A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a s

Q: Does the 30 m hill affect the final velocity?

No, as long as the intermediate hill is lower than the initial height (100 m), the ball will cross it. Since gravity is a conservative force, the final velocity depends only on the initial and final heights, not the path taken.

Q: Why didn't we use the mass of 20 kg in the calculation?

If you write out the energy conservation equation, $mgH = mgh_2 + \frac{1}{2}mv^2$, you'll notice that mass $m$ is a common factor in every term. It cancels out completely! The final speed is independent of the object's mass.

Q: What if the surface wasn't smooth?

If the surface had friction, we could no longer use the conservation of mechanical energy directly. We would need to use the Work-Energy Theorem: $W_{gravity} + W_{friction} = \Delta K$. The final velocity would be lower due to the negative work done by friction.

Q: What would happen if the intermediate hill was 120 m high?

The ball starts with a total energy equivalent to a height of 100 m. It does not have enough energy to climb a 120 m hill. It would reach a maximum height of 100 m on that hill, momentarily stop, and then roll back down.

Q: Why is the initial kinetic energy zero?

The problem states that the ball is "stationary" at the top of the hill. This means its initial velocity is zero, and therefore its initial kinetic energy ($\frac{1}{2}mv^2$) is also zero.

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

Options:

(A) 40 m/s

(B) 20 m/s

(C) 10 m/s

(D) $10\sqrt{30}$ m/s

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the top of a massive $100	ext{ m}$ high hill, holding a $20	ext{ kg}$ spherical ball. You let it go. It rolls down into a valley, climbs up a smaller $30	ext{ m}$ hill, rolls back down, and finally settles onto a horizontal base that is $20	ext{ m}$ above the ground.

Our mission is to find the exact velocity of the ball when it reaches that final $20	ext{ m}$ base. At first glance, the winding path and the intermediate $30	ext{ m}$ hill might seem like they complicate things. But there is a magical phrase in the problem statement: **"smooth surface"**.

### The Master Equation

Because the surface is perfectly smooth, there is absolutely zero friction. This means no mechanical energy is lost to heat or sound. We can confidently unleash the **Law of Conservation of Mechanical Energy**.

This law tells us that the total energy at the beginning of the journey must equal the total energy at the end. The path taken in between? Completely irrelevant! As long as the intermediate hills aren't higher than our starting point (which would stop the ball in its tracks), we can ignore them.

Let's set up our equation. At the very top, the ball is stationary, so it has zero kinetic energy. All its energy is locked up as Gravitational Potential Energy:

$$E_i = mgH$$

When the ball reaches the final base at height $h_2$, it is moving with some velocity $v$. So, its energy is a mix of both potential and kinetic energy:

$$E_f = mgh_2 + \frac{1}{2}mv^2$$

Equating the two, we get our master equation:

$$mgH = mgh_2 + \frac{1}{2}mv^2$$

### Final Calculation

Now, look closely at that equation. Do you see the mass $m$ in every single term? It beautifully cancels out! This is a profound realization: **the final velocity of the ball is completely independent of its mass**. Whether it's a $20	ext{ kg}$ iron ball or a $2	ext{ g}$ marble, it will reach the exact same speed.

Let's rearrange the equation to solve for $v$:

$$gH - gh_2 = \frac{1}{2}v^2$$

$$v^2 = 2g(H - h_2)$$

$$v = \sqrt{2g(H - h_2)}$$

Now, we just plug in our numbers. We know $g = 10	ext{ m/s}^2$, the initial height $H = 100	ext{ m}$, and the final height $h_2 = 20	ext{ m}$:

$$v = \sqrt{2(10)(100 - 20)}$$

$$v = \sqrt{20 	imes 80}$$

$$v = \sqrt{1600}$$

$$v = 40	ext{ m/s}$$

And there we have it! The ball will be cruising at a brisk **$40	ext{ m/s}$** when it hits that final plateau. The elegance of conservative forces makes what looks like a complex roller-coaster problem into a straightforward, beautiful calculation.

Select Answer:

Visualized Solution

$H = 100 m, h_{1} = 30 m, h_{2} = 20 m$

$Conservation of Mechanical Energy$

$E_{i} = E_{f}$

$v = 2 g (H - h_{2})$

$v = 40 m/s$

$Path Independence$

The Sigma Insight: Conservation of Mechanical Energy

Analyzing the Setup

The Master Equation

Final Calculation

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