$A$ bullet of mass $m$ moving with velocity $v_1$ strikes a wooden block of mass $M$ suspended as shown in the figure and gets embedded in it. If the block rises to a height $h$,what is the initial velocity of the bullet?

$A$ box of mass $3 \,kg$ moves on a horizontal frictionless table and collides with another box of mass $3 \,kg$ initially at rest on the edge of the table at height $1 \,m$. The speed of the moving box just before the collision is $4 \,m/s$. The two boxes stick together and fall from the table. The kinetic energy just before the boxes strike the floor is (Assume, acceleration due to gravity, $g=10 \,m/s^2$) (in $\,J$)

$A$ bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$,the initial velocity of the bullet will be

$A$ raindrop of mass $1.00 \, g$ falling from a height of $1 \, km$ hits the ground with a speed of $50 \, m s^{-1}$. Calculate
$(a)$ the loss of $PE$ of the drop
$(b)$ the gain in $KE$ of the drop
$(c)$ Is the gain in $KE$ equal to loss of $PE$? If not,why?
Take $g = 10 \, m s^{-2}$.

Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted into

Particle $A$ of mass $m_{1}$ moving with velocity $(\sqrt{3} \hat{i} + \hat{j}) \, m/s$ collides with another particle $B$ of mass $m_{2}$ which is at rest initially. Let $\vec{V}_{1}$ and $\vec{V}_{2}$ be the velocities of particles $A$ and $B$ after collision,respectively. If $m_{1} = 2m_{2}$ and after collision $\vec{V}_{1} = (\hat{i} + \sqrt{3} \hat{j}) \, m/s$,the angle between $\vec{V}_{1}$ and $\vec{V}_{2}$ is $......^{\circ}$.