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: {Use energy conservation} 
\[ m g h_{{initial}} = \frac{1}{2} m v^2 + mg h_{{final}} \] Cancel \( m \): \[ g h_{{initial}} = \frac{1}{2} v^2 + g h_{{final}} \] \[ (10 \times 100) = \frac{1}{2} v^2 + (10 \times 20) \] \[ 1000 = \frac{1}{2} v^2 + 200 \] \[ \frac{1}{2} v^2 = 800 \] \[ v = \sqrt{1600} = 40 { m/s} \] Thus, the correct answer is 40 m/s. 
 

Answer 2

Step 1: Given data:
- Mass \( m = 20 \) kg (not required for energy conservation as it cancels out)
- Initial height \( h_i = 100 \) m
- Final height \( h_f = 20 \) m (horizontal base is 20 m above the ground)

Step 2: Use conservation of mechanical energy:
Total mechanical energy at the top = Total mechanical energy at final position
Potential energy lost = Kinetic energy gained

Step 3: Potential energy at top: \( PE_i = mgh_i \)
Potential energy at final point: \( PE_f = mgh_f \)
Kinetic energy gained: \( KE = PE_i - PE_f = mg(h_i - h_f) \)

Step 4: Use \( KE = \frac{1}{2}mv^2 \), so:
\( \frac{1}{2}mv^2 = mg(h_i - h_f) \)
Cancel \( m \) on both sides:
\( \frac{1}{2}v^2 = g(h_i - h_f) \Rightarrow \frac{1}{2}v^2 = 9.8(100 - 20) = 784 \)
So, \( v^2 = 1568 \Rightarrow v = \sqrt{1568} \approx 39.6 \approx 40 \) m/s

Final Answer: \( 40 \) m/s