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√(30) m/s (d) 10 m/s

Answer: (b). Smooth surface: only the net height drop matters, v = √(2g Δ h). Start at 100 m, end on the base at 20 m: Δ h = 80 m (the 30 m hill is lower than 100 m, so it is crossed). v = √(2 × 10 × 80) = √(1600) = 40 m/s So the answer is (b) 40 m/s.

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Q37JEE Main 2005 · AIEEEEasy

Conservation of Energy

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

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Answer: (b)

Smooth surface: only the net height drop matters, $v = \sqrt{2g\,\Delta h}$ .
Start at 100 m, end on the base at 20 m: $\Delta h = 80\text{ m}$ (the 30 m hill is lower than 100 m, so it is crossed).
$v = \sqrt{2 \times 10 \times 80} = \sqrt{1600} = 40\text{ m/s}$
So the answer is (b) $40\text{ m/s}$ .

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