$A$ bomb of mass $9 \text{ kg}$ explodes into two pieces of mass $3 \text{ kg}$ and $6 \text{ kg}$. The velocity of mass $3 \text{ kg}$ is $16 \text{ m/s}$. The kinetic energy of mass $6 \text{ kg}$ (in joule) is

An object of mass $3\,m$ at rest splits into three equal fragments. Two fragments have velocities $v\hat j$ and $v\hat i$. The velocity of the third fragment is

$A$ bomb of $12 \ kg$ divides into two parts whose ratio of masses is $1 : 3$. If the kinetic energy of the smaller part is $216 \ J$,then the momentum of the bigger part in $kg \cdot m/s$ will be:

Two persons of mass $m_1$ and $m_2$ are standing at the two ends $A$ and $B$ respectively,of a trolley of mass $M$ as shown. When the person standing at $A$ jumps from the trolley towards the left with $u_{rel}$ with respect to the trolley,then:

$A$ gun fires a bullet of mass $50 \, g$ with a velocity of $30 \, m/s$. Because of this,the gun is pushed back with a velocity of $1 \, m/s$. The mass of the gun is .......... $kg$.

State the law of conservation of linear momentum for a system of particles.