The kinetic energy $K$ of a particle moving along a circle of radius $R$ depends on the distance covered $s$ as $K = as^2$,where $a$ is a constant. The force acting on the particle is:

$A$ projectile of mass $m$ is fired with velocity $v$ from the point $P$ at an angle $45^{\circ}$ with the horizontal. The magnitude of change in momentum,when it passes through the point $Q$ on the same horizontal line on which $P$ lies,is :

Two particles of equal masses are revolving in circular paths of radii $r_1$ and $r_2$ respectively with the same speed. The ratio of their centripetal forces is

If at any point on the path of a projectile its velocity is $u$ at an inclination $\alpha$ to the horizontal,then after what time will it move at right angles to its former direction?

Show that for a projectile,the angle between the velocity and the $x$-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{0x}}\right)$.
Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$,where the symbols have their usual meanings.

$A$ person walks in such a way that he covers equal distance in each step. The person takes $2$ steps forward towards East, then takes a right turn and walks $4$ steps towards South, then takes a right turn and walks $6$ steps towards West, and then takes a right turn and walks further. The direction of his final position after a total of $20$ steps walk with respect to his initial position is: