Question

A bomb of mass $3.0kg$ explodes in air into two pieces of masses $2.0kg$ and $1.0kg$. The smaller mass goes at a speed of $80m/s$. The total energy imparted to the two fragments in $kJ$ is

• A. $1.07$
• B. $2.14$
• C. $2.4$
• D. $4.8$

Answer 1

The correct option is D

$4.8$

Step 1: Given data:

1. Mass of body $m=3kg$
2. Mass of body $A$, $m_A=1kg$
3. Mass of body $B$, $m_B=2kg$
4. The final velocity of body $A$, $v_A$$=80m/s$

Step 2: Formula Used: Law of conservation of momentum, $Totalinitialmomentum=Totalfinalmomentumofthesystem$

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 $B$,

$m_A{v}_A+m_B{v}_B=1\times 80+2\times v_B$

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

$80+2*v_B=0$ $\Rightarrow v_B=-40m/s$……$\left(i\right)$

The velocity of the bigger part is $40m/s$

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

As we know the kinetic energy of mass $A$ = $\frac{1}{2}\times 1\times {80}^2=3200J$

$\frac{1}{2}{m}_B{v_B}^2=\frac{1}{2}2{\left(40\right)}^2=1600J$

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

Total Kinetic energy = $1600+3200J=4800J=4.8kJ$

Hence, the total kinetic energy is $4.8kJ$

Option D is correct.