Question

A bomb of mass $9kg$ explodes into two parts. One part of mass $3kg$ moves with velocity $16m/s$, then the kinetic energy of the other part is

• A. $162J$
• B. $150J$
• C. $192J$
• D. $200J$

Answer 1

The correct option is C

$192J$

Step 1: Given data:

Mass of bomb $m=9kg$

Mass of one part A, $m_A=3kg$

Mass of another part B, $m_B=6kg$

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 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,

$m_A{v}_A+m_B{v}_B=3*16+6*v_B=48+6v_B$

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

$6(8+v_B)=0$ $\Rightarrow v_B=-8m/s$……$\left(i\right)$

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

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

The kinetic energy of mass B =$\frac{1}{2}{m}_B{v_B}^2=\frac{1}{2}6{\left(v_B\right)}^2=3{\left(8\right)}^2=192J$

Hence the kinetic energy of mass B is $192J$. Option C is correct.