$A$ cricket ball of mass $0.25 \,kg$ with speed $10 \,m/s$ collides with a bat and returns with the same speed within $0.01 \,s$. The force exerted on the bat is: (in $\,N$)

$A$ soccer ball of mass $250 \text{ g}$ is moving horizontally to the left with a speed $22 \text{ ms}^{-1}$. This ball is kicked towards the right with a velocity $30 \text{ ms}^{-1}$ at an angle $53^{\circ}$ with the horizontal in the upward direction. Assuming that it took $0.01 \text{ s}$ for the collision to take place, the average force acting is $(\cos 53^{\circ} = \frac{3}{5}; \sin 53^{\circ} = \frac{4}{5})$. (in $\text{ N}$)

The momentum is most closely related to

$A$ ball of mass $0.5 \ kg$ moving horizontally at $10 \ m \ s^{-1}$ strikes a vertical wall and rebounds with speed $v$. The magnitude of the change in linear momentum is found to be $8.0 \ kg \ m \ s^{-1}$. The magnitude of $v$ is, (in $m \ s^{-1}$)

Using the variation of force and time given below, the final velocity of a particle of mass $2 \, kg$ moving with an initial velocity $6 \, m/s$ will be: (in $m/s$)

$A$ ball of mass $50 \,g$ is dropped from a height of $20 \,m$. $A$ boy on the ground hits the ball vertically upwards with a bat with an average force of $200 \,N$,so that it attains a vertical height of $45 \,m$. The time for which the ball remains in contact with the bat is ........... of a second. [Take $g = 10 \,m/s^2$]