A hollow smooth uniform sphere $A$ of mass $\mathrm{m}$ rolls without sliding on a smooth horizontal surface…
- $1: 1$
- $2: 3$
- $3: 2$
- None of these
Solution
No horizontal force on the system, so linear momentum will be conserved.
Here's the corrected text with proper LaTeX formatting:
$\begin{aligned}
& P_{i}=P_{f} \\
& m V_{0}+0=m V_{1}+m V_{2} \\
& \Rightarrow V_{0}=V_{1}+V_{2} \quad \ldots(1)
\end{aligned}$
$\begin{aligned}
& e=\frac{\text { velocity of separation }}{\text { velocity of approach }} \frac{v_{2}-v_{1}}{\mu_{1}-u_{2}} \\
& =\frac{V_{2}-V_{1}}{V_{0}-0}=1
\end{aligned}$
For elastic collision $e=1$
$\Rightarrow V_{2}-V_{1}=V_{0}$
From Eq. (i) and (ii),
$V_{2}=V_{0} ~\&~ V_{1}=0$
Since there is no torque acting on either sphere during collision, their angular velocities about respective centres remains the same.
i.e $\omega_{A}=\omega ~\&~ \omega_{B}=0$ Thus, Kinetic energy of $A$ after collision $=\frac{1}{2} m V_{1}^{2}+\frac{1}{2} \omega^{2}$
Here, \(\mathrm{V}_1=\mathrm{O}\)
\((\mathrm{KE})_A=\frac{1}{2} \times\left(\frac{2}{3} \mathrm{mR}^2\right) \times\left(\frac{\mathrm{V}_{\mathrm{g}}^2}{\mathrm{R}}=\frac{\mathrm{mV}{ }_0^2}{3}\right)\)
Kinetic energy of \(B\) after collision \((K E)_B=\frac{1}{2} m V_2^2=\frac{1}{2} \mathrm{mV}_0^2\)
\(\Rightarrow \frac{(\mathrm{KE})_{\mathrm{B}}}{(\mathrm{KE})_{\mathrm{A}}}=\frac{\frac{1}{2} \mathrm{mV} \mathrm{~V}_{\circ}^2}{\frac{\mathrm{~m} V_0^2}{3}}=\frac{3}{2}\)
Asked in: JEE Mains - Rotational Motion - Test 4