The velocity of the bullet becomes one-third after it penetrates $4 \ cm$ in a wooden block. Assuming that the bullet faces a constant resistance during its motion in the block,the bullet stops completely after traveling a total distance of $(4+x) \ cm$ inside the block. The value of $x$ is $.....$ (in $cm$)

$A$ slide with a frictionless curved surface,which becomes horizontal at its lower end,is fixed on the terrace of a building of height $3h$ from the ground,as shown in the figure. $A$ spherical ball of mass $m$ is released on the slide from rest at a height $h$ from the top of the terrace. The ball leaves the slide with a velocity $\vec{u}_0 = u_0 \hat{x}$ and falls on the ground at a distance $d$ from the building,making an angle $\theta$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_1$. The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $e = 1 / \sqrt{3}$. Which of the following statement$(s)$ is(are) correct?
$(A)$ $\vec{u}_0 = \sqrt{2gh} \hat{x}$
$(B)$ $\vec{v} = \sqrt{2gh} \hat{x} + \sqrt{2gh} \hat{z}$
$(C)$ $\theta = 60^{\circ}$
$(D)$ $d / h_1 = 2\sqrt{3}$

Consider two blocks $A$ and $B$ of masses $m_1 = 10 \ kg$ and $m_2 = 5 \ kg$ that are placed on a frictionless table. The block $A$ moves with a constant speed $v = 3 \ m/s$ towards the block $B$ kept at rest. $A$ spring with spring constant $k = 3000 \ N/m$ is attached to the block $B$ as shown in the figure. After the collision,suppose that the blocks $A$ and $B$,along with the spring in a constant compression state,move together. Then the compression in the spring is (neglect the mass of the spring). (in $m$)

$A$ spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 \,N \,m^{-1}$ and the mass of the block is $2.0 \,kg$. Ignore the mass of the spring. Initially, the spring is in an unstretched condition. Another block of mass $1.0 \,kg$ moving with a speed of $2.0 \,m \,s^{-1}$ collides elastically with the first block. The collision is such that the $2.0 \,kg$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is. . . .

$A$ 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 $x = -a$.

$A$ $2 \, kg$ block is placed on a frictionless inclined plane with an angle of $30^\circ$ and a length of $2 \, m$. After reaching the bottom of the incline,it travels on a surface with a coefficient of friction of $0.25$. What distance (in $m$) will it travel on the surface before coming to rest?