Question

A bomb of mass $9kg$ explodes into two parts. One part of mass $3kg$ moves with velocity $16m{s^{ - 1}}$ then the kinetic energy of the other part is
(A) $162J$
(B) $150J$
(C) $192J$
(D) $200J$

Answer 1

Hint: In order to solve this question, we will first find the velocity of the other part using the law of conservation of momentum and then we will use the kinetic energy formula to find the value of the kinetic energy of the other part of the bomb.

Formula Used:
According to the law of conservation of momentum:
${P_i}={P_f}$
$\dfrac {{m_1}{{u_1}^2}}{2}+\dfrac {{m_2}{{u_2}^2}}{2}=\dfrac {{m_1}{{v_1}^2}}{2}+\dfrac {{m_2}{{v_2}^2}}{2}$
Where,
${u_1},~{u_2}$-The initial velocities
${v_1},~{v_2}$-The final velocities

Complete answer:
We have given that a bomb which was initially at rest which means initial velocity of the bomb was ${u_i} = 0$ and having mass $m = 9kg$ explode into two parts such that one part has a mass of ${m_1} = 3kg$ and have a velocity of ${v_1} = 16m{s^{ - 1}}$ now, mass of remaining second part will be ${m_2} = 9 - 3 = 6kg$ and assume its velocity be ${v_2}$.

Now, the initial Momentum of the bomb system is
$
  {P_i} = m \times {u_i} \\
  {P_i} = 9 \times 0 = 0 \to (i) \\
 $

The final momentum is the sum of the individual momentum of both parts of the bomb so, final momentum is given as
$
  {P_f} = {m_1}{v_1} + {m_2}{v_2} \\
  {P_f} = 3 \times 16 + 6{v_2} \\
  {P_f} = 48 + 6{v_2} \to (ii) \\
 $

Equation equations (i) and (ii) using the law of conservation of momentum, we get
$
  0 = 48 + 6{v_2} \\
  {v_2} = - 8m{s^{ - 1}} \\
 $

Now, we have mass of other part is ${m_2} = 9 - 3 = 6kg$ and velocity is ${v_2} = - 8m{s^{ - 1}}$ using the Kinetic energy formula, we have
$
  K.E = \dfrac{1}{2}{m_2}{v_2}^2 \\
  K.E = \dfrac{1}{2}(6)(64) \\
  K.E = 192J \\
 $

Therefore, the kinetic energy of the second part of the bomb is $192J$

Hence, the correct option is (C) 192 J.

Note: It should be noted that the negative sign of the velocity of the second part of the bomb indicates that the second part must have flown in opposite direction to that of the first part of the bomb after the explosion.