Roller Coaster Worksheet Practice - Free Printable

Let’s solve this step by step. We’ll use conservation of energy and circular motion formulas. The roller coaster starts at 80 m height with speed 10 m/s. We assume no friction or air resistance, so mechanical energy is conserved.

We’ll use:
- Gravitational potential energy: PE = mgh
- Kinetic energy: KE = (1/2)mv²
- Centripetal acceleration: a_c = v² / r
- “g’s” felt = (normal force)/mg → for bottom of loop: (v²/r + g)/g; for top: (v²/r - g)/g (if upside down)

Mass = 800 kg — but it will cancel out in most cases, so we can ignore it unless needed.

---

Step 1: Speed at bottom of hill (height = 0 m)

Initial total energy at start (80 m high, 10 m/s):

E_initial = KE_initial + PE_initial
= (1/2)(800)(10)² + (800)(9.8)(80)
= (1/2)(800)(100) + (800)(784)
= 40,000 + 627,200 = 667,200 J

At bottom (h=0), all energy is kinetic:

(1/2)(800)v_bottom² = 667,200
→ 400 v_bottom² = 667,200
→ v_bottom² = 667,200 / 400 = 1668
→ v_bottom = √1668 ≈ 40.84 m/s

✔ Answer to Q1: ~40.8 m/s

---

Step 2: Speed as it enters the loop

The loop entrance is at ground level? Wait — looking at diagram: the loop is drawn such that its bottom is at ground level, and the car enters the loop at the bottom. So entering the loop = same as bottom of hill? But wait — the diagram shows the loop has radius 10 m, and the track goes up to 50 m before looping? Actually, let’s read carefully.

Wait — the diagram says: “The roller coaster car enters the loop here.” And there’s an arrow pointing to the bottom of the loop, which is at ground level. Also, the first drop is from 80 m to ground (0 m). Then it goes into the loop. So yes — entering the loop is at the bottom, same point as Q1.

But wait — question 2 says “as it enters the loop”. If the loop starts at ground level, then it’s the same as Q1. But maybe not? Let me check heights again.

Actually, looking at the diagram description:

- Start: 80 m high, 10 m/s
- Drops to bottom (0 m) → that’s where it enters the loop? The arrow says “enters the loop here” at the bottom.
- Then loop has radius 10 m → so top of loop is at 20 m above ground? But diagram also labels “50 m” and “40 m” — perhaps those are other points?

Wait — re-examining the image description:

There’s a label “50 m” on the left side of the loop — probably meaning the height of the track just before the loop? No — actually, the vertical line labeled “50 m” seems to be from ground to some point before the loop? That doesn’t make sense.

Alternatively, perhaps the “50 m” is the height of the hill after the loop? No — let’s look at the numbers given:

- Start: 80 m
- After first drop: bottom at 0 m → enters loop
- Loop radius 10 m → so top of loop is at 20 m
- Then it goes up to a hill labeled “40 m” — but 40 m from what? The dotted line is ground, so 40 m high?
- Then another hill at 35 m.

But the problem says: “How fast is the roller coaster traveling as it enters the loop?” — if it enters at the bottom (0 m), then it’s same as Q1.

But that would make Q1 and Q2 identical — unlikely.

Wait — perhaps “enters the loop” means at the beginning of the curved part, which might be slightly above ground? But typically, loops are entered at the bottom.

Another possibility: the “50 m” label might indicate the height of the track at the point just before entering the loop? But that contradicts the 80 m drop to ground.

I think there’s confusion in interpreting the diagram. Let me try to reconstruct based on standard problems.

Typically in such diagrams:

- Car starts at 80 m with 10 m/s.
- Drops to ground level (0 m) — that’s the bottom of the first hill.
- Then immediately enters a vertical loop of radius 10 m — so the bottom of the loop is at 0 m, top at 20 m.
- Then it goes up to a hill that is 40 m high? But 40 m is higher than the loop top — possible if it has enough speed.
- Then another hill at 35 m.

But the diagram has a label “50 m” — perhaps that’s a typo or mislabel? Or maybe it’s the height of the track at the entrance to the loop? That doesn’t make sense.

Wait — looking back at user’s image description: “50 m” is written next to a vertical line that seems to be from ground to a point before the loop? Perhaps it’s indicating that the track rises to 50 m before dropping into the loop? But that contradicts the 80 m start.

I think I need to rely on the text and standard interpretation.

Perhaps the “50 m” is irrelevant or a distractor. Let’s see the questions:

Q1: bottom of the hill — clearly 0 m.

Q2: enters the loop — likely also at 0 m, since loop starts at bottom.

But then Q1 and Q2 are same. Unless “bottom of the hill” refers to the very bottom after 80m drop, and “enters the loop” is after some flat section? But still same height.

Alternatively, maybe the loop is not at ground level? The diagram shows the loop with center at some height? No — usually loops are symmetric.

Another idea: perhaps the “50 m” is the height of the track at the point where it begins to curve into the loop? But that would mean the car hasn't reached ground yet.

Let’s calculate assuming that “enters the loop” is at height 0 m — same as bottom. Then Q2 = Q1.

But let’s check Q3: centripetal acceleration at bottom of loop — which requires speed at bottom, so same.

Then Q4: g’s at bottom — also same speed.

Then Q5: speed at top of loop — which is at height 2r = 20 m.

So let’s proceed with that assumption.

So for Q2: speed as it enters the loop = speed at bottom = 40.84 m/s

But that seems redundant. Perhaps “enters the loop” means at the base, but the loop has a certain shape — but still, height is 0.

Maybe the diagram intends that the car drops to 0 m, then travels horizontally, then enters the loop — but still same height.

I think we have to go with it.

So Q2: same as Q1: 40.84 m/s

But let’s double-check with energy.

From start to bottom (0 m):

ΔPE = mg(80) = 800*9.8*80 = 627,200 J

Initial KE = 0.5*800*100 = 40,000 J

Total E = 667,200 J

At bottom, KE = 667,200 J → v = sqrt(2*667200/800) = sqrt(1668) = 40.84 m/s — correct.

Now, for entering the loop — if it's at the same point, same speed.

But perhaps "enters the loop" means at the point where the track starts curving upward, which might be at a small height? But typically, it's considered at the bottom.

To resolve, let's look at Q5: speed at top of loop. Top of loop is at height 2*r = 2*10 = 20 m.

So from start to top of loop:

Initial E = 667,200 J

At top of loop, h = 20 m, so PE = mgh = 800*9.8*20 = 156,800 J

KE = E_total - PE = 667,200 - 156,800 = 510,400 J

Then (1/2)*800*v_top^2 = 510,400

400 v_top^2 = 510,400

v_top^2 = 1276

v_top = sqrt(1276) ≈ 35.72 m/s

That seems reasonable.

Now, for Q2, if "enters the loop" is at the bottom, then it's 40.84 m/s.

But let's see if there's another interpretation. The diagram has a label "50 m" — perhaps that's the height of the track at the entrance to the loop? But that would mean the car didn't drop to 0 m.

Suppose the car drops from 80 m to a point at 50 m height? But the diagram shows a drop to ground.

I think I should trust the standard interpretation: bottom of hill is 0 m, loop starts at 0 m.

So Q2: 40.84 m/s

But to confirm, let's see Q3: centripetal acceleration at bottom of loop.

a_c = v^2 / r = (40.84)^2 / 10 = 1668 / 10 = 166.8 m/s²

Which is about 17g — very high, but possible for roller coasters? Maybe.

Q4: g's felt at bottom = (a_c + g)/g = (166.8 + 9.8)/9.8 = 176.6/9.8 ≈ 18.02 g's — extremely high, unrealistic, but mathematically correct based on numbers.

Perhaps the mass is not needed, and we can work with velocities only.

Another thought: maybe the "800 kg" is a red herring, and we can use energy per unit mass.

Let me redefine using specific energy (per kg).

Initial specific energy: e_initial = (1/2)v_i^2 + g h_i = 0.5*100 + 9.8*80 = 50 + 784 = 834 J/kg

At any point, e = 0.5 v^2 + g h = 834

So for bottom, h=0: 0.5 v^2 = 834 → v^2 = 1668 → v=40.84 m/s — same.

For top of loop, h=20 m: 0.5 v^2 + 9.8*20 = 834 → 0.5 v^2 + 196 = 834 → 0.5 v^2 = 638 → v^2 = 1276 → v=35.72 m/s — same as before.

Now, for Q2, if "enters the loop" is at h=0, then v=40.84 m/s.

But perhaps "enters the loop" means at the point where the loop begins, which might be at a different height. Looking at the diagram description, there's a "50 m" label — maybe that's the height at which the car enters the loop? But that doesn't make sense because it dropped from 80 m to 0 m.

Unless the 50 m is a mistake, or it's the height of the loop's top or something.

Another idea: perhaps the "50 m" is the height of the track after the loop or before. Let's read the diagram again as described:

"80 m" at start, "10 m/s", then drop to bottom, then "50 m" is written on the left side of the loop — perhaps it's the height of the hill that the car climbs after the loop? But then it says "40 m" for another hill.

The text says: "50 m" and "40 m" are labeled, and "35 m" for the last hill.

Perhaps the car, after the first drop to 0 m, goes up a hill to 50 m? But that would require more energy than available.

Initial energy corresponds to max height if v=0: h_max = e_initial / g = 834 / 9.8 ≈ 85.1 m — so it can reach 85.1 m, so 50 m is possible.

But the diagram shows the first drop to ground, then a loop, then a hill at 40 m, then 35 m.

I think I found the issue: in many such diagrams, the "50 m" might be the height of the track at the entrance to the loop, but that would mean the car didn't drop to 0 m.

Let's assume that the bottom of the first hill is not at 0 m, but the diagram has a dotted line for reference.

The diagram has a dotted line at the bottom, labeled with arrows to 80 m, 50 m, etc.

Specifically:

- From dotted line to start: 80 m
- From dotted line to a point before the loop: 50 m? But that doesn't make sense.

Perhaps the 50 m is the height of the loop's top or something.

Let's look at the loop: it has radius 10 m, and there's a label "10 m" inside the loop, and "40 m" next to it.

Another interpretation: perhaps the "40 m" is the height of the top of the loop? But radius is 10 m, so if bottom is at 0, top is at 20 m, not 40.

Unless the loop is not at ground level.

Suppose the bottom of the loop is at height h_b, then top is at h_b + 20 m.

But the diagram shows the loop with the bottom at the dotted line, so h_b = 0.

I think I have to proceed with the initial assumption.

Perhaps "enters the loop" means at the base, and "bottom of the hill" is the same point.

So for Q2, answer is same as Q1.

But let's see the questions order:

1. bottom of the hill — 0 m

2. enters the loop — likely same point

3. centripetal acceleration at bottom of loop — same point

4. g's at bottom — same

5. speed at top of loop — h=20 m

6. centripetal acceleration at top of loop

7. g's at top of loop

8. speed at 35 m hill

So perhaps it's intentional that Q1 and Q2 are the same.

Maybe "bottom of the hill" refers to the lowest point, which is 0 m, and "enters the loop" is when it starts the loop curve, which is also at 0 m.

So I'll go with that.

So Q2: 40.84 m/s

But to be precise, let's keep more digits or use exact values.

Let me calculate exactly.

g = 9.8 m/s² (assume)

Initial energy per kg: e = 0.5*10^2 + 9.8*80 = 50 + 784 = 834 J/kg

At bottom, h=0: 0.5 v^2 = 834 => v^2 = 1668 => v = sqrt(1668) = 2*sqrt(417) ≈ 40.841 m/s

For Q2, same.

Q3: centripetal acceleration at bottom: a_c = v^2 / r = 1668 / 10 = 166.8 m/s²

Q4: g's felt at bottom. At the bottom of the loop, the normal force N provides centripetal force and counteracts gravity.

So N - mg = m v^2 / r

N = m(g + v^2/r)

So g's felt = N/(mg) = 1 + (v^2/(r g)) = 1 + 1668/(10*9.8) = 1 + 1668/98 = 1 + 17.0204 = 18.0204 g's

Q5: speed at top of loop. h = 2*10 = 20 m

e = 0.5 v^2 + g*20 = 834

0.5 v^2 + 196 = 834

0.5 v^2 = 638

v^2 = 1276

v = sqrt(1276) = 2*sqrt(319) ≈ 35.721 m/s

Q6: centripetal acceleration at top of loop: a_c = v^2 / r = 1276 / 10 = 127.6 m/s²

Q7: g's felt at top of loop. At the top, both gravity and normal force provide centripetal force downward.

So mg + N = m v^2 / r

N = m(v^2/r - g)

g's felt = N/(mg) = (v^2/(r g) - 1) = 1276/(10*9.8) - 1 = 1276/98 - 1 = 13.0204 - 1 = 12.0204 g's

But is this correct? "g's felt" usually means the normal force divided by mg, which is what we have. At the top, if N is positive, the rider feels pressed into the seat; if negative, they would fall out, but here it's positive.

Q8: speed at 35 m hill.

h = 35 m

e = 0.5 v^2 + g*35 = 834

0.5 v^2 + 343 = 834 (since 9.8*35=343)

0.5 v^2 = 834 - 343 = 491

v^2 = 982

v = sqrt(982) ≈ 31.337 m/s

Now, let's write answers with appropriate significant figures. Inputs have 2 or 3 sig figs: 800 kg (1 or 3?), 10 m/s (2), 80 m (2), 10 m radius (2), etc. So probably 2 or 3 sig figs.

Let's round to 3 sig figs where possible.

Q1: 40.8 m/s

Q2: 40.8 m/s

Q3: 167 m/s² (since 166.8 rounds to 167)

Q4: 18.0 g's (18.02 rounds to 18.0)

Q5: 35.7 m/s

Q6: 128 m/s² (127.6 rounds to 128)

Q7: 12.0 g's (12.02 rounds to 12.0)

Q8: 31.3 m/s

But for Q4 and Q7, g's are dimensionless, so 18.0 and 12.0.

Now, I recall that in some contexts, "g's felt" might be defined differently, but I think this is standard.

For Q7, at the top, if the car is upside down, the rider feels the normal force from the seat, which is downward, so "felt" might be the magnitude, but usually it's the force exerted by the seat on the rider divided by mg, which is what we have.

In this case, N = m(v^2/r - g) >0, so it's fine.

Now, to box the answers.

But let's verify Q8: 35 m hill. Is that from ground? Yes, diagram shows 35 m from dotted line.

And we have e=834 J/kg, h=35, g=9.8, so PE=9.8*35=343 J/kg, KE=834-343=491 J/kg, v=sqrt(2*491)=sqrt(982)≈31.34 m/s.

All seem consistent.

So final answers:

1. 40.8 m/s

2. 40.8 m/s

3. 167 m/s²

4. 18.0 g's

5. 35.7 m/s

6. 128 m/s²

7. 12.0 g's

8. 31.3 m/s

But for Q2, is it really the same as Q1? Perhaps "enters the loop" means at a different point.

Another interpretation: maybe the loop is not at the bottom. Looking back at the diagram description: "The roller coaster car enters the loop here." with an arrow to a point that might be at the start of the loop curve, which could be at a height.

In many diagrams, the loop is entered at the bottom, but perhaps here the "50 m" is relevant.

Suppose that the car drops from 80 m to a point at 50 m height? But the diagram shows a drop to ground.

Perhaps the 50 m is the height of the track at the entrance to the loop, and the bottom of the hill is at 0 m, but the loop is entered after rising to 50 m? That doesn't make sense.

Let's calculate the speed at 50 m height.

If h=50 m, then e = 0.5 v^2 + 9.8*50 = 834

0.5 v^2 + 490 = 834

0.5 v^2 = 344

v^2 = 688

v = sqrt(688) ≈ 26.23 m/s

Then if "enters the loop" is at 50 m, then Q2 = 26.2 m/s

But then what is "bottom of the hill"? Probably 0 m.

And the loop might be at 50 m? But the loop has radius 10 m, so if entered at 50 m, the top would be at 70 m, but initial energy allows only 85.1 m, so possible, but the diagram shows the loop near the bottom.

Moreover, the label "50 m" is on the left side, perhaps indicating the height of the hill before the loop or something.

I think there's ambiguity, but in standard problems, the loop is at the bottom.

Perhaps the "50 m" is a distractor or for another purpose.

Another idea: perhaps the 50 m is the height of the center of the loop or something.

Let's look at the diagram description: "50 m" is written next to a vertical line that seems to be from ground to a point, and "40 m" is next to the loop.

Perhaps the top of the loop is at 40 m? But radius is 10 m, so if top is at 40 m, bottom is at 20 m.

That could be!

Let me try that interpretation.

Suppose the bottom of the loop is at 20 m height, top at 40 m height (since radius 10 m, diameter 20 m).

Then "enters the loop" at bottom of loop, which is at 20 m.

"Bottom of the hill" — the first drop is from 80 m to 0 m? But then how does it get to 20 m to enter the loop?

Perhaps the "bottom of the hill" is at 0 m, then it goes up to 20 m to enter the loop.

That makes sense.

Let me check the diagram description: "80 m" at start, then drop to bottom (0 m), then it goes up to a point labeled "50 m"? No, "50 m" is on the left, perhaps for the loop area.

The text says: "50 m" and "40 m" are labeled, and "10 m" for radius.

Perhaps the 40 m is the height of the top of the loop.

Assume that the top of the loop is at 40 m above ground.

Since radius is 10 m, the bottom of the loop is at 40 - 20 = 20 m above ground.

Then "enters the loop" at the bottom of the loop, so at 20 m height.

"Bottom of the hill" — the first drop is to 0 m, so at 0 m.

Then after reaching 0 m, it goes up to 20 m to enter the loop.

That makes sense with the labels.

Also, after the loop, it goes to a hill at 40 m? But if top of loop is at 40 m, and it's a loop, after exiting, it might go to a lower hill.

The diagram has "40 m" next to the loop, and "35 m" for the last hill.

So let's adopt this interpretation.

So:

- Start: h=80 m, v=10 m/s

- Bottom of first hill: h=0 m

- Enters loop at h=20 m (bottom of loop)

- Top of loop: h=40 m

- Then goes to a hill at 40 m? Or perhaps the 40 m is the top of the loop, and the next hill is at 35 m.

The diagram says "40 m" and then "35 m", so probably after the loop, it goes to a 35 m hill.

But if top of loop is at 40 m, and it exits, it might have less speed, so it can climb to 35 m.

Yes.

So let's recalculate with this.

Initial specific energy: e = 0.5*10^2 + 9.8*80 = 50 + 784 = 834 J/kg

Q1: bottom of hill, h=0 m

0.5 v^2 = 834 => v^2 = 1668 => v = sqrt(1668) = 40.841 m/s ≈ 40.8 m/s

Q2: enters the loop at h=20 m (bottom of loop)

e = 0.5 v^2 + 9.8*20 = 834

0.5 v^2 + 196 = 834

0.5 v^2 = 638

v^2 = 1276

v = sqrt(1276) = 35.721 m/s ≈ 35.7 m/s

This is different from Q1, which makes sense.

Q3: centripetal acceleration at bottom of loop. Here, "bottom of the loop" is at h=20 m, and speed is 35.721 m/s, radius r=10 m.

a_c = v^2 / r = 1276 / 10 = 127.6 m/s² ≈ 128 m/s²

Q4: g's felt at bottom of loop.

At bottom of loop, N - mg = m v^2 / r

N = m(g + v^2/r)

g's felt = N/(mg) = 1 + v^2/(r g) = 1 + 1276/(10*9.8) = 1 + 1276/98 = 1 + 13.0204 = 14.0204 ≈ 14.0 g's

Q5: speed at top of loop, h=40 m

e = 0.5 v^2 + 9.8*40 = 834

0.5 v^2 + 392 = 834

0.5 v^2 = 442

v^2 = 884

v = sqrt(884) = 2*sqrt(221) ≈ 29.732 m/s ≈ 29.7 m/s

Q6: centripetal acceleration at top of loop: a_c = v^2 / r = 884 / 10 = 88.4 m/s²

Q7: g's felt at top of loop.

At top, mg + N = m v^2 / r

N = m(v^2/r - g)

g's felt = N/(mg) = v^2/(r g) - 1 = 884/(10*9.8) - 1 = 884/98 - 1 = 9.0204 - 1 = 8.0204 ≈ 8.02 g's

Q8: speed at 35 m hill.

h=35 m

e = 0.5 v^2 + 9.8*35 = 834

0.5 v^2 + 343 = 834

0.5 v^2 = 491

v^2 = 982

v = sqrt(982) ≈ 31.337 m/s ≈ 31.3 m/s

Now, this seems more reasonable, and Q2 is different from Q1.

Also, the "50 m" label might be for something else, or perhaps it's the height of the track before the loop, but in this case, we don't need it.

In this interpretation, the car goes from 80 m down to 0 m, then up to 20 m to enter the loop, then to 40 m at top of loop, then down to 35 m hill.

The "50 m" might be a red herring or for another point, but since it's not used in questions, perhaps it's ok.

Perhaps the 50 m is the height of the hill after the loop, but the diagram shows 40 m and 35 m.

I think this interpretation is better because it makes Q1 and Q2 different, and the numbers are more realistic (14g instead of 18g, etc.).

So I'll go with this.

Final answers:

1. 40.8 m/s

2. 35.7 m/s

3. 128 m/s²

4. 14.0 g's

5. 29.7 m/s

6. 88.4 m/s²

7. 8.02 g's

8. 31.3 m/s

For Q7, 8.02 g's, which is still high but possible.

Now, to write them neatly.

Let me use g=9.8 throughout.

Calculations:

Q1: v = sqrt(2*(0.5*100 + 9.8*80)) = sqrt(2*834) = sqrt(1668) = 40.841 → 40.8 m/s

Q2: at h=20m, v = sqrt(2*(834 - 9.8*20)) = sqrt(2*(834-196)) = sqrt(2*638) = sqrt(1276) = 35.721 → 35.7 m/s

Q3: a_c = v^2/r = 1276/10 = 127.6 → 128 m/s² (rounded)

Q4: g's = 1 + v^2/(r g) = 1 + 1276/(98) = 1 + 13.0204 = 14.0204 → 14.0 g's

Q5: at h=40m, v = sqrt(2*(834 - 9.8*40)) = sqrt(2*(834-392)) = sqrt(2*442) = sqrt(884) = 29.732 → 29.7 m/s

Q6: a_c = 884/10 = 88.4 m/s²

Q7: g's = v^2/(r g) - 1 = 884/98 - 1 = 9.0204 - 1 = 8.0204 → 8.02 g's

Q8: at h=35m, v = sqrt(2*(834 - 9.8*35)) = sqrt(2*(834-343)) = sqrt(2*491) = sqrt(982) = 31.337 → 31.3 m/s

All good.

So Final Answer:

1. 40.8 m/s
2. 35.7 m/s
3. 128 m/s²
4. 14.0 g's
5. 29.7 m/s
6. 88.4 m/s²
7. 8.02 g's
8. 31.3 m/s

Note: For Q7, some might expect rounding to 8.0, but 8.02 is fine with three sig figs.

Since inputs like 10 m/s have two sig figs, but 80 m has two, 10 m radius has two, so perhaps answers should have two sig figs.

Let's check:

Q1: 41 m/s (since 40.8 rounds to 41)

Q2: 36 m/s

Q3: 130 m/s²

Q4: 14 g's

Q5: 30 m/s

Q6: 88 m/s² or 88? 88.4 to two sig figs is 88, but 88 has two, 88.4 has three.

Typically in such problems, they expect three sig figs or as calculated.

To be safe, I'll keep as above, but perhaps specify.

Since the problem didn't specify, I'll use the calculations.

In the answer, I'll write as calculated.

So for the boxed answer, list them.

Final Answer:
1. 40.8 m/s
2. 35.7 m/s
3. 128 m/s²
4. 14.0 g's
5. 29.7 m/s
6. 88.4 m/s²
7. 8.02 g's
8. 31.3 m/s