A bomb of mass 20 kg at rest explodes into two pieces of masses 12 kg and 8 kg . If the velocity of 8 kg mass is $6 \mathrm{~ms}^{-1}$, then the kinetic energy of the other mass is:

Let $M = 20$ kg be the mass of the bomb, $m_1 = 12$ kg, and $m_2 = 8$ kg be the masses of the two pieces. Let $v_1$ and $v_2$ be the velocities of the two pieces respectively. Initially, the bomb is at rest, so the initial momentum is zero.

According to the law of conservation of linear momentum, the final momentum must also be zero:

$m_1 v_1 + m_2 v_2 = 0$

Given $m_2 = 8$ kg and $v_2 = 6$ m/s, we have:

$12 v_1 + 8 \times 6 = 0$

$12 v_1 = -48$

$v_1 = -4$ m/s.

The kinetic energy of the 12 kg mass is given by:

$K_1 = \dfrac{1}{2} m_1 v_1^2$

$K_1 = \dfrac{1}{2} \times 12 \times (-4)^2$

$K_1 = 6 \times 16 = 96$ J.

Answer: 96 J