Question

A bomb of mass 9kg explodes into the fragments of masses 3kg and 6kg. The velocity of mass 3kg is 16m/sec. The energy of explosion is equal to

• A. 384J
• B. 576J
• C. 192J
• D. 1152J

Answer 1

Answer: (B)

576J

Answer 2

Conservation of momentum just before and after the impact yields

⇒ $0 = \left| \mathrm { m } _ { 1 } \overline { \mathrm { v } } _ { 1 } + \mathrm { m } _ { 2 } \overline { \mathrm { v } } _ { 2 } \right|$

⇒ $\mathrm { m } _ { 1 } \mathrm { v } _ { 1 } = \mathrm { m } _ { 2 } \mathrm { v } _ { 2 } = \mathrm { p } ($ say $)$

⇒ Energy of explosion =

= $\frac { \mathrm { p } ^ { 2 } } { 2 \mathrm {~m} _ { 1 } } + \frac { \mathrm { p } ^ { 2 } } { 2 \mathrm {~m} _ { 2 } } = \frac { \mathrm { p } ^ { 2 } } { 2 \mathrm {~m} _ { 1 } } \left[ 1 + \frac { \mathrm { m } _ { 1 } } { \mathrm {~m} _ { 2 } } \right]$

Putting P = $\mathrm { m } _ { 1 } \mathrm { v } _ { 1 }$ , where $\mathrm { m } _ { 1 } = 3 \mathrm {~kg}$.

$\mathrm { v } = 16 \mathrm {~m} / \mathrm { sec }$ .

We obtain,

⇒ $\mathrm { KE } _ { 1 } = \frac { \mathrm { p } ^ { 2 } } { 2 \mathrm {~m} _ { 1 } } = \frac { ( 3 \times 16 ) ^ { 2 } } { 2 \times 3 } = 384 \mathrm {~J}$ ;

$\mathrm { KE } _ { 2 } = \frac { \mathrm { p } ^ { 2 } } { 2 \mathrm {~m} _ { 2 } } = \frac { ( 3 \times 16 ) ^ { 2 } } { 2 \times 3 } = 192 \mathrm {~J}$

The conclusion that kinetic energy of the system

$\mathrm { E } = \frac { 1 } { 2 } \left( \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } \right) \mathrm { v } _ { \mathrm { cm } } ^ { 2 } = 0$ is wrong