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(totally taken as KE). Neither part leaves the original line of motion finally

Moderate

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

Solution

As the body splits into two equal parts due to internal explosion therefore momentum of system remains conserved i.e.
$8 \times 2 = 4 v_{1} + 4 v_{2} \Rightarrow v_{1} + v_{2} = 4$ …(i)
By the law of conservation of energy
Initial kinetic energy + Energy released due to explosion
= Final kinetic energy of the system
$\Rightarrow \frac{1}{2} \times 8 \times {(2)}^{2} + 16 = \frac{1}{2} 4 v_{1}^{2} + \frac{1}{2} 4 v_{2}^{2}$
$\Rightarrow v_{1}^{2} + v_{2}^{2} = 16$ …(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.

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