A bomb of mass 9 kg initially at rest explodes into two pieces of masses 3 kg and 6 kg. If the kinetic energy of 3 kg mass is 216 J, then the velocity of 6 kg mass will be

The correct option is D

6 ms^-1

Step 1: Given data

The initial mass of the bomb $m=9kg$

Velocity of bomb $v=0m{s}^{-1}$

Mass of the first piece $m_1=3kg$

Mass of the second piece $m_2=6kg$

The kinetic energy of $m_1$ ($KE$)$=216J$

Step 2: Finding the velocity of the first piece

Let $v_1$be the velocity of the first piece.

$KE=\frac{1}{2}m{v_1}^2$

$216=\frac{1}{2}\times 3\times {v_1}^2$

Multiply both sides by 2.

$432=3\times {v_1}^2$

Divide both sides by 3.

$144={v_1}^2$

Take square root on both sides.

$12=v_1$

$v_1=12m{s}^{-1}$

Step 3: Finding the velocity of the 6 kg mass

By law conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion.

Let $v_2$ is the velocity of the second piece.

Total momentum before explosion$=$Total momentum after the explosion

$mv=m_1{v}_1+m_2{v}_2$

$9\times 0=3\times 12+6v_2$

$0=36+6v_2$

$-36=6v_2$

Divide both sides by 6.

$-6=v_2$

$v_2=-6m{s}^{-1}$

The negative sign represents that the velocity of the second piece is opposite to the velocity of the first piece.

Thus, option (d) is correct.