Question:
A block of mass m = 0.1 kg is released from a height of 4 m on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC is rough with a coefficient of friction, $\mu = 0.1$. If the impact of the block with the vertical wall at C is perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be:
A block of mass m = 0.1 kg is released from a height of 4 m on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC is rough with a coefficient of friction, $\mu = 0.1$. If the impact of the block with the vertical wall at C is perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be:
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Perfect elastic collision means velocity magnitude is conserved, only direction changes.
Updated On: Jun 6, 2026
- 29 m
- 59 m
- 60 m
- 90 m
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The Correct Option is B
Solution and Explanation
Step 1: Energy idea.
The block is released from height $h = 4$ m on a smooth curve, so it reaches the bottom with kinetic energy $mgh$. Only the rough path BC removes energy.
Step 2: Initial energy.
With $g = 10\ \text{m s}^{-2}$, $E = mgh = 0.1 \times 10 \times 4 = 4$ J.
Step 3: Friction and rough distance.
Friction force $f = \mu m g = 0.1 \times 0.1 \times 10 = 0.1$ N. Total length on the rough part $d_{rough} = E/f = 4/0.1 = 40$ m.
Step 4: Include the smooth path.
With $AB = 1$ m and $BC = 2$ m, the block makes 10 trips to the elastic wall and stops at B; the smooth path adds $1 + 9 \times 2 = 19$ m.
Step 5: Result.
Total distance $= 40 + 19 = 59$ m. \[ \boxed{59\ \text{m}} \]
The block is released from height $h = 4$ m on a smooth curve, so it reaches the bottom with kinetic energy $mgh$. Only the rough path BC removes energy.
Step 2: Initial energy.
With $g = 10\ \text{m s}^{-2}$, $E = mgh = 0.1 \times 10 \times 4 = 4$ J.
Step 3: Friction and rough distance.
Friction force $f = \mu m g = 0.1 \times 0.1 \times 10 = 0.1$ N. Total length on the rough part $d_{rough} = E/f = 4/0.1 = 40$ m.
Step 4: Include the smooth path.
With $AB = 1$ m and $BC = 2$ m, the block makes 10 trips to the elastic wall and stops at B; the smooth path adds $1 + 9 \times 2 = 19$ m.
Step 5: Result.
Total distance $= 40 + 19 = 59$ m. \[ \boxed{59\ \text{m}} \]
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