Question

The kinetic energy of a particle of mass m₁ moving with a speed V is same as the kinetic energy of a solid sphere of mass m₂ rolling on the plane surface. If the speed of the Centre of the sphere is also V,then m₁/m₂ is:

• A. \(\frac{7}{10}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{5}{7}\)
• D. \(\frac{7}{5}\)
• E. \(\frac{2}{3}\)

Answer 1

The Correct Option is D

Given:

• Particle KE: \( \frac{1}{2} m_1 V^2 \).
• Rolling sphere KE: Translational + Rotational = \( \frac{1}{2} m_2 V^2 + \frac{1}{2} I \omega^2 \).
• For a solid sphere: \( I = \frac{2}{5} m_2 R^2 \) and \( \omega = \frac{V}{R} \).

Step 1: Equate Kinetic Energies

\[ \frac{1}{2} m_1 V^2 = \frac{1}{2} m_2 V^2 + \frac{1}{2} \left(\frac{2}{5} m_2 R^2\right) \left(\frac{V}{R}\right)^2 \]

Simplify:

\[ \frac{1}{2} m_1 V^2 = \frac{1}{2} m_2 V^2 + \frac{1}{5} m_2 V^2 \]

\[ m_1 V^2 = m_2 V^2 \left(1 + \frac{2}{5}\right) = \frac{7}{5} m_2 V^2 \]

\[ \frac{m_1}{m_2} = \frac{7}{5} \]

Conclusion:

The mass ratio is \(\frac{7}{5}\).

Answer 2

1. Write the kinetic energy of the particle:

The kinetic energy (KE) of a particle of mass \(m_1\) moving with speed \(V\) is given by:

\[KE_1 = \frac{1}{2}m_1V^2\]

2. Write the kinetic energy of the rolling sphere:

The kinetic energy of a rolling solid sphere of mass \(m_2\) and radius \(r\) with a center-of-mass speed \(V\) is the sum of its translational and rotational kinetic energies:

\[KE_2 = \frac{1}{2}m_2V^2 + \frac{1}{2}I\omega^2\]

where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

3. Substitute the moment of inertia and angular velocity:

For a solid sphere, \(I = \frac{2}{5}m_2r^2\). Since the sphere rolls without slipping, the angular velocity is related to the linear velocity by \(\omega = \frac{V}{r}\). Substituting these values:

\[KE_2 = \frac{1}{2}m_2V^2 + \frac{1}{2}(\frac{2}{5}m_2r^2)(\frac{V}{r})^2\]

\[KE_2 = \frac{1}{2}m_2V^2 + \frac{1}{5}m_2V^2 = \frac{7}{10}m_2V^2\]

4. Set the kinetic energies equal and solve for the mass ratio:

Given that \(KE_1 = KE_2\):

\[\frac{1}{2}m_1V^2 = \frac{7}{10}m_2V^2\]

Canceling out the common terms \(V^2\):

\[\frac{1}{2}m_1 = \frac{7}{10}m_2\]

\[\frac{m_1}{m_2} = \frac{7}{10} \times 2 = \frac{7}{5}\]