A pendulum consists of a solid sphere suspended by a string from a fixed point P, as shown in Fig. 3.1. [Figure 3.1] The sphere swings from side to side. At one instant the sphere is at its lowest position X, where it has kinetic energy 0.86 J and momentum 0.72 Ns in a horizontal direction. A short time later the sphere is at position Y, where it is momentarily stationary at a maximum vertical height h above position X. The string has a fixed length and negligible weight. Air resistance is also negligible.
- (a)On Fig. 3.1, draw a solid line to represent the displacement of the centre of the sphere at position Y from position X.[1]
- (b)Show that the mass of the sphere is 0.30 kg.[3]
- (c)Calculate height h.[2]
- (d)The distance between point P and the centre of the sphere is 0.93 m. When the sphere is at position Y, the string is at an angle θ to the vertical. Show that θ is 47°.[1]
- (e)For the sphere at position Y, calculate the moment of its weight about point P.[2]
- (f)State and explain whether the sphere is in equilibrium when it is stationary at position Y.[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [B1] solid straight line drawn between centre of sphere at X and at Y
- [C1] p = mv or 0.72 = mv
- [C1] E = ½mv² or 0.86 = ½mv²
- [A1] (m =) 0.72² / (2 × 0.86) = 0.30 (kg)
- [C1] v = 2EK/p
- [A1] v = (0.86 × 2) / 0.72 = 2.4 (to 2 s.f.)
- [A1] m = 0.72/2.4 = 0.30 (kg)
- [C1] (Δ)E = mg(Δ)h
- [A1] h = 0.86 / (0.30 × 9.81)
- [A1] = 0.29 m
- [A1] cosθ = (0.93 – 0.29)/0.93 so θ = 47°
- [C1] moment = (0.30 × 9.81) × 0.93 × (sin 47° or cos 43°)
- [A1] = 2.0 N m
- [A1] moment = (0.30 × 9.81) × [0.93² – (0.93 – 0.29)²]⁰·⁵
- [A1] = 2.0 N m
- [B1] there is a resultant force (acting on sphere) or there is a resultant moment (about P acting on pendulum) (so) not in equilibrium
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