Roller Coaster: Energy Conservation

A roller coaster is the most exciting demonstration of conservation of energy. The cart starts at rest at the top of the highest hill, where it has lots of gravitational potential energy and zero kinetic energy. As it rolls down, PE converts to KE — and at the bottom of the first drop, all of the original PE has become KE, so the cart is moving at its maximum speed.

The Physics

In the absence of friction, mechanical energy is conserved:

$E_1 = E_2$

$\tfrac{1}{2}mv_1^{2} + mgh_1 = \tfrac{1}{2}mv_2^{2} + mgh_2$

If the cart starts from rest ($v_1 = 0$) at height $h_1$ and reaches the lowest point ($h_2 = 0$):

$mgh_1 = \tfrac{1}{2}mv_2^{2}$

Solving:

$v_2 = \sqrt{2g\,h_1}$

Notice the mass cancels — every cart, no matter how heavy, reaches the same speed at the bottom.

Step-by-Step Solution

Given: Initial height $h_1 = 20\;\text{m}$, $v_1 = 0$, $g = 9.81\;\text{m/s}^{2}$, frictionless.

Find: The speed $v_2$ at the bottom of the first drop.

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Step 1 — Set up the energy equation.

At the top: $E_1 = mgh_1$ (all potential, no kinetic).

At the bottom: $E_2 = \tfrac{1}{2}mv_2^{2}$ (all kinetic, no potential).

By conservation: $E_1 = E_2$, so $mgh_1 = \tfrac{1}{2}mv_2^{2}$.

Step 2 — Cancel the mass and solve for $v_2$.

$gh_1 = \tfrac{1}{2}v_2^{2}$

$v_2^{2} = 2gh_1 = 2(9.81)(20) = 392.4$

$v_2 = \sqrt{392.4} \approx 19.81\;\text{m/s}$

Step 3 — Convert to km/h for intuition.

$v_2 = 19.81 \times 3.6 \approx 71.31\;\text{km/h}$

That's about 44 mph — comparable to highway-driving speeds. From a 20-meter drop alone (about a 6-story building), you reach freeway speeds.

Step 4 — Find the speed at any intermediate height $h$.

By conservation $mgh_1 = \tfrac{1}{2}mv^{2} + mgh$:

$v(h) = \sqrt{2g(h_1 - h)}$

For example, halfway down ($h = 10$):

$v = \sqrt{2(9.81)(10)} \approx 14.01\;\text{m/s}$

That's about 70% of the maximum speed at the very bottom — not 50%, because speed depends on $\sqrt{h}$, not $h$.

Step 5 — Compare to a freely falling object.

A ball dropped from 20 m straight down would also reach $v = \sqrt{2gh} = 19.81\;\text{m/s}$ at the bottom. The roller coaster takes a longer path but reaches the same final speed — because energy depends only on height, not on the shape of the track. (Path-independence is a defining feature of conservative forces like gravity.)

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Answer: The roller coaster reaches a speed of

$\boxed{\;v_2 = \sqrt{2gh_1} = \sqrt{2(9.81)(20)} \approx 19.81\;\text{m/s}\;}$

at the bottom of the first 20-meter drop. Mass doesn't matter — every cart hits the same speed regardless of how heavy it is, as long as friction is negligible.

Try It

• Adjust the starting height — see the bottom speed grow as $\sqrt{h}$.
• Watch the energy bar graph — KE and PE trade off but their sum stays constant.
• The cart animation shows the actual speed scaling with the slope.