$A$ body of mass $0.5 \ kg$ is supplied with a power $P$ (in watt) which varies with time $t$ (in second) as $P = 3t^2 + 3$. If the velocity of the body at time $t = 0$ is zero, then the velocity of the body at time $t = 3 \ s$ is (in $ms^{-1}$)

$A$ student skates up a ramp that makes an angle $30^{\circ}$ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed $v_0$ and wants to turn around over a semicircular path $xyz$ of radius $R$ during which he/she reaches a maximum height $h$ (at point $y$) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ($g$ is the acceleration due to gravity):
$(A)$ $v_0^2 - 2gh = \frac{1}{2} gR$
$(B)$ $v_0^2 - 2gh = \frac{\sqrt{3}}{2} gR$
$(C)$ The centripetal force required at points $x$ and $z$ is zero.
$(D)$ The centripetal force required is maximum at points $x$ and $z$.

Two identical balls $A$ and $B$ are released from the positions shown in the figure. They collide elastically on the horizontal portion $MN$. All surfaces are smooth. The ratio of the maximum heights attained by $A$ and $B$ after the collision will be (Neglect energy loss at $M$ and $N$):

$A$ ball is released from a certain height. It loses $50\%$ of its kinetic energy on striking the ground. It will attain a height again equal to

In the diagram shown,there is no friction at any contact surface. Initially,the spring has no deformation. What will be the maximum deformation in the spring? Consider all the strings to be sufficiently large. Consider the spring constant to be $K$.

$A$ particle of mass $m$ moving eastward with a speed $v$ collides with another particle of same mass moving northward with same speed $v$. The two particles coalesce after collision. The new particle of mass $2m$ will move in north-east direction with a speed (in $m/s$) of: