A cylinder is suspended from the end of a string. The cylinder is stationary in water with the axis of the cylinder vertical, as shown in Fig. 2.1. [Figure 2.1] The cylinder has weight 0.84 N, height h and a circular cross-section of diameter 0.031 m. The density of the water is 1.0 × 10³ kg m⁻³. The difference between the pressures on the top and bottom faces of the cylinder is 520 Pa. The string is now used to move the cylinder in (a) vertically upwards through the water. The variation with time t of the velocity v of the cylinder is shown in Fig. 2.2. [Figure 2.2] The top face of the cylinder is at a depth of 0.32 m below the surface of the water at time t = 0. The cylinder in (b) is released from the string at time t = 4.0s. The cylinder falls, from rest, vertically downwards through the water. Assume that the upthrust acting on the cylinder remains constant as it falls.
- (a)(i)Calculate the height h of the cylinder. h = .............................................. m[2]
- (a)(ii)Show that the upthrust acting on the cylinder is 0.39N.[2]
- (a)(iii)Calculate the tension T in the string. T = .............................................. N[1]
- (b)(i)Use Fig. 2.2 to determine the acceleration of the cylinder at time t = 2.0s. acceleration = .............................................. ms⁻²[2]
- (b)(ii)Use Fig. 2.2 to determine the depth of the top face below the surface of the water at time t = 4.0 s. depth = .............................................. m[2]
- (c)(i)State the name of the force that acts on the cylinder when it is moving and does not act on the cylinder when it is stationary.[1]
- (c)(ii)State and explain the variation, if any, of the acceleration of the cylinder as it falls downwards through the water.[2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [C1] (Δ)p = ρg(Δ)h
- [A1] 520 = 1000 × 9.81 × h h = 0.053 m
- [C1] (upthrust =) (Δ)p × A
- [A1] = (Δ)p × π(d/2)² or (∆)p × πr² = 520 × π(0.031 / 2)² = 0.39 (N)
- [A1] T = 0.84 – 0.39 = 0.45 N
- [C1] a = (v – u) /t or (△)v/ (△)t or gradient
- [A1] = e.g. 8.0 × 10-2 / 2.0 = 4.0 x 10-2 m s-2
- [C1] distance = (1/2 × 2.5 × 0.10) + (1½ × 1.5 × 0.10) or (1/2 × 4.0 × 0.10) (= 0.20 (m))
- [A1] depth = 0.32 – 0.20 = 0.12 m
- [B1] viscous (force)
- [B1] viscous force increases (with speed/time/depth)
- [B1] (so) acceleration decreases
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