Oxford Revise AQA A Level Physics | Chapter P7 answers
P7: Force, energy, and momentum 2
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Question |
Answers |
Extra information |
Mark |
AO |
Spec reference |
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01.1 |
\( \begin{array}{l} v = 30\textrm{ km hr}^{-1} = \frac{130 \times 1000\rm{\ m}}{60 \times 60} = 36 (36.1) \textrm{ m s}^{-1}\\ \textrm{Use of } v = u + at \textrm{ and } u = 0\\ a = \frac{v – u}{t} = \frac{36}{1.9} = 19 \textrm{ m s}^{-2}\\ \end{array} \) |
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3.4.1.3 |
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01.2 |
\( \begin{array}{l} s = ut + \frac{1}{2}\ at^{2} \textrm{ or } v^2 = u^2 + 2as\\ s = \frac{1}{2} \times 19 \times 1.9^{2} = 34\ \rm{m}\\ \end{array} \) |
Allow e.c.f here for answer to part 01.1 |
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3.4.1.3 |
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01.3 |
\( \begin{array}{l} E_{\rm{k}} \textrm{ at the top } = E_{\rm{k}} \textrm{ at bottom } – E_{\rm{p}}\\ \frac{1}{2} mv_2^{2} = \frac{1}{2}\ mv_1^{2} – mgh\\ v_2^{2} = v_1^{2} – 2 gh = 36^{2} – (2 \times 9.81 \times 62.5)\\ v_2 = 8.4 \textrm{ m s}^{-1}\\ \end{array} \) |
Must use change in energy not equations of motion |
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3.4.1.7 |
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01.4 |
Any three from:
Safety:
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For full marks answer must include relevant comment about safety |
max 4 |
3 |
3.7.5.4 |
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02.1 |
Kinetic energy is conserved |
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3.4.1.6 |
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02.2 |
Momentum before = mu = 2.0 × 10−26 kg × 500 m s−1 OR Velocity after collision equal but opposite direction Δmv = mv – (−mu) = 2 × 10−23 kg m s−1 |
allow N s |
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3.4.1.6 |
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02.3 |
\( \begin{array}{l} \textrm{Distance } = 2 \times 0.02 = 0.04\ \rm{m}\\ \textrm{Time } = \frac{0.04}{500} = 8 \times 10^{-5}\ \rm{s}\\ \end{array} \) |
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02.4 |
\( \begin{array}{l} F = \frac{\Delta mv}{\Delta t}\\ F = \frac{2 \times 10^{-23}}{8 \times 10^{-5}}\ \rm{s} = 2.5 \times 10^{-19}\ \rm{N}\\ \end{array} \) |
Allow e.c.f from 02.2 and 02.3 |
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3.6.2.3 |
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02.5 |
\( \begin{array}{l} P = \frac{F}{A} = \frac{2.5 \times 10^{-19}\ \rm{N}}{0.02^2} = 6.25 \times 10^{-16} (\textrm{N m}^{-2})\\ \textrm{Number of molecules } = \frac{101\ 000}{6.25 \times 10^{-16}} = 6.6 \times 10^{20}\\ \end{array} \) |
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3.6.2.3 |
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03.1 |
GWh is a unit of energy/power × time = 11 × 109 W × 60 × 60 = 3.96 × 1013 J |
2 marks for calculation in joules |
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3.1.1 |
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03.2 |
\( \begin{array}{l} p = \frac{m}{V}\\ m = \rho V\\ \textrm{Mass per second } = 390\ 000\ \textrm{kg s}^{-1}\\ P = \frac{\Delta w}{\Delta t} = \frac{mgh}{\Delta t} = 9.81 \times 390\ 000 \times 500 = 1.9 \times 10^{9}\ \rm{W}\\ \end{array} \) |
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3.4.1.7 |
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03.3 |
\( \begin{array}{l} \textrm{Capacity } = 3.96 \times 10^{13}\ \rm{J}\\ \textrm{Time } = \frac{3.96 \times 10^{13}\ \rm{J}}{1.9 \times 10^9\ \rm{W}} = 20\ 842\ \rm{s}\\ = \textrm{5 hours 50 mins OR 350 mins}\\ \end{array} \) |
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03.4 |
\( \begin{array}{l} \textrm{Efficiency } = \frac{\textrm{useful power output}}{\textrm{input power}}\\ \textrm{Efficiency } = \frac{1.7 \times 10^9}{1.9 \times 10^9}\\ = 0.89\ \rm{or}\ 89\%\\ \end{array} \) |
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3.4.1.7 |
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03.5 |
The generators work less efficiently as pumps/example of named transfer of energy |
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3.4.1.8 |
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04.1 |
Same shape graph Inverted |
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3.4.5.1 |
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04.2 |
Area under graph=change in momentum (0.6 × 10−3 × 0.5 × 2.2 × 103) + (2.2 × 103 × 0.3 × 10−3) + (2.2 × 103 × 0.5 × 0.6 × 10−3) = 1.98 N s |
\( \begin{array}{l} \textrm{OR trapezium method } A = \frac{h\left( a + b \right)}{2}\\ = \frac{2.2 \times 10^{-3} \times \left( 1.5 + 0.3 \right) \times 10^{-3}}{2}\\ \end{array} \) |
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3.4.1.6 |
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04.3 |
Impulse = change in momentum 1.98 = 0.14 × v v = 14 m s−1 |
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3.4.1.6 |
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04.4 |
Velocity would be lower Any one from:
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3 |
3.4.1.6 |
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05.1 |
Accept any reliable method for measuring acceleration:
For accuracy:
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For full marks must have at least one accuracy point |
max 4 |
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PS 4.1 |
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05.2 |
\( \begin{array}{l} \textrm{If force } \propto \textrm{ acceleration, then since } F = W = mg,\\ \textrm{falling mass } \propto \textrm{ acceleration}\\ \frac{\textrm{falling mass}}{\textrm{acceleration}} = \text{ constant}\\ \frac{100}{0.69} = 145\\ \frac{200}{1.38} = 145\\ \frac{300}{2.07} = 145\\ \end{array} \) |
1 mark for explanation 1 mark for at least two tests to see if constant |
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3.4.1.5 |
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05.3 |
\( \begin{array}{l} F = \frac{\Delta mv}{\Delta t}\\ \textrm{If m constant}\\ F = \frac{m\Delta v}{\Delta t} \textrm{ or } a = \frac{\Delta v}{\Delta t}\\ F = ma\\ \end{array} \) |
1 mark for factorising out m with explanation that m is constant 1 mark for explanation/recall of \(a = \frac{\Delta v}{\Delta t}\) |
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3.4.1.6 |
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05.4 |
The crumple zone/seatbelts/airbags increase the time over which the momentum changes (Δt)/increase impact time This decreases the impact force |
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3.4.1.6 |
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06.1 |
No it does not obey Hooke’s law Hooke’s law states that force is proportional to extension Would expect to see a straight line through the origin (wtte) |
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3.4.2.1 |
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06.2 |
Work done is area under graph Counting squares = 69 squares ±3 or 1 square = 0.2 N × 0.02 m = 4 × 10−3 J Work done = 0.28 (0.276) J |
Allow 0.26 J to 0.29 J |
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3.4.2.1 |
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06.3 |
Work done stretching the spring = kinetic energy of pellet \(\begin{array}{l} E_{\rm{k}} = \frac{1}{2}\ mv^{2}\\ 0.28\ \rm{J} = \frac{1}{2} \times 0.01\ \rm{kg} \times v^2\\ v = 7.4\ \textrm{m s}^{-1}\\ \end{array} \) |
Range of velocity is 7.6 to 7.2 |
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3.4.1.8 |
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06.4 |
The unloading curve would be lower/less force required for each extension/hysteresis energy has been transferred to the internal energy of the elastic as it was stretched |
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3.4.2.1 |
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07.1 |
\({}_{88}^{222}{\rm{Ra}} + {}_2^4\alpha\) |
1 mark for correct mass numbers 1 mark for correct atomic numbers |
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3.2.1.2 |
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07.2 |
Alpha particle is moving faster because its mass is smaller/radium slowest because it has largest mass They have to have same magnitude of momentum since momentum before zero Have to move in opposite directions |
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3.4.1.6 |
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07.3 |
Alpha decay followed by two beta decays |
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3.2.1.2 |
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08.1 |
P = Fv = 10 × 5 = 50 J each second |
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3.4.1.7 |
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08.2 |
\( \begin{array}{l} P = Fv\\ F = \frac{250}{5} = 50\ \textrm{N}\\ \end{array} \) |
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3.4.1.7 |
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08.3 |
Bike will accelerate – work done per second by cyclist greater than work done against resistance forces OR Frictional force of the road pushing the bike forward is greater than the resistance forces pushing the bike back |
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3.4.1.7 |
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08.4 |
Weight arrow acting vertically down Normal reaction arrow perpendicular to slope Drag/resistance arrow parallel to slope Frictional force between bike wheel and road acting forwards |
All four arrows labelled The arrows must be on the object, not near it |
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3.4.1.5 |
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08.5 |
Attempt to find resultant of forces parallel to the slope: (95 × 9.81) \(\sin\theta\) or F = \(\frac{250}{5}\) = 50 N or drag = 10 N Equating forces parallel to slope because moving at a constant speed: (95 × 9.81) \(\sin\theta\) + 10 = 50 N \(\begin{array}{l} \sin\theta = \frac{40}{95 \times 9.81}\\ \theta = 2.5^{\circ}\\ \end{array} \) |
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3.4.1.1 |
Skills box answers
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Question |
Answer |
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1 |
Impulse is area under graph. Split the graph into three shapes (2 triangles and a rectangle) and work out the area of these individually. Area = (\(\frac{1}{2}\) × 0.6 × 10–3 s × 2.2 × 103 N) + (0.3 × 10–3 s × 2.2 × 103 N) + (\(\frac{1}{2}\) × 0.6 × 10–3 s × 2.2 × 103 N) = 0.66 N s + 0.66 N s + 0.66 N s = 1.98 N s Therefore, impulse = 1.98 N s |
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2 |
force × time = change in momentum (mv) = 420 N × 0.01 s = 4.2 N s |
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3 |
The change in momentum in both cases will be the same. However, if you move your hands back as you catch the ball, you increase the time that the change in momentum occurs over and so decrease the force required to stop the ball. This means that the ball does not hit your hands as hard as it would do if you just caught it without moving your hands. |