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

Work Power and Energy PYQ — JEE Main 2005 · Shift 0 · Q180 with Solution

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Correct answer: B

Interpret the motion

The ball starts from rest at the top of the first hill at height $100\,\text{m}$ above the ground.

It then:

rolls down to the ground,
climbs a second hill of height $30\,\text{m}$ ,
and finally comes down to a horizontal base at height $20\,\text{m}$ above the ground.

We need the speed of the ball at the final level $20\,\text{m}$ .

Use conservation of mechanical energy

Since the surface is smooth, no energy is lost. Hence,

\text{Initial mechanical energy} = \text{Final mechanical energy}

Initially, the ball is at rest, so:

E_i = mgh_i = mg(100)

Finally, at height $20\,\text{m}$ , the ball has potential energy plus kinetic energy:

E_f = mg(20) + \frac12 mv^2

Equating:

mg(100) = mg(20) + \frac12 mv^2

Cancel $m$ :

g(100-20) = \frac12 v^2

g(80) = \frac12 v^2

Taking $g = 10\,\text{m/s}^2$ ,

10 \times 80 = \frac12 v^2

800 = \frac12 v^2

v^2 = 1600

v = 40\,\text{m/s}

Why the intermediate hill of $30\,\text{m}$ does not matter

Mechanical energy depends only on initial and final heights (since no loss occurs). The ball can climb the $30\,\text{m}$ hill because it had enough energy after descending from $100\,\text{m}$ . The final speed depends only on the net drop from $100\,\text{m}$ to $20\,\text{m}$ .

Check options

A: $20\,\text{m/s}$
B: $40\,\text{m/s}$
C: $10\sqrt{30}\,\text{m/s}$
D: $10\,\text{m/s}$

So the correct option is B.

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View written solutionFree

Correct answer: B

Interpret the motion

The ball starts from rest at the top of the first hill at height $100\,\text{m}$ above the ground.

It then:

rolls down to the ground,
climbs a second hill of height $30\,\text{m}$ ,
and finally comes down to a horizontal base at height $20\,\text{m}$ above the ground.

We need the speed of the ball at the final level $20\,\text{m}$ .

Use conservation of mechanical energy

Since the surface is smooth, no energy is lost. Hence,

\text{Initial mechanical energy} = \text{Final mechanical energy}

Initially, the ball is at rest, so:

E_i = mgh_i = mg(100)

Finally, at height $20\,\text{m}$ , the ball has potential energy plus kinetic energy:

E_f = mg(20) + \frac12 mv^2

Equating:

mg(100) = mg(20) + \frac12 mv^2

Cancel $m$ :

g(100-20) = \frac12 v^2

g(80) = \frac12 v^2

Taking $g = 10\,\text{m/s}^2$ ,

10 \times 80 = \frac12 v^2

800 = \frac12 v^2

v^2 = 1600

v = 40\,\text{m/s}

Why the intermediate hill of $30\,\text{m}$ does not matter

Check options

A: $20\,\text{m/s}$
B: $40\,\text{m/s}$
C: $10\sqrt{30}\,\text{m/s}$
D: $10\,\text{m/s}$

So the correct option is B.

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