JEE Main · 2005 · 08 May 2005 · Offline · Work, Power & Energy

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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) $40 m / s$ ✓ Correct
(B) $20 m / s$
(C) $10 m / s$
(D) $1030 m / s$

Answer: (A)

Solution

On a smooth track only the net change in height matters, because the intermediate hill is climbed and descended without loss. Equating the potential energy lost between the start at

100

m and the final level at

20

m to the kinetic energy gained,

m g (100 - 20) = \frac{1}{2} m v^{2}

, so

v = 2 \times 10 \times 80 = 1600 = 40

m/s.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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