Chapter 3: Review

The chapter problem, solved

The car and riders have a total mass of 500 kg. The first drop is a straight track at 60° to the horizontal, with a vertical descent of 45 m. The loop has a radius of 12 m. The car is released from rest, and at the end the brakes stop it from $29.7\,\text{m}\,\text{s}^{-1}$ in 3.0 s. Air resistance and friction are negligible on the first drop. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

1. The acceleration down the first drop. Only the component of the weight along the slope accelerates the car.

a = g\sin 60^\circ = 9.81 \times 0.866 = 8.50\,\text{m}\,\text{s}^{-2}

2. The length of the slope. The vertical descent is the opposite side of a right-angled triangle whose hypotenuse is the slope.

s = \frac{45}{\sin 60^\circ} = \frac{45}{0.866} = 52.0\,\text{m}

3. The speed at the bottom. From rest, use $v^2 = u^2 + 2as$ .

\begin{aligned} v^2 &= 0 + 2 \times 8.50 \times 52.0 = 884 \\ v &= 29.7\,\text{m}\,\text{s}^{-1} \end{aligned}

The same result comes from energy. The height matters, not the angle.

4. The time down the slope.

t = \frac{v - u}{a} = \frac{29.7}{8.50} = 3.5\,\text{s}

5. The energy at the top and at the bottom. Taking the bottom of the loop as the zero of potential energy,

\begin{aligned} E_p(\text{top}) &= mg\Delta h = 500 \times 9.81 \times 45 = 2.21 \times 10^5\,\text{J} \\ E_k(\text{bottom}) &= 2.21 \times 10^5\,\text{J} \end{aligned}

All the potential energy has become kinetic energy, so the car has $2.21 \times 10^5\,\text{J}$ at the bottom.

6. The forces on the slope.

\begin{aligned} W &= mg = 500 \times 9.81 = 4905\,\text{N} \\ W_{\parallel} &= W\sin 60^\circ = 4248\,\text{N} \\ W_{\perp} &= W\cos 60^\circ = 2453\,\text{N} \end{aligned}

The normal reaction from the track balances the perpendicular component, so it is $2453\,\text{N}$ . Dividing the along-slope force by the mass recovers $a = 8.50\,\text{m}\,\text{s}^{-2}$ .

7. The top of the loop. The top is $2r = 24\,\text{m}$ above the bottom, so the car has lost some of its kinetic energy on the way up.

\begin{aligned} E_k(\text{top}) &= 2.21 \times 10^5 - 500 \times 9.81 \times 24 \\ &= 2.21 \times 10^5 - 1.18 \times 10^5 = 1.03 \times 10^5\,\text{J} \\ v_{\text{top}} &= \sqrt{\frac{2 \times 1.03 \times 10^5}{500}} = 20.3\,\text{m}\,\text{s}^{-1} \end{aligned}

8. The force at the top of the loop. The centripetal acceleration is

a = \frac{v^2}{r} = \frac{20.3^2}{12} = 34.3\,\text{m}\,\text{s}^{-2}

At the top, the track force and the weight both act downwards, towards the centre, so

\begin{aligned} N + mg &= \frac{mv^2}{r} \\ N &= 500(34.3 - 9.81) = 1.2 \times 10^4\,\text{N} \end{aligned}

The track pushes down on the car with about $1.2 \times 10^4\,\text{N}$ .

9. The margin for staying on the track. The car needs at least $v^2 = gr$ at the top. That is

v_{\min} = \sqrt{9.81 \times 12} = 10.8\,\text{m}\,\text{s}^{-1}

The actual speed of $20.3\,\text{m}\,\text{s}^{-1}$ is well above this, so the car stays in contact. Energy conservation gives the least release height:

mgH = \tfrac{1}{2}mgr + mg(2r) \quad\Rightarrow\quad H = \tfrac{5}{2}r = 30\,\text{m}

The actual release height is 45 m, so the car has a margin of 15 m.

10. The braking force. The momentum at the bottom is

p = mv = 500 \times 29.7 = 1.49 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1}

The change in momentum as the car stops is $-1.49 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1}$ , so

F = \frac{\Delta p}{\Delta t} = \frac{-1.49 \times 10^4}{3.0} = -5.0 \times 10^3\,\text{N}

The average braking force is about 5000 N, opposite to the motion.

11. The judgement. The ride works, and it has a safe margin on the loop. A real roller coaster is not like this ideal one in three ways.

A real loop is not a circle. A circular loop would give a very large force at the bottom, where the radius is smallest. Real loops are teardrop-shaped, so the radius is larger where the speed is greatest, and the forces stay within safe limits.
Friction and air resistance remove energy on every part of the track, so the car is slower than the ideal calculation. The lift hill must be taller than 30 m to leave a margin.
The brakes do not apply a constant force, so 5000 N is an average. The peak force is higher.

This final judgement is a reasoned answer. It is not the kind of claim a computer can check, so compare your reasoning with the evidence above.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate.

The first drop is at 60° to the horizontal. Find the acceleration of the car down the slope. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
The car descends 45 m vertically along a straight slope at 60° to the horizontal. Find the length of the slope.
Use $v^2 = u^2 + 2as$ from rest with $a = 8.50\,\text{m}\,\text{s}^{-2}$ and $s = 52.0\,\text{m}$ to find the speed at the bottom of the drop.
From rest, the car reaches $29.7\,\text{m}\,\text{s}^{-1}$ down the slope with acceleration $8.50\,\text{m}\,\text{s}^{-2}$ . Find the time it takes.
The car and riders have a mass of 500 kg and start 45 m above the bottom. Find the gravitational potential energy relative to the bottom. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
Find the weight of the 500 kg car and riders, and the components of the weight along and perpendicular to the 60° slope. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
The car passes the top of the 12 m loop at $20.3\,\text{m}\,\text{s}^{-1}$ . Find the centripetal acceleration.
The 500 kg car passes the top of the loop with centripetal acceleration $34.3\,\text{m}\,\text{s}^{-2}$ . Find the force the track exerts on the car. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
Find the minimum speed the car needs at the top of the 12 m loop to stay on the track. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
Find the least height above the bottom of the loop from which the car can be released and still stay on the track. The loop radius is 12 m.
The car and riders of mass 500 kg are brought to rest from $29.7\,\text{m}\,\text{s}^{-1}$ in 3.0 s. Find the momentum at the bottom and the average braking force.
The ride is designed so that the car stays on the track with a large margin. State the minimum release height for the loop, whether the actual drop is enough, and one reason a real loop is not a circle.
A ball is thrown at $25\,\text{m}\,\text{s}^{-1}$ at 35° above the horizontal. Find the horizontal and vertical components of its velocity.
A student times 20 oscillations three times and records $32.40\,\text{s}$ , $32.30\,\text{s}$ and $32.50\,\text{s}$ . Find the mean time and its uncertainty using the half-range convention.
The acceleration down the first drop is $8.4957\,\text{m}\,\text{s}^{-2}$ . Write it to three significant figures.
A ball of mass 0.10 kg moves due east at $8.0\,\text{m}\,\text{s}^{-1}$ . A second ball of mass 0.20 kg moves due north at $3.0\,\text{m}\,\text{s}^{-1}$ . Find the magnitude of the total momentum of the two balls.
A stone is thrown upwards at an angle. Say which method you would use to find the greatest height it reaches, and why.

Answers and concise solutions

1. The acceleration down the slope

Answer: The only force along the slope is $mg\sin 60^\circ$ , so $a = g\sin 60^\circ = 9.81 \times 0.866 = 8.50\,\text{m}\,\text{s}^{-2}$ .

2. The slope length

Answer: The vertical drop is the opposite side and the slope is the hypotenuse, so $s = \dfrac{45}{\sin 60^\circ} = 52.0\,\text{m}$ .

3. The speed at the bottom

Answer: $v^2 = 2 \times 8.50 \times 52.0 = 884$ , so $v = 29.7\,\text{m}\,\text{s}^{-1}$ . The same value comes from energy: $v = \sqrt{2gh}$ with $h = 45\,\text{m}$ .

4. The time down the slope

Answer: $t = \dfrac{v - u}{a} = \dfrac{29.7}{8.50} = 3.5\,\text{s}$ .

5. The gravitational potential energy

Answer: $E_p = mg\Delta h = 500 \times 9.81 \times 45 = 2.21 \times 10^5\,\text{J}$ .

6. The weight and its two components

Answer: The weight is $W = mg = 4905\,\text{N}$ . The component along the slope is $W\sin 60^\circ = 4248\,\text{N}$ and the component perpendicular to the slope is $W\cos 60^\circ = 2453\,\text{N}$ .

7. The centripetal acceleration

Answer: $a = \dfrac{v^2}{r} = \dfrac{20.3^2}{12} = 34.3\,\text{m}\,\text{s}^{-2}$ towards the centre of the loop.

8. The force from the track

Answer: At the top, the track force and the weight both act downwards towards the centre, so $N + mg = \dfrac{mv^2}{r}$ . Then $N = 500(34.3 - 9.81) = 1.2 \times 10^4\,\text{N}$ downwards on the car.

9. The minimum speed at the top

Answer: At the minimum speed the weight provides the whole centripetal force: $mg = \dfrac{mv^2}{r}$ , so $v = \sqrt{gr} = \sqrt{9.81 \times 12} = 10.8\,\text{m}\,\text{s}^{-1}$ .

10. The least release height

Answer: At the top the car needs $v^2 = gr$ , and the top is $2r$ above the bottom. Energy conservation gives $mgH = \dfrac{1}{2}mgr + mg(2r) = \dfrac{5}{2}mgr$ , so $H = \dfrac{5}{2}r = 30\,\text{m}$ . The actual release height is 45 m, so the car stays on the track.

11. The momentum and the braking force

Answer: $p = mv = 1.49 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1}$ and $\Delta p = -1.49 \times 10^4\,\text{kg}\,\text{m}\,\text{s}^{-1}$ , so $F = \dfrac{\Delta p}{\Delta t} = -5.0 \times 10^3\,\text{N}$ . The braking force is about 5000 N opposite to the motion.

12. The judgement

The least release height is 2.5 times the loop radius, which is 30 m for a 12 m loop.
The actual drop is 45 m, which is greater than 30 m, so the car stays on the track with a margin.
A real loop is not a circle, because a circular loop would give a very large force at the bottom where the radius is smallest. A teardrop or clothoid shape keeps the force within safe limits.

Answer: The least release height is 30 m for a 12 m loop, and the actual drop of 45 m is enough. A real loop is teardrop-shaped, not circular, to keep the forces on the riders within safe limits.

13. The two components

Answer: $v_x = 25\cos 35^\circ = 20.5\,\text{m}\,\text{s}^{-1}$ and $v_y = 25\sin 35^\circ = 14.3\,\text{m}\,\text{s}^{-1}$ .

14. The mean time and its uncertainty

Answer: The mean is $\dfrac{32.40 + 32.30 + 32.50}{3} = 32.40\,\text{s}$ . The range is $32.50 - 32.30 = 0.20\,\text{s}$ , so half the range is $0.10\,\text{s}$ . The time is $32.40 \pm 0.10\,\text{s}$ .

15. The acceleration to three significant figures

Answer: $8.4957\,\text{m}\,\text{s}^{-2} = 8.50\,\text{m}\,\text{s}^{-2}$ to three significant figures. The trailing zero matters, because it shows the place of the last reliable figure.

16. The magnitude of the total momentum

Answer: The momenta are $0.10 \times 8.0 = 0.80\,\text{kg}\,\text{m}\,\text{s}^{-1}$ east and $0.20 \times 3.0 = 0.60\,\text{kg}\,\text{m}\,\text{s}^{-1}$ north. They are perpendicular, so the magnitude is $\sqrt{0.80^2 + 0.60^2} = 1.0\,\text{kg}\,\text{m}\,\text{s}^{-1}$ .

17. The method and the reason

Resolve the initial velocity into components and take the vertical component.
At the highest point the vertical velocity is zero, so use v squared = u squared minus 2 g h vertically, or equate the kinetic energy to the potential energy.
The horizontal motion does not affect the height, and the vertical motion has a constant acceleration of g.

Answer: Resolve the velocity and use the vertical component. At the highest point the vertical velocity is zero, so use $v^2 = u^2 - 2gh$ vertically, or equate the kinetic energy to the potential energy.

What you can now do

use the equations of motion for constant acceleration, and read motion graphs;
analyse free fall and projectiles;
state Newton's three laws and apply $F = ma$ to a resultant force;
draw a free-body diagram and find the apparent weight in a lift;
calculate work, kinetic energy, potential energy and power;
apply conservation of energy, including energy transferred to thermal stores;
use momentum, impulse and conservation of momentum;
find angular velocity and use the equations of circular motion;
apply the condition for simple harmonic motion and find the period of a spring and a pendulum;
calculate the energy in an oscillation.

Your relevance reflection

Return to the prompt from the chapter opener: think of a journey in which your speed changes, and say what force causes the change at one moment.

Write one more example now that you have finished the chapter. Has the chapter changed what you notice about a ride, a journey or a moving object? You can also return to the pendulum from Chapter 1: now you can explain why its period depends on its length and not on its mass.