A particle of mass $m$ is fixed to one end of a light rigid rod of length $L$ and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle $\mathrm{\theta }$ with the vertical:

The elastic collision between two bodies, $A$ and $B$, can be considered using the following model. $A$ and $B$ are free to move along a common line without friction. When their distance is greater than $d=1 \mathrm{m}$, the interacting force is zero; when their distance is less than $d$, a constant repulsive force $F=6 \mathrm{N}$ is present. The mass of body $A$ is $m_A=1\mathrm{kg}$ and it is initially at rest; the mass of body $B$ is $m_B=3\mathrm{kg}$ and it is approaching body $A$ head-on with a speed $v_0=2 \mathrm{m} {\mathrm{s}}^{-1}$. Find the minimum distance between $A$ and $B$:-

Two particles of mass $\mathrm{m}$, constrained to move along the circumference of a smooth circular hoop of equal mass $\mathrm{m}$, are initially located at opposite ends of a diameter and given equal velocities $v_0$ shown in the figure. The entire arrangement is located in gravity free space. Their velocity just before collision is:-