A-LevelPhysicsWork, energy and power9702/22 · Oct/Nov 2020 · Q313 Marks

A spring is fixed at one end and is compressed by applying a force to the other end. The variation of the force F acting on the spring with its compression x is shown in Fig. 3.1. [Figure 3.1] A compression of 0.045m is produced when a force F₁ acts on the spring. The spring has a spring constant of 800 Nm⁻¹. A child's toy uses the spring in (a) to launch a ball of mass 0.020 kg vertically into the air. The ball is initially held against one end of the spring which has a compression of 0.045m. The spring is then released to launch the ball. The kinetic energy of the ball as it leaves the toy is 0.72 J. The ball in (b) leaves the toy at point A and moves vertically upwards through the air. Point B is the position of the ball when it is at maximum height h above point A, as illustrated in Fig. 3.2. [Figure 3.2] The gravitational potential energy of the ball increases by 0.60 J as it moves from A to B.

(a)(i)Determine F₁. F₁ = .............................................. N[2]
(a)(ii)Use Fig. 3.1 to show that, for a compression of 0.045m, the elastic potential energy of the spring is 0.81 J.[2]
(b)(i)The toy converts the elastic potential energy of the spring into the kinetic energy of the ball. Use the information in (a)(ii) to calculate the percentage efficiency of this conversion. efficiency = .............................................. %[1]
(b)(ii)Determine the initial momentum of the ball as it leaves the toy. momentum = .............................................. N s[3]
(c)(i)Calculate h. h = .............................................. m[2]
(c)(ii)Determine the average force due to air resistance acting on the ball for its movement from A to B. average force = .............................................. N[2]
(c)(iii)When there is air resistance, the ball takes time T to move from A to B. State and explain whether the time taken for the ball to move from A to its maximum height will be more than, less than or equal to time T if there is no air resistance.[1]

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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] F = kx
[A1] F₁ = 800 × 0.045 = 36 N
[C1] (E =) ½kx² or ½Fx or area under graph
[A1] ½ × 800 × (0.045)² or ½ × 36 × 0.045 = 0.81 (J)
[A1] efficiency = (0.72 / 0.81) × 100 = 89%
[C1] E = ½mv²
[C1] p = mv
[A1] 0.72 = ½ × 0.020 × v² and p = 0.020 × v p = 0.17 N s
[C1] (Δ)Ε = mg(△)h
[A1] h = 0.60 / (0.020 × 9.81) = 3.1 m
[C1] F = (0.72 – 0.60) / 3.1
[A1] = 0.039 N
[B1] resultant force on ball is less (than that with air resistance) so time (taken) is more (than T)

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