A moving body with mass ' $\mathrm{m}_1$ ' strikes a stationary mass ' $\mathrm{m}_2$ '. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of first by (1.5) times the velocity after the collision?

A metal rod of weight ' $W$ ' is supported by two parallel knife-edges A and B . The rod is in equilibrium in horizontal position. The distance ' between two knife-edges is ' $r$ '. The centre of mass of the rod is at a distance ' $x$ ' from $A$. The normal reaction on A is

In the system of two particles of masses ' $\mathrm{m}_1$ ' and ' $\mathrm{m}_2$ ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

In projectile motion two particles of masses $\mathrm{m}_1$ and $m_2$ have velocities $\vec{V}_1$, and $\vec{V}_2$ respectively at time $t=0$. Their velocities become $\overline{V_1^{\prime}}$ and $\overrightarrow{V_2^{\prime}}$ at time 2 t while still moving in air. The value of $\left[\left(m_1 \overrightarrow{V_1^{\prime}}+m_2 \overrightarrow{V_2^{\prime}}\right)-\left(m_1 \vec{V}_1+m_2 \vec{V}_2\right)\right]$ is ( $\mathrm{g}=$ acceleration due to gravity)