$A$ body of mass $m$ is projected at an angle of $45^{\circ}$ with the horizontal with an initial velocity $v$. If air resistance is negligible,then the total change in momentum when it strikes the ground is:

Assertion: In a circular motion,work done by centripetal force is not always zero.
Reason: If the speed of the particle increases or decreases in circular motion,then the net force acting on the particle does not remain towards the centre.

Three particles start simultaneously from a point on a horizontal smooth plane. The first particle moves with speed $v_1$ towards the east,the second particle moves towards the north with speed $v_2$,and the third one moves towards the north-east. The velocity of the third particle,such that the three particles always lie on a straight line,is

The figure shows the velocity and the acceleration of a point-like body at the initial moment of its motion. The direction and the absolute value of the acceleration remain constant. Find the time when the speed becomes minimum. (Given: $a = 4\, m/s^2$,$v_0 = 40\, m/s$,$\phi = 143^o$)

$A$ particle is moving in the $xy$-plane and crosses the origin at time $t=0$. The equation of motion of the particle is $y=4x^2$. If the velocity of the particle is $\vec{v}=(2\hat{i}+2\hat{j}) \text{ m s}^{-1}$ and acceleration is $\vec{a}=(a\hat{j}) \text{ m s}^{-2}$, then the magnitude of $a$ is

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a$ varies with time $t$ as $a = k^{2} r t^{2}$,where $k$ is a constant. The power delivered to the particle by the force acting on it is given as: