$A$ cricket ball of mass $50 \ g$ having velocity $50 \ cm \ s^{-1}$ is stopped in $0.5 \ s$. The force applied to stop the ball is (in $N$)

$A$ force $(2 \hat{i} + \hat{j} - \hat{k}) \text{ N}$ acts on a body, which is initially at rest. At the end of $20 \text{ s}$, the velocity of the body is $(4 \hat{i} + 2 \hat{j} - 2 \hat{k}) \text{ ms}^{-1}$. The mass of the body is: (in $\text{ kg}$)

$A$ constant force acting on a body of mass $3.0 \, kg$ changes its speed from $2.0 \, m/s$ to $3.5 \, m/s$ in $25 \, s$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

At a place where the acceleration due to gravity is $10\,m/s^2$,a force of $5\,kg-wt$ acts on a body of mass $10\,kg$ initially at rest. The velocity of the body after $4\,s$ is .......... $m/s$.

An object with a mass of $10\, kg$ moves at a constant velocity of $10\, m/s$. $A$ constant force then acts for $4\, s$ on the object and gives it a speed of $2\, m/s$ in the opposite direction. The force acting on the object is ......... $N$.

$A$ cricket player catches a ball of mass $120 \,g$ moving with $25 \,m/s$ speed. If the catching process is completed in $0.1 \,s$, then the magnitude of force exerted by the ball on the hand of the player will be (in $SI$ unit):