Question

A bomb of the mass $16Kg$ at rest explodes into two pieces of mass $4Kg$ and $12Kg$. The velocity of the $12Kg$ mass is $4m{s^{ - 1}}$. The kinetic energy of the other mass is :
A. $144J$
B. $288J$
C. $192J$
D. $96J$

Answer 1

Hint: This numerical can be approached by law of conservation of linear momentum. It states that a system or body retains the total momentum unless any external force is applied to the body or system. Momentum is calculated by taking the product of mass and velocity of a body.

Complete step by step answer:
We know that it is given in the problem statement that mass of first piece $({m_1}) = $ $12Kg$ , velocity $({v_1}) = $ $4m{s^{ - 1}}$ and mass of second piece $({m_2}) = $ $4Kg$. Now we can apply conservation of linear momentum in this case. According to conservation of momentum the total momentum remains constant thus,
Momentum of bomb before explosion = Momentum of first piece + Momentum of second piece
 $ \Rightarrow mu = {m_1}{v_1} + {m_2}{v_2}$ , Here m is the mass of the bomb and $u = 0$ because before explosion bomb was at rest so the velocity of the bomb will be zero.
 $
   \Rightarrow 0 = 12 \times 4 + 4 \times {v_2} \\
   \Rightarrow {v_2} = - 12m/s \\
$
Now we have calculated the velocity of second piece which has negative sign which shows that it will move with $12m/s$ in opposite direction so the kinetic energy of the $4Kg$ mass will be as follows;
 $
  K = \dfrac{1}{2}m{v^2} = \dfrac{1}{2} \times 4 \times 12 \times 12 \\
  K = 288J \\
$

So, the correct answer is “Option B”.

Note: We have approached this numerical with conservation of momentum. The total momentum of a body never changes means it remains constant and this is known as conservation of momentum. The required quantities are given in the problem statement in order to calculate the velocity of another piece and then we have calculated its kinetic energy with the help of formula $K = \dfrac{1}{2}m{v^2}$.