30 marks · 10 question parts · Question paper, mark scheme and examiner's report on Pearson's site
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Part (a) was well answered with most candidates using one equation but a few found the maximum height first. There were a variety of methods seen for the second part, with the most common being use of v = u + at, but a significant
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second part, with the most common being use of v = u + at, but a significant number made a sign error in their equation and 1.4 was a very common wrong answer. In part (c), a significant minority failed to understand that s = 1.2 was
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answer. In part (c), a significant minority failed to understand that s = 1.2 was required with many using 15.6. Those using one equation often made a sign error. Calculators were usually used to solve the resulting quadratic equation. The other method which seemed to produce fewer errors was to find the time to the top and the time down and then add. In part (d), very few correct graphs seen. Many
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the time down and then add. In part (d), very few correct graphs seen. Many started from the origin to produce an upside down V-shape or a trapezium. In
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part (e), it was very rare not to see lesser, with the vast majority not realising that for the ball to hit the ground with the given speed of 19.6 ms-1, it would need to be projected with a greater speed to overcome air resistance. In the final part,
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to be projected with a greater speed to overcome air resistance. In the final part, correct refinements were just in the majority. A more accurate value of g was the most common correct response with weight/mass being the most common incorrect answer. It should be remembered that, when only one refinement of the model is required, any extra answers given would all have to be correct for the mark to be awarded.
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In part (a), the equations of motion were mostly correct when attempted. A significant number of candidates obviously did not understand the model or that T was required and stopped at that point. There were a few with missing m’s but this was quite rare as were sign errors. A very small number of candidates obtained the equations but then completely misunderstood the situation and replaced a by g and lost all the marks. Only a handful realised that 2T was required to find the force exerted on the pulley by the string. The second part was
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to find the force exerted on the pulley by the string. The second part was reasonably well answered but, as in question 1(f), it should be remembered that, when only one limitation of the model is required, any extra answers given would all have to be correct for the mark to be awarded.
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In part (a), a significant minority misunderstood the question and solved v = 0 instead of a = 0. The majority, however, did multiply out and differentiate correctly to find a before then equating it to zero. A few errors in multiplying out the brackets were seen and a significant number had changed the sign of v which then had consequences in part (b).The majority, for part (b), realised that the particle the answer. A few, however, used the t value that had been found in part(a) and lost the method mark.
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had consequences in part (b).The majority, for part (b), realised that the particle would change its direction of motion when v = 0 and solved this equation to give t = 3.5. They were then able to integrate v correctly and use their t value to obtain the answer. A few, however, used the t value that had been found in part(a) and
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