The displacement $x$ of a particle moving in one dimension under the action of a constant force is related to the time $t$ by the equation $t = \sqrt{x} + 3$,where $x$ is in meters and $t$ is in seconds. The work done by the force in the first $6$ seconds is.....$J$

Two bodies of different masses $m_1$ and $m_2$ have equal momenta. Their kinetic energies $E_1$ and $E_2$ are in the ratio:

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-$1$: $A$ point particle of mass $m$ moving with speed $u$ collides with a stationary point particle of mass $M$. If the maximum energy loss possible is given as $f \left( \frac{1}{2} m u^2 \right)$, then $f = \left( \frac{m}{M + m} \right)$.
Statement-$2$: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Equal force $F (> mg)$ is applied to the string in all the $3$ cases. Starting from rest,the point of application of force moves a distance of $2 \ m$ down in all cases. In which case does the block have maximum kinetic energy?

$A$ body of mass $2 \,kg$ is moving with a constant acceleration of $(2 \hat{i}+3 \hat{j}-\hat{k}) \,ms^{-2}$. If the displacement made by the body is $(3 \hat{i}-\hat{j}+2 \hat{k}) \,m$, then the work done is: (in $\,J$)

$A$ ball of mass $0.2 \ kg$ is at rest at a height of $5 \ m$. $A$ bullet of mass $0.01 \ kg$ moving horizontally with a velocity $V \ m/s$ strikes the center of the ball. After the collision,the ball and the bullet move independently. The ball hits the ground at a distance of $20 \ m$ and the bullet hits the ground at a distance of $100 \ m$ from the base of the pillar. What is the initial velocity $V$ of the bullet in $m/s$?