Exampro a-level Physics (7407/7408)

Part (b) (ii) proved to be too difficult for many candidates.

E8. (a) (i) Most candidates completed this successfully.
(ii) The majority did this correctly using mgh = ½ mv². The situation was not one of free fall with constant acceleration so the use of v² = u² + 2as, although giving the same answer, was inappropriate.
(iii) There was a good proportion of correct responses but many did not know the relationship between energy and power. Others failed to get the correct answer because they ignored the fact that there were four generators or because they could not convert correctly from MW to W.

(b) It was disappointing that many candidates failed to identify losses due to physical factors in the pumping, generating and transmission systems. When friction was mentioned, answers lacked depth and commonly included phrases such as ‘energy is lost to friction’. Relatively few stated clearly where the frictional force existed. Inefficiency in the pump or generator was often mentioned without any detail of the causes. Weak responses included evaporation of water, leakage though soil or animals drinking some of the water.

E9. (a) Use of p.e. = mgh was common here but the majority of candidates did use ½kx² or ½Fx to calculate the correct elastic potential energy. A significant minority of candidates were penalised for quoting the answer to a single significant figure (0.03J).

(b) (i) Although most candidates appeared to be able to distinguish the amplitude from twice its value, many failed to gain this mark be marking a vague distance that clearly was not equivalent to the distance between two adjacent arrow lines.

(ii) Too often candidates were vague regarding use of the terms elastic potential energy and gravitational potential energy – calling them both “potential energy”. Other candidates made no mention of the conversion between potential energy and kinetic energy. A few weaker candidates made no mention of energy at all – simply writing about the period of oscillation and damping.

E10.Candidates failing to equate kinetic energy with change in potential energy in part (a) made little useful headway. Such candidates attempted to use the expression for the velocity of a particle performing harmonic motion in terms of amplitude and angular frequency. Since the amplitude of the oscillator is equal to 90° the mass does not perform harmonic motion and such a calculation is invalid.

In part (b) the velocity of the mass is easily determined from the kinetic energy.

Most candidates were able to determine the magnitude of the centripetal force in part (c).

In part (d) the magnitude of the tension in the string is equal to the sum of the centripetal force and the weight of the oscillating mass. Many candidates assumed, incorrectly, that the tension in the string was equal to the weight of the mass.

E11.Candidates had problems with this question. The majority did not resolve the weight and simply quoted the actual weight. There was some confusion as to the meaning of 1.0 m in 15 m. Those candidates who did resolve the weight often gave the component at right angles to the slope rather than along the slope.

The calculation of power in part (a)(ii) proved easier for most candidates and, with the allowance of an error carried forward, the majority of candidates picked up marks here.

Part (b) also caused much confusion. Many candidates were not sure how to begin. Those candidates who had some idea often used the equations of motion with the acceleration equal to 9.81 m s^–2.

E12.Only a few candidates scored high marks on this question. In parts (a)(i) and (a)(ii) the more able candidates drew the three graphs correctly, taking more care than has been the case in the past. It was often not possible to establish exactly which of the three graphs was being referred to in the many confused answers to part (a)(iii).

Part (b)(i) caused much confusion and many candidates did not know how to include the effects of the circular track. Attempting to work out angular velocity was a common error. Parts (b)(ii) and (iii) were much belier. Most candidates found part (b)(iv) to be conceptually very difficult.

E13.The responses to this question were generally good and candidates were able to obtain the correct answer to part (a) and conduct a meaningful discussion in part (b) in terms of energy changes. Unfortunately a minority of candidates did use the equations of uniform acceleration in an incorrect context in this question.

E14.The calculations in parts (a) and (b) were well done although the unit for momentum produced the usual problems. Explaining the crumple zone in part (c) was often well answered although some candidates’ answers did tend to lack focus. The idea that the time duration of the collision was increased and that this was important, seemed to be well understood.

E15.The majority of candidates gained maximum marks in both parts (a)(i) and (a)(ii), but a few candidates, who scored reasonably well elsewhere, lost marks in part (ii) as a result of attempting unnecessarily complicated solutions involving expressions for loss of potential energy and gain of kinetic energy. In part (iii), candidates often knew the correct equation for efficiency but used incorrect data.
Candidates often gained both marks in part (b)(i) but gave confused and incorrect answers in part (ii), with many of the weaker candidates attempting to convert from rms to peak values or vice versa. Some candidates failed to recognise that the output current from the generators was the input current to the transformer.

E16. Answers to part (a) caused no serious difficulty and usually gained both marks by correct substitution of values into the well-known equation. Part (b) provided a greater challenge, but was usually met with partial success by the use of v_max = 2πfA. Many candidates attempted to produce the required two values by using this equation twice, once for v_max (by substituting 0.18 m) and then for A (by substituting 0.63 m s^–1, which was also given in the question). This gained only two marks. It was necessary to break into the circular argument, either by energy conservation (giving v_max) or by use of Pythagoras (giving A), to access all four marks.

Most candidates were unable to marry oscillatory motion with the circular motion content of Unit 4 in order to solve part (c). In the vast majority of the work submitted this was treated as an equilibrium problem, with the tension equated to mg. A small minority of candidates, realising that centripetal force was involved, introduced mω²r rather than mv²/r. This approach was seldom successful, because of confusion between ω as the angular frequency of the SHM (which is constant) and ω as the angular velocity of the circular motion of the mass (which is not constant in this case).

E17. This question produced many high marks. The usual error was failing to calculate the area correctly in m², usually through multiplying by 10³ instead of 10⁶. Only the very best candidates realised in part (ii) that the centre of gravity of the water dropped by 5 m and not 10 m. In part (iii) most candidates knew how to calculate the average loss of gravitational potential energy per second and how to use the efficiency correctly.

E18. A significant proportion of weaker candidates were not able to select the correct equations to calculate the loss in potential energy and gain in kinetic energy, which is a little surprising given that these have been tested frequently in the past and do not usually cause problems. Significant figure errors were very common in this question with many candidates quoting the changes in kinetic and potential energies to five and six significant figures respectively. Part (b) proved to be successful with a pleasing number of candidates able to calculate the force of air resistance from the energy differences.

E19. This question provided excellent discrimination and more able candidates analysed the situation successfully. Less able candidates were confused with the distinction between mass and weight and consequently calculated the acceleration as 0.50 m s^–2. Calculating the stopping distance proved more straightforward, although there was the usual confusion of signs for acceleration and whether the given velocity of 18 m s^–1 was the initial or final velocity. Most candidates were able to identify friction as a reason for the stopping distance being shorter in practice, but were less confident when explaining why.

E20. Parts (a) and (b) of this question involved a familiar theme of energy conversion that has been assessed before. For the most part, candidates identified the energy changes correctly. The calculation of the horizontal velocity did however, cause problems. This was because many candidates applied the equations of uniform acceleration which were not appropriate in this situation as the acceleration changed.
Parts (c) and (d) proved to be challenging. Only a few of the more able candidates were able to identify the component of the weight parallel to the slope as the cause of the acceleration. A significant proportion of candidates concentrated on the changing frictional forces as an explanation of decreased acceleration. Part (d) proved to be too difficult for less able candidates who became very confused by the horizontal velocity and tried to use it in calculations for motion in the vertical plane.
Projectile questions of a similar type have been asked before but the impression given was that candidates found the analysis more difficult this time round. The diagram did, unfortunately, include an incorrect label of a vertical height of 1.0 m. There was, however, no evidence that this caused confusion and candidates were given full credit if they used this value for vertical displacement.