State two conditions for an object to be in equilibrium. 1. .................................................................................................................................... 2. ....................................................................................................................................
- (a)State two conditions for an object to be in equilibrium. 1. .................................................................................................................................... 2. ....................................................................................................................................[2]
- (b)(i)A sphere of weight 2.4N is suspended by a wire from a fixed point P. A horizontal string is used to hold the sphere in equilibrium with the wire at an angle of 53° to the horizontal, as shown in Fig. 3.1. [Figure 3.1] Calculate: 1. the tension T in the wire T = ..................................................... N 2. the force F exerted by the string on the sphere. F = ..................................................... N[2]
- (b)(ii)The wire has a circular cross-section of diameter 0.50 mm. Determine the stress σ in the wire. σ = ..................................................... Pa[3]
- (c)(i)The string is disconnected from the sphere in (b). The sphere then swings from its initial rest position A, as illustrated in Fig. 3.2. [Figure 3.2] The sphere reaches maximum speed when it is at the bottom of the swing at position B. The distance between P and the centre of the sphere is 75cm. Air resistance is negligible and energy losses at P are negligible. Show that the vertical distance h between A and B is 15 cm.[1]
- (c)(ii)Calculate the change in gravitational potential energy of the sphere as it moves from A to B. change in gravitational potential energy = ..................................................... J[2]
- (c)(iii)Use your answer in (c)(ii) to determine the speed of the sphere at B. Show your working. speed = ..................................................... m s⁻¹[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [B1] resultant force (in any direction) is zero
- [B1] resultant torque/moment (about any point) is zero
- [C1] T sin 53° = 2.4
- [A1] T = 3.0 N
- [C1] F = T cos 53° or F² = T² – 2.4²
- [A1] F = 1.8 N
- [C1] σ= ΤΙΑ or σ= FIA
- [C1] Α = πd²/4 or A = πr²
- [A1] σ = 3.0 × 4 / [π × (0.50 × 10⁻³)²] = 1.5 × 10⁷ Pa
- [A1] h = 75 – 75 sin 53° = 15 cm
- [C1] (Δ)Ε = mg(△)h or (△)E = W(A)h
- [A1] (Δ)Ε = 2.4 × 15 × 10⁻² = 0.36 J
- [B1] E = ½mv²
- [C1] 0.36 = ½ × (2.4 / 9.81) × v²
- [A1] v = 1.7 m s⁻¹
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →