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

$A$ large block of wood of mass $M = 5.99 \, kg$ is hanging from two long massless cords. $A$ bullet of mass $m = 10 \, g$ is fired into the block and gets embedded in it. The (block $+$ bullet) then swing upwards,their centre of mass rising a vertical distance $h = 9.8 \, cm$ before the (block $+$ bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is: (Take $g = 9.8 \, ms^{-2}$) (in $m/s$)

$A$ man of mass $80 \ kg$ goes to the market on a scooter of mass $100 \ kg$ with a certain speed. On application of brakes, the stopping distance is $S_1$. The man returns home on the same scooter, with the same speed with a $60 \ kg$ bag of rice. If $S_2$ is the new stopping distance when the brakes are applied with the same force, then:

If a skater of mass $3 \,kg$ has an initial speed of $32 \,m/s$ and a second skater of mass $4 \,kg$ has a speed of $5 \,m/s$. After a perfectly inelastic collision,they move together with a speed of $5 \,m/s$. Calculate the loss in kinetic energy.

In two separate collisions,the coefficients of restitution $e_1$ and $e_2$ are in the ratio $3:1$. In the first collision,the relative velocity of approach is twice the relative velocity of separation. Find the ratio between the relative velocity of approach and the relative velocity of separation in the second collision.

$A$ particle is moving in a circle of radius $r$ under the action of a force $F = \alpha r^2$ which is directed towards the centre of the circle. The total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $= 0$ for $r = 0$).