A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{{m{v^2}}}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

A body of mass $2\, kg$ slides down a curved track which is quadrant of a circle of radius $1$ $meter$ as shown in figure. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is …………. $\mathrm{m}/ \mathrm{s}$

$A$ & $B$ are blocks of same mass $m$ exactly equivalent to each other. Both are  placed on frictionless surface connected by one spring. Natural length of spring is $L$  and force constant $K$. Initially spring is in natural length. Another equivalent block $C$  of mass $m$ travelling at speed $v$ along line joining $A$ & $B$ collide with $A$. In  ideal condition maximum compression of spring is :-

If the potential energy of a gas molecule is

$U = \frac{M}{{{r^6}}} – \frac{N}{{{r^{12}}}}$,

$M$ and $N$ being positive constants, then the potential energy at equilibrium must be

Answer carefully, with reasons :

$(a)$ In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact) ?

$(b)$ Is the total linear momentum conserved during the short time of an elastic collision of two balls ?

$(c)$ What are the answers to $(a)$ and $(b)$ for an inelastic collision ?

$(d)$ If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic ?

(Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).