An explosion blows a stationary rock into three parts. Two parts of masses $1 \,kg$ and $2 \,kg$ move at right angles to one another with velocities $12 \,ms^{-1}$ and $8 \,ms^{-1}$, respectively. If the velocity of the third part is $4 \,ms^{-1}$, the mass of the rock is (in $\,kg$)

$A$ ball of mass $200\,g$ rests on a vertical post of height $20\,m$. $A$ bullet of mass $10\,g$,travelling in a horizontal direction,hits the centre of the ball. After the collision,both travel independently. The ball hits the ground at a distance of $30\,m$ and the bullet at a distance of $120\,m$ from the foot of the post. The value of the initial velocity of the bullet will be $............m/s$ (if $g = 10\,m/s^2$).

The figure shows a man of mass $m$ standing at the end $A$ of a trolley of mass $M$ placed at rest on a smooth horizontal surface. The man starts moving towards the end $B$ with a velocity $u_{rel}$ with respect to the trolley. The length of the trolley is $L$. The velocity of the man with respect to ground $v_1$ will be

Three particles of masses $10 \ g, 20 \ g$ and $40 \ g$ are moving with velocities $10 \hat{i}, 10 \hat{j}$ and $10 \hat{k} \ m/s$ respectively. If due to some mutual interaction,the first particle comes to rest and the velocity of the second particle becomes $(3 \hat{i} + 4 \hat{j}) \ m/s$,then the velocity of the third particle is:

$A$ bomb of mass $12 \ kg$ initially at rest explodes into two fragments with a mass ratio of $1:3$. The kinetic energy of the smaller fragment is $216 \ J$. What is the momentum of the larger (heavier) fragment in $kg \cdot m/s$?

You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface?