$A$ spherical bowl of radius $R = 2 \ m$ is placed on a horizontal plane. $A$ particle of mass $m = 1 \ g$ oscillates on its inner surface. If the particle starts from a point on the bowl at a height of $h = 1 \ cm$ from the horizontal plane and the coefficient of friction is $\mu = 0.01$,find the total distance $s$ traveled by the particle before it comes to rest at the bottom of the bowl. (in $m$)

$A$ particle of mass $m$ is projected with a speed $u$ from the ground at an angle $\theta = \frac{\pi}{3}$ w.r.t. horizontal ($x$-axis). When it has reached its maximum height,it collides completely inelastically with another particle of the same mass and velocity $u \hat{i}$. The horizontal distance covered by the combined mass before reaching the ground is

$A$ block of mass $m$ starts at rest at height $h$ on a frictionless inclined plane. The block slides down the plane,travels across a rough horizontal surface with coefficient of kinetic friction $\mu$,and compresses a spring with force constant $k$ a distance $x$ before momentarily coming to rest. Then the spring extends and the block travels back across the rough surface,sliding up the plane. The block travels a total distance $d$ on the rough horizontal surface. The correct expression for the maximum height $h'$ that the block reaches on its return is

Consider the following two statements: $[A]$ The linear momentum of a system is zero. $[B]$ The kinetic energy of the particles of the system is zero.

$A$ $0.5 \, kg$ block moving at a speed of $12 \, ms^{-1}$ compresses a spring through a distance of $30 \, cm$ when its speed is halved. The spring constant of the spring in $N m^{-1}$ is:

The linear momentum of a body of mass $8 \ kg$ is $24 \ kg \ m \ s^{-1}$. If a constant force of $24 \ N$ acts on the body in the direction of motion of the body for a time of $3 \ s$,then the increase in the kinetic energy of the body is (in $J$)