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 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
Both parts continue to move in the same direction as that of the original body
One part comes to rest and the other moves in the same direction as that of the original body
One part comes to rest and the other moves in the direction opposite to that of the original body
One part moves in the same direction and the other in the direction opposite to that of the original body
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
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}4v_1^2 + \frac{1}{2}4v_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.