Applications of Circular Motion: 40-Mark Practice Exam

This original tutorgum practice paper targets the supplied Physics 2025 scope for applications of circular motion. It assesses force identification, resolving tension and weight for a mass moving in a horizontal circle on a string, and analysing friction on a car taking a flat circular bend. This is a focused topic paper, not a full-course exam, and it is not an official, endorsed, past, or predicted NESA paper. For another paper in this topic cluster, see Applications of Circular Motion: HSC Physics Practice Exam.

01Exam instructions

• Suggested reading time: 5 minutes.
• Suggested working time: 55 minutes.
• Total marks: 40.
• Permitted equipment: scientific calculator, ruler, and writing materials. Do not use notes.
• Use \(g=9.80\ \text{m s}^{-2}\) throughout.
• Give numerical answers to three significant figures, unless stated otherwise. Include units and show your working.
• Assume that tyres roll without slipping unless a question asks you to determine whether skidding will occur. Treat the coefficient of static friction as the maximum available friction coefficient.
• These are practice-paper conditions, not claims about official NESA examination timing or equipment rules.

Useful relationships supplied for this paper:

\[ F_{\text{in}}=\frac{mv^2}{r} \]

\[ a_c=\frac{v^2}{r}, \qquad W=mg \]

\[ f_{s,\max}=\mu_s N \]

Here, \(F_{\text{in}}\) is the inward resultant force, \(a_c\) is centripetal acceleration, \(m\) is mass, \(v\) is speed, \(r\) is the radius of the circular path, \(W\) is weight, \(g\) is gravitational field strength, \(N\) is the normal force, and \(\mu_s\) is the coefficient of static friction.

Assessed scope

• Identifying forces acting on a physical system.
• Resolving tension and weight for a mass moving in a horizontal circle on a string.
• Analysing friction as the source of the inward resultant force for cars on flat circular bends.
• Calculating inward resultant forces and evaluating whether the available friction is sufficient.

Excluded scope: banked bends, vertical circles, and other circular-motion applications outside the supplied scope.

02Questions

Question 1 (4 marks)

A car travels at constant speed around a level circular bend. Assume its tyres roll without slipping.

(a) Identify the horizontal force responsible for providing the inward resultant force. (1 mark)

(b) State the directions of the normal force and weight. Explain their relationship to the vertical resultant force. (2 marks)

(c) The driver increases the car’s speed while following the same circular path. State how the required inward resultant force changes, assuming the car’s mass and the bend’s radius remain constant. (1 mark)

Question 2 (6 marks)

A \(0.400\ \text{kg}\) mass moves at constant speed in a horizontal circle of radius \(0.750\ \text{m}\). A taut string makes an angle of \(25.0^\circ\) with the vertical. Assume only tension and weight act on the mass.

(a) Resolve the forces vertically and calculate the tension in the string. (2 marks)

(b) Calculate the inward resultant force acting on the mass. (2 marks)

(c) Calculate the speed of the mass. (2 marks)

Question 3 (6 marks)

A \(1.10\times10^3\ \text{kg}\) car travels at \(15.0\ \text{m s}^{-1}\) around a level circular bend of radius \(50.0\ \text{m}\). Assume tyre-road friction is the only horizontal force.

(a) Calculate the inward resultant force acting on the car. (2 marks)

(b) Calculate the minimum coefficient of static friction required for the car to maintain its circular path. (2 marks)

(c) A second car has twice the mass of the first car but travels at the same speed around the same bend. For this part, assume the coefficient of static friction is \(0.500\) for both cars. Compare the required inward force and maximum available static friction for the second car with those for the first. State whether the mass change alone makes the second car more likely to skid. (2 marks)

Question 4 (7 marks)

A \(0.250\ \text{kg}\) mass moves at \(2.40\ \text{m s}^{-1}\) in a horizontal circle of radius \(0.900\ \text{m}\), attached to a taut string. The angle \(\theta\) between the string and the vertical is unknown. Assume only tension and weight act on the mass.

(a) Calculate the inward resultant force acting on the mass. (2 marks)

(b) Determine the angle \(\theta\) between the string and the vertical. (3 marks)

(c) Calculate the tension in the string. (2 marks)

Question 5 (7 marks)

A \(1.25\times10^3\ \text{kg}\) car travels at \(18.0\ \text{m s}^{-1}\) around a level bend of radius \(65.0\ \text{m}\). The coefficient of static friction between the tyres and the road is \(0.450\).

(a) Calculate the inward resultant force required for the car to follow the bend. (2 marks)

(b) Calculate the maximum available static friction force. (2 marks)

(c) Determine whether the car can maintain its circular path at this speed without skidding. Justify your decision using your calculated forces. (1 mark)

(d) Calculate the maximum speed at which the car can take this bend without skidding, assuming the coefficient of static friction remains constant. (2 marks)

Question 6 (10 marks)

A driver can take either of two level circular paths at a constant speed of \(14.0\ \text{m s}^{-1}\). The tighter path has a radius of \(32.0\ \text{m}\), while the wider path has a radius of \(48.0\ \text{m}\).

The car has a mass of \(1.00\times10^3\ \text{kg}\). The coefficient of static friction is \(0.500\) on both paths.

(a) Calculate the inward resultant force required for each path. (4 marks)

(b) Calculate the maximum available static friction force. (2 marks)

(c) Determine which path the car can follow at \(14.0\ \text{m s}^{-1}\) without skidding. Justify your choice by comparing the required force with the maximum available friction. (2 marks)

(d) A student claims, “The wider bend requires more friction because the car travels farther around it.”

Assess this claim using the relationship between inward resultant force and radius. (2 marks)

03Worked solutions and marking guide

Solution 1

(a) Static friction between the tyres and the road provides the inward resultant force.

Marking: 1 mark for identifying static friction as the horizontal force directed towards the centre of the bend.

(b) The normal force acts vertically upwards, perpendicular to the road. Weight acts vertically downwards.

Because the car has no vertical acceleration, these forces balance. The vertical resultant force is zero.

Marking:

• 1 mark for correctly identifying the directions of the normal force and weight.
• 1 mark for stating that they balance, giving zero vertical resultant force.

(c) The required inward resultant force increases as the car’s speed increases.

From

\[ F_{\text{in}}=\frac{mv^2}{r} \]

the force is proportional to the square of the speed when mass and radius are constant. For example, doubling the speed would require four times the inward resultant force.

Marking: 1 mark for stating that the inward resultant force increases with the square of the speed.

Total: 4 marks.

Solution 2

(a) The mass moves in a horizontal circle, so it has no vertical acceleration. The vertical component of tension must balance the weight.

\[ T\cos25.0^\circ=mg \]

Rearranging,

\[ \begin{aligned} T&=\frac{mg}{\cos25.0^\circ}\\ &=\frac{(0.400)(9.80)}{\cos25.0^\circ}\\ &=4.33\ \text{N} \end{aligned} \]

The tension is \(4.33\ \text{N}\).

Marking:

• 1 mark for the correct vertical force relationship.
• 1 mark for calculating the tension with the correct unit.

(b) The horizontal component of tension supplies the inward resultant force.

\[ \begin{aligned} F_{\text{in}}&=T\sin25.0^\circ\\ &=(4.33)\sin25.0^\circ\\ &=1.83\ \text{N} \end{aligned} \]

The inward resultant force is \(1.83\ \text{N}\).

Marking:

• 1 mark for resolving tension horizontally.
• 1 mark for the correct numerical result with units.

(c) Use the circular-motion relationship:

\[ F_{\text{in}}=\frac{mv^2}{r} \]

Rearrange to make speed the subject:

\[ v=\sqrt{\frac{F_{\text{in}}r}{m}} \]

Substituting,

\[ \begin{aligned} v&=\sqrt{\frac{(1.83)(0.750)}{0.400}}\\ &=1.85\ \text{m s}^{-1} \end{aligned} \]

The mass moves at \(1.85\ \text{m s}^{-1}\).

Marking:

• 1 mark for correctly rearranging and substituting into the circular-motion equation.
• 1 mark for the correct speed with units.

Total: 6 marks.

Solution 3

(a) The inward resultant force is

\[ F_{\text{in}}=\frac{mv^2}{r} \]

Substituting the given values,

\[ \begin{aligned} F_{\text{in}}&=\frac{(1.10\times10^3)(15.0)^2}{50.0}\\ &=4.95\times10^3\ \text{N} \end{aligned} \]

The required inward resultant force is \(4.95\times10^3\ \text{N}\).

Marking:

• 1 mark for the correct equation and substitution.
• 1 mark for the correct force with units.

(b) On a level road, the vertical forces balance, so

\[ N=mg \]

Therefore,

\[ N=(1.10\times10^3)(9.80) =1.078\times10^4\ \text{N} \]

At the minimum coefficient, the maximum available friction equals the required inward force:

\[ \mu_{s,\min}N=F_{\text{in}} \]

Hence,

\[ \begin{aligned} \mu_{s,\min}&=\frac{F_{\text{in}}}{N}\\ &=\frac{4950}{10780}\\ &=0.459 \end{aligned} \]

The minimum coefficient of static friction is \(0.459\).

Marking:

• 1 mark for relating the normal force and minimum friction coefficient to the required inward force.
• 1 mark for the correct coefficient.

(c) The required inward force is proportional to mass:

\[ F_{\text{in}}=\frac{mv^2}{r} \]

The maximum available friction is also proportional to mass when the coefficient of friction is unchanged:

\[ f_{s,\max}=\mu_s mg \]

Doubling the mass therefore doubles both forces. For the first car, the maximum friction is

\[ f_{s,\max}=(0.500)(10780)=5390\ \text{N} \]

For the second car, the required inward force is \(9.90\times10^3\ \text{N}\), and the maximum available friction is \(1.078\times10^4\ \text{N}\).

Both forces double, so the ratio of required friction to maximum available friction remains unchanged. The mass change alone does not make the second car more likely to skid.

Marking:

• 1 mark for recognising that both the required inward force and maximum available friction double.
• 1 mark for explaining that their ratio remains unchanged, so mass alone does not increase the likelihood of skidding.

Total: 6 marks.

Solution 4

(a) Calculate the inward resultant force using

\[ F_{\text{in}}=\frac{mv^2}{r} \]

Substituting,

\[ \begin{aligned} F_{\text{in}}&=\frac{(0.250)(2.40)^2}{0.900}\\ &=1.60\ \text{N} \end{aligned} \]

The inward resultant force is \(1.60\ \text{N}\).

Marking:

• 1 mark for the correct equation and substitution.
• 1 mark for the correct result with units.

(b) Resolve tension into vertical and horizontal components.

The vertical component balances the weight:

\[ T\cos\theta=mg \]

\[ T\cos\theta=(0.250)(9.80)=2.45\ \text{N} \]

The horizontal component supplies the inward resultant:

\[ T\sin\theta=F_{\text{in}}=1.60\ \text{N} \]

Divide the horizontal relationship by the vertical relationship to eliminate tension:

\[ \begin{aligned} \tan\theta&=\frac{1.60}{2.45}\\ &=0.653 \end{aligned} \]

Therefore,

\[ \theta=\tan^{-1}(0.653)=33.1^\circ \]

The string makes an angle of \(33.1^\circ\) with the vertical.

Marking:

• 1 mark for the correct resolved force relationships.
• 1 mark for obtaining the tangent relationship and substituting correctly.
• 1 mark for the correct angle.

(c) The vertical and horizontal components of tension are perpendicular. The magnitude of tension can therefore be calculated using Pythagoras’ theorem:

\[ \begin{aligned} T&=\sqrt{(T\cos\theta)^2+(T\sin\theta)^2}\\ &=\sqrt{(2.45)^2+(1.60)^2}\\ &=2.93\ \text{N} \end{aligned} \]

The tension is \(2.93\ \text{N}\).

Marking:

• 1 mark for correctly combining the perpendicular components of tension.
• 1 mark for the correct tension with units.

Total: 7 marks.

Solution 5

(a) The inward resultant force required is

\[ F_{\text{in}}=\frac{mv^2}{r} \]

Substituting,

\[ \begin{aligned} F_{\text{in}}&=\frac{(1.25\times10^3)(18.0)^2}{65.0}\\ &=6.23\times10^3\ \text{N} \end{aligned} \]

The car requires an inward resultant force of \(6.23\ \text{kN}\).

Marking:

• 1 mark for the correct equation and substitution.
• 1 mark for the correct force with units.

(b) On the level bend, the normal force balances the weight:

\[ N=mg=(1.25\times10^3)(9.80) =1.225\times10^4\ \text{N} \]

The maximum available static friction is

\[ \begin{aligned} f_{s,\max}&=\mu_sN\\ &=(0.450)(12250)\\ &=5.51\times10^3\ \text{N} \end{aligned} \]

The maximum available static friction is \(5.51\ \text{kN}\).

Marking:

• 1 mark for calculating the normal force and applying the maximum-friction relationship.
• 1 mark for the correct maximum friction with units.

(c) The required inward force is \(6.23\ \text{kN}\), but the maximum available static friction is only \(5.51\ \text{kN}\).

The car cannot maintain the circular path at \(18.0\ \text{m s}^{-1}\) without skidding.

Marking: 1 mark for the correct decision supported by comparing the two forces.

(d) At the maximum speed, the required inward force equals the maximum available friction:

\[ \frac{mv_{\max}^2}{r}=\mu_s mg \]

Cancel \(m\) and rearrange:

\[ v_{\max}=\sqrt{\mu_sgr} \]

Substituting,

\[ \begin{aligned} v_{\max}&=\sqrt{(0.450)(9.80)(65.0)}\\ &=16.9\ \text{m s}^{-1} \end{aligned} \]

The maximum speed is \(16.9\ \text{m s}^{-1}\).

Marking:

• 1 mark for equating the required force to maximum friction and correctly rearranging.
• 1 mark for the correct maximum speed with units.

Total: 7 marks.

Solution 6

(a) Use

\[ F_{\text{in}}=\frac{mv^2}{r} \]

For the tighter path, \(r=32.0\ \text{m}\):

\[ \begin{aligned} F_{32}&=\frac{(1.00\times10^3)(14.0)^2}{32.0}\\ &=6125\ \text{N}\\ &=6.13\ \text{kN} \end{aligned} \]

For the wider path, \(r=48.0\ \text{m}\):

\[ \begin{aligned} F_{48}&=\frac{(1.00\times10^3)(14.0)^2}{48.0}\\ &=4083\ \text{N}\\ &=4.08\ \text{kN} \end{aligned} \]

The tighter path requires \(6.13\ \text{kN}\), while the wider path requires \(4.08\ \text{kN}\).

Marking:

• 1 mark for the correct equation and substitution for the tighter path.
• 1 mark for the correct tighter-path force with units.
• 1 mark for the correct equation and substitution for the wider path.
• 1 mark for the correct wider-path force with units.

(b) The car’s normal force balances its weight:

\[ N=mg=(1.00\times10^3)(9.80)=9800\ \text{N} \]

The maximum available static friction is

\[ \begin{aligned} f_{s,\max}&=\mu_sN\\ &=(0.500)(9800)\\ &=4900\ \text{N} \end{aligned} \]

The maximum available friction is \(4.90\ \text{kN}\) on either path.

Marking:

• 1 mark for calculating the normal force and applying the friction relationship.
• 1 mark for the correct maximum friction with units.

(c) The car can follow the wider path without skidding at \(14.0\ \text{m s}^{-1}\).

The tighter path requires \(6.13\ \text{kN}\), which exceeds the available \(4.90\ \text{kN}\). The wider path requires only \(4.08\ \text{kN}\), which is below the available friction.

Marking:

• 1 mark for selecting the wider path.
• 1 mark for correctly comparing the required force with the maximum available friction.

(d) The claim is incorrect.

At constant mass and speed,

\[ F_{\text{in}}=\frac{mv^2}{r} \]

shows that the inward resultant force is inversely proportional to the radius. A larger radius means less inward force is required to maintain circular motion at the same speed. The distance travelled around the bend does not change this relationship.

Marking:

• 1 mark for identifying the inverse relationship between inward force and radius at constant mass and speed.
• 1 mark for explaining that the wider bend requires less inward force, not more.

Total: 10 marks.

End of paper – 40 marks.