kg}$ mass is $16\text{m/s}$, the total energy released in the explosion is

• A. $192m\text{J}$
• B. $96m\text{J}$
• C. $384m\text{J}$
• D. $768m\text{J}$

Answer 1

The correct option is A $192m\text{J}$

Given that,
Initial velocity of mass $3m\text{kg},\left(v_i\right)=0\text{m/s}$
after explosion


By the conservation of momentum,
initial momentum of the systme = final momentum of the system
initial momentum of the system = 0
$0=m_A{v}_A-m_B{v}_B$
$\Rightarrow m_A{v}_A=m_B{v}_B$
$\Rightarrow m\times 16=2m\times v_B$
$\Rightarrow v_B=8m/s$
Energy release during explosion is kinetic energy of system
=$\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2=192m\text{J}$
$\therefore$(a).