Question:
A solid sphere of mass $2\, kg$ rolls up a $ 30^{\circ} $ incline with an initial speed of $10 \,m/s$. The maximum height reached by the sphere is $(g = 10\,m/s^2)$
A solid sphere of mass $2\, kg$ rolls up a $ 30^{\circ} $ incline with an initial speed of $10 \,m/s$. The maximum height reached by the sphere is $(g = 10\,m/s^2)$
Updated On: Jun 14, 2022
- 3.5 m
- 7.0 m
- 10.5 m
- 14.0 m
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The Correct Option is B
Solution and Explanation
$m g h =\frac{1}{2} m v^{2}\left(1+\frac{K^{2}}{R^{2}}\right) $
$\Rightarrow \frac{7}{10} m v^{2} =m g h $
$\left( \because \frac{K^2}{R^2} = \frac{2}{5} \right)$
$\Rightarrow h =\frac{7}{10}\left(\frac{v^{2}}{g}\right)$
Given, $ v=10 \,m / s , g=10 \,m / s ^{2} . $
$\therefore h=\frac{7}{10} \times \frac{10 \times 10}{10} $
$\Rightarrow h=7 \,m$
$\Rightarrow \frac{7}{10} m v^{2} =m g h $
$\left( \because \frac{K^2}{R^2} = \frac{2}{5} \right)$
$\Rightarrow h =\frac{7}{10}\left(\frac{v^{2}}{g}\right)$
Given, $ v=10 \,m / s , g=10 \,m / s ^{2} . $
$\therefore h=\frac{7}{10} \times \frac{10 \times 10}{10} $
$\Rightarrow h=7 \,m$
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Concepts Used:
System of Particles and Rotational Motion
- The system of particles refers to the extended body which is considered a rigid body most of the time for simple or easy understanding. A rigid body is a body with a perfectly definite and unchangeable shape.
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- The few common examples of rotational motion are the motion of the blade of a windmill and periodic motion.