Question

A body slides down a smooth inclined plane having angle \( \theta \) and reaches the bottom with velocity \( v \). If a body is a sphere then its linear velocity at the bottom of the plane is

• A. \( \sqrt{\frac{2}{7}} v \)
• B. \( \sqrt{\frac{3}{7}} v \)
• C. \( \sqrt{\frac{5}{7}} v \)
• D. \( \sqrt{\frac{9}{7}} v \)

Hint:

In rolling motion, the total kinetic energy includes both translational and rotational kinetic energy. The fraction of energy going into rotation depends on the shape of the object.

Answer 1

The Correct Option is C

Step 1: Energy analysis.
When the body slides down the inclined plane, its potential energy is converted into kinetic energy. The total kinetic energy is divided into translational and rotational parts. The total energy at the bottom is: \[ mgh = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] For a sphere, \( I = \frac{2}{5} m r^2 \) and \( \omega = \frac{v}{r} \), so: \[ mgh = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m v^2 = \frac{7}{10} m v^2 \]
Step 2: Conclusion.
The velocity at the bottom of the plane is \( v_{\text{final}} = \sqrt{\frac{5}{7}} v \), so the correct answer is (C).