$A$ body of mass $1 \,kg$ is suspended by a weightless string which passes over a pulley of mass $2 \,kg$ as shown in the figure. The mass is released from a height of $1.6 \,m$ from the ground. With what velocity does it strike the ground?

$A$ particle executes uniform circular motion with angular momentum $L$. Its rotational kinetic energy becomes half,when the angular frequency is doubled. Its new angular momentum is

$A$ solid sphere $(A)$ of mass $5m$ and a spherical shell $(B)$ of mass $m$, both having the same radius $R$, are placed on a rough surface. When a force $F$ of the same magnitude is applied tangentially at the highest points of $A$ and $B$, they start rolling without slipping with accelerations $a_A$ and $a_B$, respectively. The ratio of $a_A$ to $a_B$ is . . . . . . .

$A$ block of mass $m = 1 \, kg$ slides with velocity $v = 6 \, m/s$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about $O$ and swings as a result of the collision,making an angle $\theta$ before momentarily coming to rest. If the rod has mass $M = 2 \, kg$ and length $l = 1 \, m$,the value of $\theta$ is approximately (Take $g = 10 \, m/s^2$) (in $^{\circ}$)

$A$ ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. The kinetic energy of the system is: (Slipping is absent) (in $, mv_0^2$)

$A$ spherical rigid ball is released from rest and starts rolling down an inclined plane from height $h=7 \, m$,as shown in the figure. It hits a block at rest on the horizontal plane (assume elastic collision). If the mass of both the ball and the block is $m$ and the ball is rolling without sliding,then the speed of the block after collision is close to ............. $m/s$.