A student does an experiment to find the mass of a metre rule. (a) A fixed mass is placed on the metre rule at the 5.0 cm mark as shown in Fig. 3.1. [Figure 3.1] The student: • places the pivot below the 25.0 cm mark on the metre rule • places a mass of mass m = 20g on the metre rule • adjusts the position of the 20g mass until the rule is as close to balance as possible.
- (a)(i)The position of the 20g mass when the rule is as close to balance as possible is shown in Fig. 3.2. [Figure 3.2] Using Fig. 3.2, determine the distance d of the centre of the 20g mass from the pivot when the metre rule is as close to balance as possible. d = .................................................... cm[1]
- (a)(ii)Describe the technique the student uses to balance the metre rule with the 20g mass.[1]
- (a)(iii)• The student repeats the procedure for values of mass m = 30g, 40g, 50 g and 60 g. Table 3.1 shows the results. Add your value of d for mass m = 20 g in (a)(i) to Table 3.1. Calculate 1/d for each mass m, and record all values in Table 3.1. Give your answers to an appropriate number of significant figures. [Table 3.1][2]
- (a)(iv)Suggest why a value of d cannot be found for mass m = 10g.[1]
- (b)(i)Using the grid provided in Fig. 3.3 on page 13, plot a graph of 1/d on the y-axis against m on the x-axis. Start your axes from the origin (0, 0). Draw the straight line of best fit.[4]
- (b)(ii)Calculate the gradient G of your line. Show all your working, and indicate on the graph the values you use. G = ....................................................[2]
- (c)The mass M of the metre rule can be calculated using the equation: M = 160 - 0.040 / G Use your value of G in (b)(ii) to calculate M. M = .................................................... g[1]
- (d)The student is given a piece of modelling clay. He places it on the metre rule as shown in Fig. 3.4. He finds that the metre rule is balanced when the modelling clay is a distance of 40.0 cm from the pivot. [Figure 3.4] Using your graph in Fig. 3.3 on page 13, find the mass of the piece of modelling clay. Show your working. mass of piece of modelling clay = .................................................... g[2]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.
- [1] 3(a)(i) d = 57 cm
- [1] 3(a)(ii) Moves the 20 g mass backwards and forwards or slowly along the rule until the balance achieved.
- [1] 3(a)(iii) 1/d calculated for all values (0.018, 0.027, 0.036, 0.045, 0.056)
- [1] 3(a)(iii) All to 2 or 3sf
- [1] 3(a)(iv) d greater than 75 cm / off the end of the rule / no balance point / ruler not long enough
- [1] 3(b)(i) Axes labelled, quantity and unit
- [1] 3(b)(i) Scales linear, not awkward, start from (0,0)
- [1] 3(b)(i) Points plotted accurately, to the nearest ½ square
- [1] 3(b)(i) Thin best-fit straight line drawn
- [1] 3(b)(ii) values indicated, ≥ ½ the line used (m > 25)
- [1] 3(b)(ii) G = 9.6 (± 1.0) ×10-4 or 0.00096 ± 0.0001
- [1] 3(c) 110g < M < 130g
- [1] 3(d) Correct reading from candidate graph at 1 / d = 0.025
- [1] 3(d) mass = 28.0 ± 3.0g
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