Question

A spherical ball of mass \( 20 \) kg is stationary at the top of a hill of height \( 100 \) m. It rolls down a smooth surface to the ground, then climbs up another hill of height \( 30 \) m and finally rolls down to a horizontal base at a height of \( 20 \) m above the ground. The velocity attained by the ball is:

• A. \( 20 \) m/s
• B. \( 40 \) m/s
• C. \( 10\sqrt{30} \) m/s
• D. \( 10 \) m/s

Hint:

For energy conservation in rolling motion, consider both translational and potential energy.

Answer 1

The Correct Option is B

Step 1: Apply conservation of energy. \[ m g h_{{initial}} = \frac{1}{2} m v^2 + mg h_{{final}} \] Cancel \( m \): \[ g h_{{initial}} = \frac{1}{2} v^2 + g h_{{final}} \] Substitute known values: \[ (10 \times 100) = \frac{1}{2} v^2 + (10 \times 20) \] \[ 1000 = \frac{1}{2} v^2 + 200 \] Isolate \( \frac{1}{2} v^2 \): \[ \frac{1}{2} v^2 = 800 \] Solve for \( v \): \[ v = \sqrt{1600} = 40 { m/s} \] The velocity is 40 m/s.