$A$ bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$,the initial velocity of the bullet will be

Three objects $A$,$B$,and $C$ are kept in a straight line on a frictionless horizontal surface. These have masses $m$,$2m$,and $m$,respectively. The object $A$ moves towards $B$ with a speed of $9 \ m/s$ and makes an elastic collision with it. Thereafter,$B$ makes a completely inelastic collision with $C$. All motions occur on the same straight line. Find the final speed (in $m/s$) of the object $C$.

$A$ motor drives a body along a straight line with a constant force. The power $P$ developed by the motor must vary with time $t$ according to

$A$ force of $10 \ N$ acting at an angle on a particle produces a displacement of $(3 \hat{i} - 4 \hat{\jmath}) \ m$. Due to this force,if the kinetic energy of the particle is decreased by $25 \ J$,then the angle between the force and the displacement is:

$A$ bullet of mass $0.01 \, kg$ travelling at a speed of $500 \, m/s$ strikes a block of mass $2 \, kg$ which is suspended by a string of length $5 \, m$. The center of gravity of the block is found to rise a vertical distance of $0.1 \, m$. What is the speed of the bullet after it emerges from the block?

$A$ particle of unit mass is moving along the $x$-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column $I$ ($a$ and $U_0$ are constants). Match the potential energies in column $I$ to the corresponding statement$(s)$ in column $II$.

Column $I$	Column $II$
$(A) U_1(x) = \frac{U_0}{2} \left[1 - \left(\frac{x}{a}\right)^2\right]^2$	$(P)$ The force acting on the particle is zero at $x = a$.
$(B) U_2(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2$	$(Q)$ The force acting on the particle is zero at $x = 0$.
$(C) U_3(x) = \frac{U_0}{2} \left(\frac{x}{a}\right)^2 \exp \left[-\left(\frac{x}{a}\right)^2\right]$	$(R)$ The force acting on the particle is zero at $x = -a$.
$(D) U_4(x) = \frac{U_0}{2} \left[\frac{x}{a} - \frac{1}{3}\left(\frac{x}{a}\right)^3\right]$	$(S)$ The particle experiences an attractive force towards $x = 0$ in the region $|x| < a$.
	$(T)$ The particle with total energy $\frac{U_0}{4}$ can oscillate about the point $x = -a$.