M8.3 Weight & gravity

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), there were many correct answers seen for the equation of motion of Q. However, a significant number of candidates did not seem to know what constituted an equation of motion, sometimes making an attempt at a suvat equation, despite then using it correctly in part (b).

Common mistakes

• Q: 3g − T = 3a (two terms on the left, weight in g)
• a suvat equation is not an equation of motion
• including m (3mg − T = 3ma) loses the A mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The most common error in part (b) was to include a weight term in the horizontal equation of motion for P i.e. writing T-2g = 2a rather than T = 2a. Sometimes the direction of the acceleration was not treated consistently in the two equations resulting in a sign error in one of them. A relatively small number of candidates used a whole system approach but the requirement to use a combined mass in the ‘ma’ term ruled out some attempts.

Common mistakes

• P on the horizontal table: T = 2a (no weight term: T − 2g = 2a was the most common error)
• or whole system 3g = 5a with combined mass 5
• a = 3g/5 = 5.9 m s⁻² (2 or 3 s.f.)
• inconsistent directions in the two equations

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (c) was generally well attempted with the majority of candidates using v2 = u2+2as to find the required speed. Although most used their acceleration found in part (b) as required, a small minority substituted a = g despite it not following from previous work. A few confused u and v by assuming the final velocity when Q hit the ground was zero, but such instances were rare. The answer was required to 2 or 3 significant figures following the use of g = 9.8 ms-2.

Common mistakes

• v² = 2 × their a × 0.4 gives 2.2 m s⁻¹ (2 or 3 s.f.)
• use their a from (b), not g
• v is not 0 at the floor

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was generally well done with some making good use of the diagram. After identifying the three forces, successful responses resolved vertically to form an equation with the vertical component of 5 N clearly found. This was a ‘Show that…’ question and so those who just wrote ‘3’ down, with no supporting evidence of how it had been obtained, received no reward. Very occasionally cosine was used instead of sine and a few learners used the value of g as 9.81, which meant that the accuracy mark was not awarded.

Common mistakes

• resolve vertically: 5 sin α + R = 2g
• the vertical component 5 × 3/5 = 3 must be shown, just writing 3 scores nothing
• sin/cos confusion
• g = 9.8 not 9.81

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (c) was well answered and the correct result of 2g or 19.6 (N) was stated for the magnitude of the reaction once the force of 5 N had been removed. Common incorrect answers were 2 or 16.6.

Common mistakes

• R = 2g = 19.6 N once the 5 N force is removed
• common wrong answers 2 and 16.6

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (d), learners needed to plan their method carefully. Most responses gained the mark for finding the value of the friction as 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚= 𝜇𝜇𝜇𝜇= 0.2 x 19.6. Successful responses then applied Newton’s 2nd law, ±𝐹𝐹= 𝑚𝑚𝑚𝑚= 2𝑎𝑎 and then used a suvat approach to arrive at the distance as 4.1 or 4.08(m) to 2 or 3 s.f. Leaving the final answer to more than 3 s.f resulted in the final mark not being awarded. Errors were mainly in the 𝐹𝐹= 𝑚𝑚𝑚𝑚 equation when the force was incorrectly formed. Use of a = 0.4g or a mu = - were common mistakes. Fractional answers, after the use of g = 9.8, were not acceptable.

Common mistakes

• F = 0.2 × 19.6 (use the new reaction, not 16.6)
• −F = 2a gives a = −0.2g
• wrong a = μ or 0.4g
• then v² = u² + 2as
• answer 4.1 or 4.08 m to 2 or 3 s.f., not 40/g

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a) there were many excellent responses seen. Most responses gave an expression or an actual value for t by considering the horizontal motion and then substituted it into their equation for vertical motion. Many gained full marks, but some used a heavily rounded value for t and lost accuracy. Others were not clear whether they were taking upwards or downwards as positive in the vertical direction and this led to two common errors. The first was that both terms in their expression for vertical displacement had the same sign. The second was, if they were taking upwards as positive as most did, they replaced ‘s’ with H in their equation and erroneously reached the result H = 30 - and then just wrote H = 30 which resulted in not gained the final accuracy mark. Other errors included poor substitution into formulae, forgetting to resolve the velocity into its components and sin/cos confusion. Quoting and using the equation of the path was rarely seen. Some responses erroneously used half of their time from part (a), treating the motion as if it were symmetrical and gained no marks. The energy solution was rarely seen.

Common mistakes

• sin θ = 1/√5, cos θ = 2/√5
• t from horizontal motion 14cos θ × t = 40, then vertical motion
• keep a consistent positive direction (H = −30 must become 30 correctly)
• do not halve the time (not symmetrical)
• premature rounding of t

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a) most wrote down the correct answer of 3Mg for the reaction force using vertical resolution. The most common error was omitting the weight term for the rod and a few included trigonometric functions in their resolutions.

Common mistakes

• resolve vertically: R = Mg + 2Mg = 3Mg
• omitting the weight of the rod
• no trig needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a very friendly starter. (a) The vast majority of candidates were able to score the mark, with answers of 0.5g or 4.9 being the most common. The most common error was to omit the g whilst a few used g = 9.81 or gave a fractional answer, both of which were penalised once for the whole question. A few, seeing the word ‘magnitude’, squared and/or square rooted their answer. for (a) by 2 7 which earned the M mark and then realised that X had to be the same value. A significant number lost the A mark after having found the correct value of F for not stating a value for X, as the question required.

Common mistakes

• R = weight = 0.5g = 4.9 N (vertical equilibrium)
• omitting g
• using g = 9.81 or a fraction 49/10 is penalised once
• squaring or square-rooting because of "magnitude"

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), there were surprisingly few diagrams seen and this would have helped candidates to understand the situation. The most common error was to set v = 0 leading to a height of 78.4 m. A few used t = 5 s or t = 10 s for both the up part and the down part of the motion. The most successful approach was the “whole motion” one, possibly because it is only the most able students who ever really grasp this. A significant number of candidates gave a negative answer. Many candidates, in the second part, managed to stumble across 2.5 s for half the time m s- or to treat this speed as a height. Some candidates 1 distance from part (a) along with 24.5 didn’t seem to realise that the speed of the stone reduced as it went upwards. In part (c), “air

Common mistakes

• whole-motion s = ut + ½at² with u = 39.2, a = −9.8, t = 10 gives s = −98, height 98 m (must be positive)
• setting v = 0 gives 78.4 (height above O only)
• up-and-down method needs 78.4 and 176.4 combined
• a diagram helps

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

answer. Many candidates, in the second part, managed to stumble across 2.5 s for half the time but then failed to either find the other half or to double it. Some of the better candidates found t =1.5 s and t = 6.5 s and then subtracted which showed a clear understanding of the situation. Some successfully combined the various elements of times up and times down but the most successful approach was to find half the time and double it. A common error was to use the m s- or to treat this speed as a height. Some candidates 1 distance from part (a) along with 24.5

Common mistakes

• speed ≤ 24.5 either side of the top: 0 = 24.5 − 9.8t gives 2.5 s, double to 5 s (or times 1.5 and 6.5 s and subtract)
• forgetting to double
• treating 24.5 as a height

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

didn’t seem to realise that the speed of the stone reduced as it went upwards. In part (c), “air resistance” was the most common response, followed by “use a more accurate value of g”. A common error was not to reference the model at all, so answers about how accurately the time was measured or throwing the stone indoors to stop the wind having an effect were seen, as well as “consider the mass/weight of the stone”. Some wrongly gave the negative “there is no resistance”.

Common mistakes

• include air resistance, or a more accurate value of g, spin, shape or size of the stone
• must refer to the model
• negatives ("there is no air resistance") and mass/weight of the stone score 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The vast majority were able to score the mark in part (a), with answers of 5g or 49 being the most common. A few candidates used g = 9.81 but were not penalised here. There were very many correct answers for the second part also, but some ignored the 28N force

Common mistakes

• resolve vertically: R = 5g = 49 N
• must be positive, no m in the answer

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates who had not read the question and used g as 9.8 or 9.81 were few and far between. Part (a) was generally well done with most candidates using the figures 10, 10 and 1.8 in a correct constant acceleration formula to obtain the given ‘U = 8’. Occasionally there was a sign error and although the correct answer was quoted, it did not actually follow from the working. A few used 0 as opposed to 10 for the final velocity. Some considered the motion ‘up’ and ‘down’ separately and used the distances to successfully derive the value of ‘U’. The most common approach in part (b) was to write down a quadratic equation in t and to solve it using

Common mistakes

• v² = u² + 2as with v = 10, s = 1.8 down and g = 10 (not 9.8)
• consistent sign convention
• using 0 for the final velocity
• U = 8 only

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates who had not read the question and used g as 9.8 or 9.81 were few and far between. answer. In part (c), correct refinements seemed to be in the majority but only just. A more accurate value of gravity (9.8) was the most common correct response but wind, spin and dimensions of the stone were also seen. Many reasoned that due to air resistance the acceleration was smaller making the velocity smaller. Very few considered how changing the value of a in the equation they used would affect their answer for U. Weight/mass and force were the most common incorrect answers, often being given as an incorrect extra answer to what would have been a correct one, and as a result losing the B mark. In the final part, correct

Common mistakes

• more accurate value of g (9.8), wind, spin, size/shape of stone (not a particle)
• weight, mass or force as an extra loses the mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although fully correct answers were seen, it was clear that many candidates had little experience of answering this type of question. In particular, there were many equations seen helped some candidates, but these were not often forthcoming. In part (a), many candidates gained all 3 marks. The most common error occurred where the direction of travel had been mixed up. For those who set up an equation of motion for the cage and block together, the most common mistake was ' 50g - T = 50 x 0.2' (as if it was accelerating downwards). Some students attempted 'F = ma' for the cage, but omitted 'R', scoring zero marks as there were an incorrect number of terms. There were instances of numerical mistakes in calculating '50 x 0.2' ('25' being a common one). Most students used the correct value of 9.8 for g. There were a few cases where g was omitted from the weight or added incorrectly in the 'ma' term. In part (b), few students who had made a good attempt in part (a). For students who knew what to do here, there were a few instances of '10g - R = 10 x 0.2'. Some students attempted 'F = ma' for the cage rather than the block, but omitted 'T' or they tried to combine the cage and the block. For those students scoring zero marks for Q4, their responses were either completely blank, or they tried to equate forces, ignoring the fact that the cage and block were accelerating.

Common mistakes

• whole system: T − 50g = 50 × 0.2 gives T = 500 N (g = 9.8)
• direction of motion mixed up (50g − T = 50 × 0.2)
• g omitted from weight or put in the ma term

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although fully correct answers were seen, it was clear that many candidates had little experience of answering this type of question. In particular, there were many equations seen with an incorrect number of terms, particularly in part (b). Clear force diagrams would have helped some candidates, but these were not often forthcoming. In part (a), many candidates where g was omitted from the weight or added incorrectly in the 'ma' term. In part (b), few candidates gained the full 3 marks. Some failed to identify the forces which were acting on the block; others considered the forces acting on the cage rather than the block. A common error was to get the direction of motion mixed up. There were quite a few blanks here, even by students who had made a good attempt in part (a). For students who knew what to do here,

Common mistakes

• block alone: R − 10g = 10 × 0.2 gives 100 N (or cage: T − 40g − R = 40 × 0.2)
• wrong number of terms, e.g. omitting R or including T for the block
• draw a force diagram

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(a) Many candidates just found the time to the top, losing both marks.

Common mistakes

• displacement 0 at time T: 0 = 14.7T − 4.9T² gives T = 3
• finding only the time to the top (1.5 s) scores nothing
• use g = 9.8

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(b) There were very few fully correct answers because most candidates did not realise that they had to consider the change in direction and just put t = 4 into a suvat equation. Some found the distance to the top and then used t = 2.5 to find the remaining distance but then did not round the answer to 2 or 3 sf following use of g = 9.8. A few used g = 9.81 or 10 despite the clear instruction on the front of the paper.

Common mistakes

• distance, not displacement: the stone changes direction at t = 1.5, so add 11.025 up and 30.625 down = 41.7 m
• substituting t = 4 into one suvat equation gives the displacement (19.6) only
• 2 or 3 s.f. after g = 9.8

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(c) Candidates had very little idea – considering the mass was the most frequent incorrect response. Some quoted air resistance even though that had been specifically excluded in the question.

Common mistakes

• e.g. take account of the size of the stone (spin), do not model it as a particle, use a more accurate value of g
• air resistance is excluded
• "mass" is wrong
• incorrect extra refinements score B0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

(a) There were many completely correct answers and those who took the acceleration to be positive in a different direction could also score full marks. Some had no idea where g should go so answers like ‘T – 2m = 2mga’ were sometimes seen. The mass m was occasionally left out also. A few quoted suvat equations, connecting time and distance, instead of equations of motion for P and Q but if they went on to use correct equations of motion in part (b) they scored the marks for part (a). Some had no idea at all and T never appeared.

Common mistakes

• T − 2mg = 2ma for P and 5mg − T = 5ma for Q
• g placed wrongly (e.g. T − 2m = 2mga)
• m omitted
• suvat equations are not equations of motion

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

motion for P and Q but if they went on to use correct equations of motion in part (b) they scored the marks for part (a). Some had no idea at all and T never appeared. (b)Those who had correct equations of motion generally went on to find the acceleration but some missed g in their answer. Not many progressed successfully after this - some found T because that is what is usually required but in this question, it was not necessary. Some calculated the time taken for Q to reach the ground and used it in an equation for P, not realising that they had to find the velocity that P had acquired. Of those who did make progress, some missed the h in their answer for v, and others only added 2h instead of 3h in the final step.

Common mistakes

• a = 3g/7
• P's speed when Q lands from v² = 2ah (P has risen h)
• then P moves freely under gravity: H = 3h/7
• total height 2h + h + H = 24h/7
• T is not needed
• adding only 2h, or using the time Q takes to fall, are common errors

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Candidates were instructed to use g = 10 in this question but were only penalised once if theorem and calculate the actual speed of 75 m s-. For part (c), there were several 1 possible answers, and candidates should always focus their attention on refining the model that has been used, in this case g = 10 and the stone being modelled as a particle i.e. the dimensions of the particle ignored (not the mass of the particle, which was a common wrong answer). Candidates should note that they should concentrate on giving a single correct answer to this type of question as extra incorrect answer(s) will lose the mark. The most common correct answer was ‘to use a more accurate value for g’.

Common mistakes

• e.g. g = 10 is approximate (use a more accurate g), dimensions/shape/spin of the stone, wind
• mass of the stone is a common wrong answer
• incorrect extras score B0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• v² = u² + 2as with s = 16.8 down and g = 10 (not 9.8)
• sign convention for U and 19 consistent
• allow verification

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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.

Common mistakes

• more accurate value of g, spin, wind, not a particle
• weight/mass is wrong
• any incorrect extra loses the mark

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

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

Common mistakes

• 4mg − T = 4ma and T − 3mg = 3ma give T = 24mg/7
• force on the pulley is 2T = 48mg/7 (few realised)
• omitting m loses the M mark
• do not replace a by g

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The first part was well done by the majority of candidates who used a suvat equation to find the required speed after 3 seconds.

Common mistakes

• free fall from rest with g = 10: v = 0 + 10 × 3 = 30 m s⁻¹

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (e) required a suggestion for a further refinement of the model. There were many possible acceptable answers with using a more accurate value for g, considering wind effects and allowing time for the parachute to open being some of the more popular options. A common suggestion which was not credited was to include the mass of the parachutist. If two or more answers were offered they all had to be correct to achieve the mark; this was to avoid rewarding candidates who were ‘hedging their bets’ by listing everything they could think of.

Common mistakes

• e.g. more accurate g, wind, time for the parachute to open, dimensions of the parachutist, deceleration not constant
• mass of the parachutist, upthrust, air resistance or terminal velocity are B0
• all suggestions offered must be correct

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although the first part of this question was well done by some candidates, a significant minority failed to write down an equation of motion of the correct structure for each mass. Some included a weight term in the equation for the particle moving horizontally whilst others showed no real understanding of what was required and confused equations of motion with suvat equations or omitted this part completely. Those who managed to produce two valid equations generally had no difficulty in deriving the given answer for the acceleration. Since it was a ‘show that’ question, it was important that the final answer followed from sufficient and entirely correct working; those who went straight from the two simultaneous equations to writing down the answer were not awarded the final mark. Such instances were, however, fairly rare. Some attempted a ‘whole system’ approach which was equivalent to writing down just one equation with the tension eliminated. This was deemed not sufficient for a complete method and scored a maximum of two marks.

Common mistakes

• separate equations of motion: Q: 0.6g − T = 0.6a, P: T = 0.8a (no weight term for P on the horizontal table)
• solve to a = 4.2 showing working
• a whole-system equation alone is incomplete

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a reasonably accessible starter to the Mechanics section with a modal mark of 4 out of 4, achieved by just over a third of the students. There were various possible approaches to this question which involved a ball moving up and down freely under gravity. Those who attempted a solution using one suvat equation generally included appropriate terms, but sign errors were very common. If the resulting quadratic equation was incorrect an explicit method (such as factorising or using the quadratic formula) was required for the second method mark. Some split the motion into two or three separate stages with a fair degree of success. Again, however, sign errors (showing a lack of understanding of how to incorporate directions) were often seen. Those who just found the time to the greatest height or to the same horizontal level received no credit, and use of g = 9.8 m s-2 rather than 10 m s-2 as specified in the question was penalised as an accuracy error.

Common mistakes

• one suvat for the whole flight with up positive: −2 = 9t − 5t², so 5t² − 9t − 2 = 0 and T = 2 only (reject −0.2)
• sign errors are common
• use g = 10 as given (9.8 is penalised)
• time to greatest height alone scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a), most students had some idea that it was necessary to write down an equation of motion for P in an attempt to find the value of the tension in this pulley question. Some wrote down equations for both particles at this stage. Sign errors and missing ‘m’ from terms were fairly common. Those who reached the correct expression ‘ T = 2mg – 10mg/7 ‘ often failed to simplify further or made errors when attempting to do so. An answer in terms of m and g was required and so use of g = 9.8 m s-2 lost the final mark. The only acceptable answer for part (b) was that the accelerations are the same because the string is not just in part (a). Attempts to eliminate T and solve for k often revealed some poor algebraic processing. There were several possible acceptable answers to part (d). The assumptions of the model were listed

Common mistakes

• equation of motion for P: 2mg − T = 2m × 5g/7, so T = 4mg/7 in terms of m and g (substituting g = 9.8 loses the A mark)
• sign errors and missing m are common