2022 RSS Phy Prelims P3 Ans

1 2022 Physics Prelim P3 – Mark Scheme Deduct 1 m if wrong dp/sf (per section) Deduct 1 m if wrong/no units (per section) Separate scoring for tabulation (sf and units). 1ai V0 = 4.50 V 1 Following results for Q1 are based on old motor ii V1 = 2.50 V I1 = 0.13 A 1 1 Old motors: 2.50 V – 2.70 V 0.12 A – 0.13 A (tested 5 pcs) New motors: 2.35 V – 2.50 V All 0.14 A (tested 6 pcs) (V1 = 1.85 V – 2.90 V) iii P1 = V1I1 = 2.50 x 0.13 = 0.33 W 1 bi V2 = 0.25 V I2 = 0.30 A 1 V2 = 0.10 V – 0.50 V I2 = 0.17 A accepted ii R = 0.25 0.30 = 0.83 Ω 1 c Q = 0.13 × 0.83 2.50 = 0.043 J 1 d While the shaft is spinning, raise the height of the boss until the shaft just stops spinning. Measure the length of the spring and find its extension, e. Remove the wooden rod and motor. Hang a known weight from the spring and find the extension of the spring. Repeat for 5 further different values of extension and known weights. Plot a graph of the extension, e against the weight. From the graph, read the value of the force that corresponds to the extension e. 1 1

2 OR Find k by hanging the 5.0 N weight on the spring. Measure the extension e. Calculate k using k = 5.0 e . Raise the boss till the motor just stops spinning. Measure the extension of the spring, e. Calculate with F = ke. 1 1 e Replace the spring with a spring balance. 1 2a h = 7.0 cm j = 32.2 cm k = 12.0 cm 1 h minimum 6.4 cm 8.5 cm NOK j = 26.8 cm NOK b t1 = 5.87 s, t2 = 5.99 s, t3 = 5.99s <t> = 5.87+5.99+5.99 3 = 5.95 s 1 1 Average of 2 timings OK c P1 = 5 × 0.010 6.0 = 0.0083 W 1 1 d t1 = 5.32 s, t2 = 5.20 s, t3 = 5.22s <t> = 5.32+5.20+5.22 3 = 5.24 s P2 = 2 × 0.010 5.24 = 0.00382 W 1 (after removing 3 weights, h = 3.7 cm) Working for <t> must be evident.

3 e Constant variables:  mass-hangar and weights,  change of height ∆h  length k  length j Independent variable:  weight W Dependent variable:  time taken to unwind fully, t Set up the apparatus as shown in Fig. 2.1. Carry out steps (b) and (c) to obtain a set of values of W, t and P. Repeat steps (b) and (c) to obtain 5 further sets of values of W, t and P by removing one weight from the mass-hangar for each set of reading. Tabulate the results of W, t and P. The relationship between P and W can be found by plotting a graph of P against W: 1 1 1 1 Type of weights and string used must be consistent. h, j and k must be constant as they affect the tension in the string. Dependent variable: OK if students write t and P. Graph with positive gradient (when W increases, P decreases). Accept curve graphs. P W

4 3a L0 = 0.022 m 1 bi L = 0.147 m x = 0.147-0.022 = 0.125 m x = 0.125 m 1 1 Need to show working for x. ii 1. Avoid parallax error by reading the mark on the scale of the half-metre rule at eye level when taking measurements. 2. Ensure the metre rule was vertical by aligning it to a vertical feature (eg. door frame) or the use of set square placed between the bench and the metre rule. 3. The readings of x was taken 1 1 “accuracy of x” refers to taking the readings of x, not the experimental set up. iii k = 0.300 kg ×10 N / kg 0.125 m = 24 N/m 1 1 iv E = 0.300 kg × 10 N kg × 0.125 m 2 = 0.19 J 1 c m / kg L / m x / m x2 / m2 E / J 0.050 0.035 0.013 0.00017 0.0033 0.100 0.058 0.036 0.0013 0.018 0.150 0.079 0.057 0.0032 0.043 0.200 0.100 0.078 0.0061 0.078 0.250 0.123 0.101 0.010 0.13 0.300 0.147 0.125 0.016 0.19 (accept 2dp for m/kg, need to be consistent) 1 – headings 1 – 6 sets of readings 1 – correct dp for m, L, and x. Correct sf for x2 and E 1 – correct calculations of x2 and E Mass m can be recorded in ascending or descending order. E and x2 can be 2 sf or 3sf, but must be consistent in the column. Allow 1 mistake for calculations. d See attached 1 – Axis 1 – Scale 1 – Plotting 1 – Best fit ei Gradient = 0.180−0.052 0.0148−0.0040 = 12 (2 sf) or 11.9 (3 sf) 1 1 Gradient triangle in graph must be shown as part of working

5 If no x and y labels in gradient triangle, OK. ii G = 12 and 0.5k = 0.5 x 24 = 12 N/m, which is within the limits of experimental error. Therefore G = 0.5k is supported. 1 f Method 1: Equation of line Equation of line is: E = 13x2 + 0.001 (y-intercept is 0.001) When m = 0.600 kg, the extension will be 0.125 x 2 = 0.250 m. Therefore, the energy stored will be: E = 13 (0.250)2 + 0.001 = 0.81 J Method 2: Trend From data, when the mass m is doubled, the value of E will be on average 4.9 times larger. Therefore, the predicted energy stored will be 4.9 x 0.19 = 0.93 J. Method 3: Calculation based on k Taking k from student’s results, and sub m = 0.600 kg, k = mg x , so x = mg k = …………m E = mgx 2 = …………J Method 4: Calculation based on x When m = 600 g, x will be twice for m = 300g. So x = ……….. E = mgx 2 = …………J 1 1 1 1 NO marks if students state their answer without working or reasoning. Students need to show line equation and working.