$A$ $100 \, g$ iron ball having a velocity of $10 \, m/s$ collides with a wall at an angle of $30^\circ$ with the wall surface and rebounds with the same angle. If the period of contact between the ball and the wall is $0.1 \, s$,then the force experienced by the ball is ............. $N$.

$Assertion$ : $A$ quick collision between two bodies is more violent than a slow collision,even when the initial and final velocities are identical.
$Reason$ : The rate of change of momentum is greater in the first case.

In an hour-glass,approximately $100$ grains of sand fall per second (starting from rest); and it takes $2 \, s$ for each sand particle to reach the bottom of the hour-glass. If the average mass of each sand particle is $0.2 \, g$,then the average force exerted by the falling sand on the bottom of the hour-glass is close to .......... $N$.

$A$ ball of mass $0.1 \,kg$ strikes a wall normally with a speed of $30 \,ms^{-1}$ and rebounds with a speed of $20 \,ms^{-1}$. The impulse of the force exerted by the wall on the ball is (in $\,N-s$)

The figure below shows the position-time graph of a body of mass $0.04 \; kg$. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

$A$ force is applied to a body of mass $3 \ kg$ at rest on a frictionless horizontal surface, as shown in the force-time $(F-t)$ graph. The speed of the body after $1 \ s$ is: (in $m/s$)