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

$A$ particle moves in the $X-Y$ plane under the influence of a force such that its linear momentum is $\vec{p}(t)=A[\hat{i} \cos (kt)-\hat{j} \sin (kt)]$,where $A$ and $k$ are constants. The angle between the force and the momentum is (in $^{\circ}$)

$A$ particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha(y \hat{i} + 2x \hat{j})$,where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

The centripetal acceleration of a particle in uniform circular motion is $18 \,m/s^2$. If the radius of the circular path is $50 \,cm$, the change in velocity of the particle in a time of $\frac{\pi}{18} \,s$ is (in $\,m/s$)

Fill in the blanks:
$(a)$ For vectors $\vec{A} = 3\hat{i} + 2\hat{j}$ and $\vec{B} = \hat{i} + \hat{j} - 2\hat{k}$,the $y$-component of the vector $(\vec{A} - \vec{B})$ is .....
$(b)$ The formula for the total time of flight of a projectile projected vertically upwards with an initial velocity $u$ is ..........
$(c)$ The formula for the centripetal acceleration of a particle moving in a circular path of radius $R$ with a constant angular velocity $\omega$ is ......
$(d)$ The component of any vector is always .........

$A$ train is moving along a straight line with a constant acceleration '$a$'. $A$ boy standing in the train throws a ball forward with a speed of $10 \ m/s$,at an angle of $60^{\circ}$ to the horizontal. The boy has to move forward by $1.15 \ m$ inside the train to catch the ball back at the initial height. The acceleration of the train,in $m/s^2$,is