Two bodies A and B of masses 2 m and m are projected vertically upwards from the ground with velocities $u$…
Two bodies A and B of masses 2 m and m are projected vertically upwards from the ground with velocities $u$ and 2 u respectively. The ratio of the kinetic energy of body A and the potential energy of body B at a height equal to half of the maximum height reached by body A is
$8: 1$
$1: 1$
$4: 1$
$2: 1$
Solution
For body A,
$\mathrm{H}_{\max }=\frac{\mathrm{u}^2}{2 \mathrm{~g}}, \mathrm{~m}_1=2 \mathrm{~m}, \mathrm{u}_1=\mathrm{u}, \mathrm{H}_1=\frac{\mathrm{H}_{\max }}{2}=\frac{\mathrm{u}^2}{4 \mathrm{~g}}$
By principle of energy conservation,
$\frac{1}{2} m_1 u_1^2=m_1 \mathrm{gH}_1+(\mathrm{KE})_{\mathrm{A}}$
$\therefore(\mathrm{KE})_{\mathrm{A}}=\frac{1}{2}(2 \mathrm{~m}) \mathrm{u}^2-(2 \mathrm{~m}) \mathrm{g}\left(\frac{\mathrm{u}^2}{4 \mathrm{~g}}\right)=\frac{1}{2} \mathrm{mu}^2$
For body $\mathrm{B}, \mathrm{H}_2=\frac{\mathrm{u}^2}{4 \mathrm{~g}}, \mathrm{~m}_2=\mathrm{m}, \mathrm{u}_2=2 \mathrm{u}$
$(\mathrm{P.E})_{\mathrm{B}}=\mathrm{m}_2 \mathrm{gH}_2=\mathrm{mg}\left(\frac{\mathrm{u}^2}{4 \mathrm{~g}}\right)=\frac{1}{4} \mathrm{mu}^2$
$\therefore \frac{(\mathrm{KE})_{\mathrm{A}}}{(\mathrm{PE})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{mu}^2}{\frac{1}{4} \mathrm{mu}^2}=2: 1$