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

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.

Tardigrade Admin · Accepted Answer

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

K E = \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 k J

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

1.07 kJ

2.14 kJ

2.4 kJ

4.8 kJ.

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