AQA AS Level Physics Paper 1, November 2021: Question 4

7 marks · Hard difficulty · Extended Answer

Determine angle theta and the magnitude of force U using a scale diagram for a uniform vaulting pole in equilibrium, and discuss differences in magnitude and direction between forces U and V when the pole is held at different positions.

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Question

Figure 7 shows an athlete holding a vaulting pole at an angle of 40 degrees to the horizontal, with forces D = 53 N applied at 90 degrees to point X, a 31 N weight acting downwards at the center, and force U applied at point Y at an angle theta to the vertical. Figure 8 shows the forces acting on the pole in equilibrium. Question 4.1 asks to determine, using a scale diagram, theta and the magnitude of U. Question 4.2 asks to discuss the differences in magnitudes and directions between force U in Figure 7 and force V applied at Y in Figure 9, where the pole is horizontal and held by a vertical downward force S at X and V at Y.
Question text

04 Figure 7 shows an athlete holding a vaulting pole at an angle of 40° to the horizontal.

Figure 7

Forces D and U are exerted on the pole by the athlete’s right and left hands

respectively.

U is applied at point Y at an angle θ to the vertical.

The magnitude of D is 53 N and is applied at 90° to the pole at X.

The uniform pole is in equilibrium. It has a weight of 31 N.

Figure 8 shows the forces acting on the pole.

Figure 8

04.1 Determine, using a scale diagram, θ and the magnitude of U.

[4 marks]

θ = °

magnitude of U = N

04.2 The athlete now moves the pole to a horizontal position. The pole is held stationary in

this position.

The athlete’s right hand applies a force S vertically downwards at X as shown

in Figure 9. The athlete’s left hand applies a force V at Y.

Figure 9

Discuss the differences between the magnitudes and directions of force U in Figure 7

and force V applied at Y in Figure 9.

[3 marks]

Mark scheme

Show the mark scheme The mark scheme outlines the answers for question 4.1 requiring a closed triangle of forces, an appropriate scale, theta between 23 and 27 degrees, and U between 77 and 81 N for 4 marks. For question 4.2, it awards 3 marks for discussing that V is vertical without a horizontal component, has greater magnitude than U due to increased moments of weight, and relating the forces to vertical components and moments about Y.

Question Answers Additional Comments/Guidance Mark AO

04.1 Closed triangle of forces drawn Accept scale where 10 N is represented by at 4 AO1.1a

least 1 cm. AO1.1b

Appropriate scale AO1.1b

AO1.1b

θ = 23 to 27 (°)

U = 77 to 81 (N)

Treat each marking point independently.

Do not accept answers for U and θ without a

scale diagram.

Maximum of 3 marks for a free-body

diagram where forces have been drawn to

scale. (Check figure 8)

04.2 3 AO3.1a

V is vertical / Force at Y is now vertical / V does not have a AO3.1a

horizontal component / V = S + 31 / V is perpendicular to the

pole / V is of greater magnitude than U / Force at Y has AO3.1a

increased in magnitude

(Because) S and weight (or mg) are both vertical (in Fig 9)

(Because) greater moment of weight (about Y) in Fig 9 /

smaller moment of weight (about Y) in Fig 7 / (Because) S is

larger in magnitude than D (to produce a greater moment

(about Y because they are equal distances from Y)

Total 7

How to answer it

Vaulting Pole Equilibrium Analysis

What this question tests

This question assesses your mastery of static equilibrium in rigid bodies under the action of non-concurrent coplanar forces. Key testing objectives include constructing and interpreting closed vector triangles for concurrent force systems, applying the principle of moments, and analyzing how changing the physical orientation of a system alters force magnitudes and directional components.

Question 04.1 [4 Marks]

Scale Diagram Determination of Angle and Force Magnitude

✅ Correct Answers

  • Closed triangle of forces: Drawn accurately with vector arrows following head-to-tail.
  • Appropriate scale: At least 1 cm representing 10 N (ensuring the diagram is large enough to read precisely).
  • Angle θ : 23° to 27°
  • Magnitude of U: 77 N to 81 N

💡 Key Knowledge

  • When three coplanar forces keep an object in translational equilibrium, placing them tip-to-tail forms a closed vector triangle.
  • The vector loop must account for all three known forces: Weight ( 31 N downwards), Force D ( 53 N along its line of action), and Force U .

🧠 Exam Technique

  • Use a sharp pencil and a ruler with clear millimeter markings.
  • Start by drawing the known vertical weight vector, attach the known force D at the correct angle, and close the triangle using force U .
  • Measure angles directly using a protractor and scale lengths with absolute precision to stay within the permitted tolerance window.

❌ Common Errors

  • Attempting to calculate values purely through trigonometry without providing a scale diagram (results in 0 marks as the question explicitly mandates a scale diagram).
  • Drawing open vectors or reversing arrow directions so they do not form a continuous loop.
  • Using an overly cramped scale (e.g., 1 cm representing 50 N) which leads to reading inaccuracies outside the mark scheme tolerances.

📐 Step-by-Step Construction Method

  1. Step 1: Establish your scale (e.g., 1 cm = 10 N ). Therefore, the 31 N weight is drawn as 3.1 cm vertically downwards, and the 53 N force D is drawn as 5.3 cm along its directional line.
  2. Step 2: Connect the vectors head-to-tail to form a closed triangle.
  3. Step 3: Measure the length of the side representing force U , then convert it back using your scale factor to find the magnitude in Newtons.
  4. Step 4: Measure angle θ relative to the vertical reference line using a protractor.
Question 04.2 [3 Marks]

Discussion of Force Differences (Slanted vs. Horizontal Pole)

✅ Correct Answers (Mark Scheme Points)

  • Point 1: V is vertical (or force at Y has no horizontal component / V = S + 31 / V is perpendicular to the pole / V is greater in magnitude than U ).
  • Point 2: Because S and weight ( mg ) are both strictly vertical in Figure 9.
  • Point 3: Because the greater moment of weight about Y in Figure 9 requires a larger corrective force/ V to maintain rotational equilibrium compared to Figure 7.

💡 Key Knowledge

  • Re-orienting a beam from an angle to the horizontal changes the perpendicular distance from the pivot to the line of action of the weight, altering the gravitational moment.
  • When forces are strictly parallel (all vertical in Figure 9), resolving forces vertically simplifies to: upward forces equal downward forces ( V = S + 31 ).

🧠 Exam Technique

  • Structure your answer systematically: compare directions first, then magnitudes, and finally justify using moments or equilibrium conditions.
  • Be explicit: explicitly contrast "Figure 7" against "Figure 9" to ensure you earn communication and comparative credit.

❌ Common Errors

  • Vagueness: stating "forces are bigger" without specifying *which* force ( V vs U ) and *why*.
  • Ignoring directional aspects: failing to mention that V is purely vertical whereas U had a directional angle θ to the vertical.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.