Question

A body of mass 16 kg explodes into two pieces of masses 4 kg and 12 kg. The velocity of the 12 kg mass is 4 $ms^{-1}$. The kinetic energy of the second piece is

• A. 96 J
• B. 144 J
• C. 192 J
• D. 288 J

Hint:

Momentum is conserved in explosions. If the masses are in a 3:1 ratio, their velocities will be in a 1:3 ratio.

Answer 1

The Correct Option is D

Step 1: Concept
According to the law of conservation of linear momentum, the total momentum before the explosion equals the total momentum after the explosion.

Step 2: Meaning
Let $m_{1} = 12$ kg and $m_{2} = 4$ kg. Initially, the body is at rest, so initial momentum is zero. After explosion, $m_{1}v_{1} + m_{2}v_{2} = 0$.

Step 3: Analysis
Given $m_{1} = 12$ kg and $v_{1} = 4$ $ms^{-1}$, we find $v_{2}$: $12 \times 4 + 4 \times v_{2} = 0 \implies 48 + 4v_{2} = 0 \implies v_{2} = -12$ $ms^{-1}$. The kinetic energy of the 4 kg piece is $KE = \frac{1}{2} m_{2} v_{2}^{2}$.

Step 4: Conclusion
$KE = \frac{1}{2} \times 4 \times (-12)^{2} = 2 \times 144 = 288$ J. Final Answer: (D)