$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground,the magnitude of the change in its momentum will be

$A$ sportsman runs around a circular track of radius $r$ such that he traverses the path $ABAB$. The distance travelled and displacement,respectively,are

$A$ particle $A$ moves along the line $y=30 \ m$ with a constant velocity $v$ parallel to the $x$-axis. At the moment particle $A$ passes the $y$-axis, a particle $B$ starts from the origin with zero initial speed and a constant acceleration $a=0.40 \ m/s^2$. The angle between $a$ and the $y$-axis is $60^{\circ}$. If the particles $A$ and $B$ collide after some time, then the value of $|v|$ will be (in $m/s$)

$A$ projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20 \; m/s$. After $10 \; s$,its inclination with the horizontal is $\beta$. The value of $\tan \beta$ will be: $(g = 10 \; m/s^2)$

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

$A$ 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: