Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted into

$A$ $2 \, kg$ 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?

$A$ block $C$ of mass $m$ is moving with velocity $v_0$ and collides elastically with block $A$ of mass $m$ which is connected to another block $B$ of mass $2m$ through a spring of spring constant $k$. What is $k$ if $x_0$ is the compression of the spring when the velocity of $A$ and $B$ is the same?

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the particle starts moving up the inclined face of the wedge and rises to a height $h$. The final velocity of the wedge $v_2$ is

$A$ 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:

$A$ $0.5 \, kg$ 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: