Consider two carts,of masses $m$ and $2m$,at rest on an air track. If you push both the carts for $3\,s$ exerting equal force on each,the kinetic energy of the light cart is

Two solid rubber balls $A$ and $B$ have masses $200 \ g$ and $400 \ g$ respectively. They are moving towards each other,with ball $A$ having a velocity of $0.3 \ m/s$. After the collision,both balls come to rest. What is the velocity of ball $B$ in $m/s$?

$A$ body of mass $100 \ gm$ moves with a constant speed along a circular path of radius $r$. What is the work done during one complete revolution?

Two bodies of masses $m_{1}$ and $m_{2}$ are acted upon by a constant force $F$ for a time $t$. They start from rest and acquire kinetic energies,$E_{1}$ and $E_{2}$ respectively. Then $\frac{E_{1}}{E_{2}}$ is

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.)

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. Then,the ratio between the relative velocity of approach and the relative velocity of separation in the second collision is: