Question

A free body of mass 8 kg is travelling at 2 meter per second in a straight line. At a certain instant, the body splits into two equal parts due to internal explosion which releases 16 joules of energy. Neither part leaves the original line of motion finally

• A. Both parts continue to move in the same direction as that of the original body
• B. One part comes to rest and the other moves in the same direction as that of the original body
• C. One part comes to rest and the other moves in the direction opposite to that of the original body
• D. One part moves in the same direction and the other in the direction opposite to that of the original body

Answer 1

Answer: (B)

One part comes to rest and the other moves in the same

direction as that of the original body

Answer 2

As the body splits into two equal parts due to internal explosion therefore momentum of system remains conserved i.e. $8 \times 2 = 4 v _ { 1 } + 4 v _ { 2 }$⇒ $v _ { 1 } + v _ { 2 } = 4$ …(i)

By the law of conservation of energy

Initial kinetic energy + Energy released due to explosion

= Final kinetic energy of the system

⇒$\frac { 1 } { 2 } \times 8 \times ( 2 ) ^ { 2 } + 16 = \frac { 1 } { 2 } 4 v _ { 1 } ^ { 2 } + \frac { 1 } { 2 } 4 v _ { 2 } ^ { 2 }$

⇒ $v _ { 1 } ^ { 2 } + v _ { 2 } ^ { 2 } = 16$ …(ii)

By solving eq. (i) and (ii) we get $v _ { 1 } = 4$ and $v _ { 2 } = 0$

i.e. one part comes to rest and other moves in the same direction as that of original body.