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6.System of Particles and Rotational Motion
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The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$; then the ratio of their masses will be
A
$2:1$
B
$1:8$
C
$1:7$
D
$2\sqrt 2 :1$
Solution
$K=K_{t r}+K_{r o t}=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$$=\frac{1}{2} m v^{2}+\frac{1}{2} \times\left(\frac{2}{5} m r^{2}\right) \times\left(\frac{v}{r}\right)^{2}=\frac{7}{10} m v^{2}$
$\frac{K_{1}}{K_{2}}=\frac{m_{1}}{m_{2}} \times\left(\frac{v_{1}}{v_{2}}\right)^{2}$
Given, $v_{1}=v_{2}$
$\Rightarrow \frac{K_{1}}{K_{2}}=\frac{m_{1}}{m_{2}}$
$\Rightarrow \frac{2}{1}=\frac{m_{1}^{m_{1}}}{m_{2}}$
Thus if the ratio of kinetic energy is $2:1$ then their ratio of masses will also be $2:1$
Standard 11
Physics
