A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is
$mg\,\left( {h - d} \right) + \frac{1}{2}\,K{d^2}$
$mg\,\left( {h + d} \right) + \frac{1}{2}\,K{d^2}$
$mg\,\left( {h + d} \right) - \frac{1}{2}\,K{d^2}$
$mg\,\left( {h - d} \right) - \frac{1}{2}\,K{d^2}$
Two bodies of masses $m_1$ and $m_2$ are moving with same kinetic energy. If $P_1$ and $P_2$ are their respective momentum, the ratio $\frac{P_1}{P_2}$ is equal to
The force acting on a body moving along $x-$ axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at
In the non-relativistic regime, if the momentum, is increased by $100\%$, the percentage increase in kinetic energy is
Two identical particles are moving with same velocity $v$ as shown in figure. If the collision is completely inelastic then
A body of mass $2\,kg$ makes an elastic collision with another body at rest and continues to move in the original direction with one fourth of its original speed, The mass of the second body which collides with the first body is ............... $\mathrm{kg}$