$A$ block moving horizontally on a smooth surface with a speed of $40\, m/s$ splits into two parts with masses in the ratio of $1:2$. If the smaller part moves at $60\, m/s$ in the same direction,then the fractional change in kinetic energy is:

$A$ ball falls freely from a height of $180 \,m$ onto a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is $0.5$, the average speed and average velocity of the ball before it ceases to rebound are respectively (acceleration due to gravity $= 10 \,ms^{-2}$)

Two balls,having linear momenta $\vec{p}_1 = p \hat{i}$ and $\vec{p}_2 = -p \hat{i}$,undergo a collision in free space. There is no external force acting on the balls. Let $\vec{p}_1^{\prime}$ and $\vec{p}_2^{\prime}$ be their final momenta. Which of the following option$(s)$ is (are) $NOT ALLOWED$ for any non-zero value of $p, a_1, a_2, b_1, b_2, c_1$ and $c_2$?
$(A)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_2 \hat{j}$
$(B)$ $\vec{p}_1^{\prime} = c_1 \hat{k}$,$\vec{p}_2^{\prime} = c_2 \hat{k}$
$(C)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_2 \hat{j} - c_1 \hat{k}$
$(D)$ $\vec{p}_1^{\prime} = a_1 \hat{i} + b_1 \hat{j}$,$\vec{p}_2^{\prime} = a_2 \hat{i} + b_1 \hat{j}$

$A$ ball of mass $m_1$ falls from height $h_1$ from rest to strike a spring of force constant $K$,which forces another ball of mass $m_2$ to jump on a horizontal floor at a height $h_2$ below it. Find the horizontal distance at which the ball of mass $m_2$ strikes the floor from the position of the start. [Assume the spring is fixed and the collision is elastic].

$A$ pendulum of length $1 \,m$ and having a bob of mass $1 \,g$ is pulled aside through an angle $60^{\circ}$ with the vertical and then released. The power delivered by all the forces acting on the bob when the pendulum makes $30^{\circ}$ with the vertical is . . . . . . $\left(g=10 \,ms^{-2}\right)$ (in $\,mW$)

Two bodies $A$ and $B$ of masses $20 \ kg$ and $5 \ kg$ respectively are at rest. Due to the action of a force of $40 \ N$ separately,if the two bodies acquire equal kinetic energies in times $t_A$ and $t_B$ respectively,then $t_A: t_B=$