Question

A bomb of mass $3.0\, kg$ explodes in air into two pieces of masses $2.0\, kg$ and $1.0 \,kg$. The smaller mass goes at a speed of $80 \,m / s$. The total energy imparted to the two fragments is

• A. 1.07 kJ
• B. 2.14 kJ
• C. 2.4 kJ
• D. 4.8 kJ.

Answer 1

Answer: (D)

From law of conservation of momentum, when no external force acts upon a system of two (or more) bodies then the total momentum of the system remains constant.

Momentum before explosion $=$ momentum after explosion
Since bomb $v$ at rest, its velocity is zero, hence
$m v =m_{1} v_{1}+m_{2} v_{2} $
$3 \times 0 =2 v_{1}+1 \times 80 $
$v_{1} =-\frac{80}{2}=-40\, m / s$
Total energy imparted is
$KE =\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}$
$=\frac{1}{2} \times 2 \times(-40)^{2}+\frac{1}{2} \times 1 \times(80)^{2}$
$=1600+3200=4800 \,J$
$=4.8\, kJ$