Oxford Revise AQA GCSE Physics | Chapter P12 answers
P12: Newton's laws of motion
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Answers |
Extra information |
Mark |
AO / Specification reference |
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01.1 |
suitable method e.g., attach trolley to string with hanging mass over pulley or diagram keep force constant use motion sensor/light gates to measure velocities and times/acceleration change mass measure acceleration repeat several times and find mean |
accept diagram illustrating answer |
5 |
AO1 |
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01.2 |
Newton’s second law says that the acceleration of an object is proportional to the (net) force and inversely proportional to the mass |
accept F = ma |
1 |
AO1 4.5.6.2.2 |
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01.3 |
inertia is the tendency of objects to continue in their state of rest or in uniform motion/a measure of how difficult it is to change the velocity of an object |
1 |
AO1 4.5.6.2.2 |
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01.4 |
if the acceleration is inversely proportional to mass, then doubling the mass will halve the acceleration the acceleration for a mass of 0.1 kg is 40 and half of 40 is 20 |
1 |
AO3 |
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01.5 |
yes friction would reduce the resultant force which would produce an acceleration smaller than predicted by Newton’s second law |
1 |
AO2 |
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02.1 |
no there is no resultant force acting on the puck |
1 1 |
AO1 AO2 4.5.6.2.1 |
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02.2 |
no the puck does not carry the force/you need to apply a resultant force in the opposite direction (for it to stop) |
1 1 |
AO1 AO2 |
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02.3 |
no the force of the puck on the stick is the same magnitude as the force of the stick on the puck |
1 1 |
AO1 AO2 4.5.6.2.3 |
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02.4 |
yes the speed does not change, but the direction does, so the velocity changes and it accelerates |
1 |
AO3 4.5.6.2.2 |
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03.1 |
\( \begin{array}{lr} \rm{resultant\ force} = \rm{mass} \times \rm{acceleration}\\ 10\ 000 – 2000 = 8400 \times \rm{acceleration}\\ \rm{acceleration} = \frac{8000}{8400}\\ = 0.95\ \rm{m/s}^2 \end{array} \) no, it does not exceed the expected value |
1 1 1 1 1 |
AO2 4.5.6.2.2 |
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03.2 |
mass with half a load = 8400 – 1600 = 6800 kg mass with no load = 8400 – 3200 = 5200 kg \(\textrm{acceleration when half full} = \frac{8000}{6800} = 1.18\ \rm{m/s}^2\) \(\textrm{acceleration when empty} = \frac{8000}{5200} = 1.54\ \rm{m/s}^2\)it would be safe to do so with the lorry half empty but not when the lorry is completely empty. |
1 1 1 1 |
AO2 AO3 4.5.6.2 |
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04.1 |
075N, 0.4m/s2 |
1 |
AO3 |
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04.2 |
line of best fit drawn |
1 |
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04.3 |
systematic |
1 |
AO3 |
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04.4 |
the graph does not go through (0,0)/there is an intercept on the x-axis 0.15 N |
1 |
AO3 |
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04.5 |
(net) force = mass × acceleration |
accept F = ma |
1 |
AO1 4.5.6.2.2 |
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04.6 |
using a point such as (1.0, 0.89) force = 1.0 – 0.15 = 0.85 N, you need to subtract the zero error acceleration = 0.89 m/s2 0.85 = mass × 0.89 mass = 0.96 (kg) |
accept 0.85 – 1.05 with no working for two calculation marks |
1 1 1 |
AO2 4.5.6.2.2 |
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05.1 |
\( \begin{array}{lr} 40\ \rm{miles} = 40 \times 1609 = 64\ 360 \times \rm{m}\\ \rm{speed} = \frac{\rm{distance}}{\rm{time}}\\ = \frac{64\ 360}{3600}\\ = 17.9\ \rm{m/s} \end{array} \) |
1 1 1 |
AO1 AO2 |
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05.2 |
the direction |
1 |
AO1 |
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05.3 |
cyclist speed – 6m/s it is about three times faster than a cyclist |
1 |
AO1 AO2 4.5.6.1.2 |
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05.4 |
\( \begin{array}{lr} (\rm{final\ velocity})^2 – (\rm{initial\ velocity})^2 = 2\ \times\ \rm{acceleration \times distance}\\ (17.9)^2 – (0)^2 = 2\ \times\ \rm{acceleration}\ \times\ 30\\ \rm{acceleration} = \frac{17.9^2}{2 \times 30}\\ = \frac{320}{60}\\ = 5.34\\ = 5.3\ \rm{m/s}^2 \end{array} \) |
answer given to two significant figures |
1 1 |
AO2 |
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05.5 |
one mark for initial section with constant slope one mark for having this followed by horizontal section maximum velocity = 17.9 m/s labelled |
1 1 |
AO2 4.5.6.1.5 |
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06.1 |
B |
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AO2 4.5.6.2.1 |
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06.2 |
C |
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AO2 4.5.6.2.1 |
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06.3 |
force = mass × acceleration = 3500 × (-4) = -14 000 N |
two marks only if no minus sign |
1 1 1 |
AO2 4.5.6.2.2 |
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06.4 |
the track is not straight/the track is a curved/it’s rough so more friction |
1 |
AO3 4.5.6.2.1 |
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07.1 |
15 N |
1 |
AO2 4.5.1.3 |
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07.2 |
weight = mass × gravitational field strength |
1 |
AO1 4.5.1.3 |
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07.3 |
\( \begin{array}{l} 1.4 = \rm{mass \times 9.8}\\ \rm{mass} = \frac{1.4}{9.8}\\ \rm{mass} = 0.143\ \rm{kg} \end{array} \) |
1 1 1 |
AO2 4.5.1.3 |
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07.4 |
\( \begin{array}{lr} \textrm{frictional force on block} = 3.5 \textrm{N (from graph)}\\ \textrm{applied force} = 5 \textrm{N, so net force} = 5 – 3.5 = 1.5\ \textrm{N}\\ \textrm{force} = \textrm{mass} \times \textrm{acceleration}\\ 1.5 = 0.143 \times \textrm{acceleration}\\ \textrm{acceleration} = \frac{1.5}{0.143}\\ = 10.5\ \textrm{m/s}^2 \end{array} \) |
1 1 1 |
AO1 AO2 4.5.1.3 |
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07.5 |
the frictional force is less than that shown on the graph when the object is moving |
1 |
AO3 4.5.1.3 |
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08.1 |
\(\rm{acceleration} = \frac{\rm{change\ in\ velocity}}{\rm{time\ taken}}\) | \(\rm{accept\ a} = \frac{\rm{v-u}}{\rm{t}}\) |
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AO1 4.5.6.1.5 |
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08.2 |
\( \begin{array}{lr} \rm{a} = \frac{10\ 000}{5} \times 60\\ = 33.3\ \rm{m/s}^2 \end{array} \) |
1 1 1 |
AO1 4.5.6.1.5 |
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08.3 |
\( \begin{array}{lr} (\textrm{final velocity})^2 – (\textrm{initial velocity})^2 = 2 \times \textrm{acceleration} \times \textrm{distance}\\ (10\ 000)^2 – (0)^2 = 2 \times 33.3 \times \textrm{distance}\\ \textrm{distance} = \frac{10\ 000^2}{2 \times 33.3}\\ = 1\ 501\ 502\ \rm{m}\\ = 1.5 \times 10^6\ \rm{m} \end{array} \) |
answer is given in standard form |
1 |
AO2 4.5.6.1.5 |
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08.4 |
distance = speed × time 3.8 × 105 × 103 = 10 000 × time time = 38 000 s = 10.5 hours |
accept d = s × t |
1 |
AO1 AO2 4.5.6.1.2 |
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08.5 |
sensible suggestion for less time e.g., as you get close to the Moon, its gravity will accelerate you, increasing the speed and decreasing the time sensible suggestion for more time e.g., as you move away from the Earth, the Earth’s gravity still acts on you, decreasing the speed, increasing the time |
1 1 |
AO3 4.5.1.3 |
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09.1 |
weight = mass × gravitational field strength of object = 110 × 9.8 = 1078 N |
1 1 |
AO1 AO2 |
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09.2 |
force = mass × acceleration = 110 × 2.0. = 220 N |
1 1 1 |
AO1 AO2 4.5.6.2.2 |
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09.3 |
total force = 1078 + 220 N = 1298 N |
1 1 |
AO2 4.5.6.2.2 |
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09.4 |
the object will not move the upwards force is equal to the weight. there is no resultant force |
1 1 |
AO3 |
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10.1 |
place one light gate at the top of the ramp place the other light gate at the bottom place the trolley at the top of the ramp, immediately before the first light gate and release the trolley |
accept any sensible description |
1 1 |
AO1 4.5.6.2.2 |
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10.2 |
the force acting on the trolley is the component of the weight acting down the slope the magnitude of the component depends on the angle of the slope, which depends on the height as the height increases the force increases |
must include ‘component’ must include ‘angle’ |
1 1 1 |
AO2 4.5.6.2.2 |
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10.3 |
no there could be uncertainty in the positioning/height of the top of the ramp |
do not accept changes in acceleration |
1 1 |
AO2 4.5.6.2.2 |
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10.4 |
Points plotted at: (0, 0.00), (15, 2.59), (20, 3.42), (25, 4.22), (30, 5.00), (35, 5.73), (40, 6.43) |
one mark for up to four points plotted correctly one mark for rest of points correctly one mark for line of best fit |
3 |
AO2 AO3 |
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10.5 |
eventually the ramp will be vertical/the trolley will just fall it will be a value equal to the acceleration due to gravity |
accept values 9.81/10 m/s2 |
1 1 |
AO3 |
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11.1 |
force = mass × acceleration |
1 |
AO1 4.5.6.2.2 |
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11.2 |
conversion of both masses to the same units leafhopper = 2 × 10–6 × 1000 = 2 × 10–3 N cheetah = 50 × 5.0 = 250 N \(\textrm{the force produced by the cheetah is } \frac{250}{2 \times 10^{-3}} = 125\ 000 \textrm{ times bigger}\) |
1 1 + 1 1 + 1 1 |
AO1 AO2 4.5.6.2.2 |
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11.3 |
\( \begin{array}{l} \textrm{if acceleration is proportional to top speed then }\frac{\rm{acceleration}}{\rm{top\ speed}} = \rm{constant}.\\ \textrm{for the leafhopper }\frac{\rm{acceleration}}{\rm{top\ speed}} = \frac{1000}{4} = 250\\ \textrm{for the cheetah }\frac{\rm{acceleration}}{\rm{top\ speed}} = \frac{5}{30} = 0.17 \end{array} \) no, they are not directly proportional |
1 1 |
AO3 4.5.6.2.2 |
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11.4 |
F = 70 × 1000 = 70 000 N car = 40 kN = 40 000 N the suit has a force nearly twice that of a car. |
1 1 1 1 |
AO2 AO3 4.5.6.2 |
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12.1 |
force = mass × acceleration Ftesla = 1611 × 4.79 = 7717 N Faudi = 1565 × 3.78 = 5916 N FBMW = 1864 × 5.59 = 10 420 N no, the forces produced by the engines are not the same |
1 1 1 1 1 |
AO2 AO3 4.5.6.2.2 |
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12.2 |
3000 N/3 kN |
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AO2 4.5.6.2.1 |
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12.3 |
\( \begin{array}{lr} \rm{net\ force} = – 3000\ \rm{N}\\ -3000 = 1565 \times\ \rm{acceleration}\\ \rm{acceleration} = – \frac{3000}{1565}\\ = -1.92\ \rm{m/s}^2 \end{array} \) |
1 1 1 |
AO1 AO2 4.5.6.2 |
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13.1 |
when two objects interact, the forces they exert on each other are equal and opposite |
1 |
AO1 4.5.6.2.3 |
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13.2 |
the exhaust gases pushing downwards produces force of equal magnitude acting in the opposite direction |
one mark for equal magnitude one mark for opposite direction |
2 |
AO2 4.5.6.2.3 |
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13.3 |
no Newton’s Third Law applies to two different objects interacting, and the student is talking about one object. |
1 1 |
AO2 4.5.6.2.3 |
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14 |
Level 3: Describes effect of the forces in each case, and that the ball on track B ends up with the higher average speed due to the additional acceleration at the start. |
5-6 |
AO1 AO2 AO3 |
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Level 2: Describes effect of the forces in each case in accelerating and decelerating the ball, acknowledging that the ball on track B travels faster. |
3-4 |
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Level 1: Describes how the force of gravity produces an acceleration of each ball and deduces that the balls reach the end of the track at the same time. |
1-2 |
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