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OCR Physics A

Question / Answer / Marks / Guidance /
1 a / Work done is force multiplied by the distance moved in the direction of the force. /
B1 / It must be clear that there is movement in the same direction as the force is acting.
1 b i / Ep of ball = 200 ´ 10–3 ´ 9.81 ´ 0.65 = 1.275 J
Distance = = 1.77 m / C1
A1 / Mass must be converted to kg.
1 b ii / GPE is converted to KE as ball rolls down ramp.
KE is converted into thermal energy as ball slows down along the floor. / B1
B1 / Since ramp is smooth all GPE is converted to KE down the ramp, which is then used to do work
1 c i / F = 120 ´ cos 34 = 99.4 N / A1 / Many students muddle sin and cos.
1 c ii / Work done = 120 ´ 75 ´ cos 34 = 7.46 ´ 10³ J / A1 / Only the horizontal component of T does work on the trolley.
1 c iii / Trolley possesses kinetic energy and will continue to move until this is all dissipated as thermal energy (work done against F).
Ek = ´ 80 ´ (0.85)² = 28.9 J
Extra distance = = 0.29 m
Assume F is constant/independent of speed /
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B1 / Because it is necessary to use the answer to part i here, allow an e.c.f. provided all working is shown.
2 a / Resultant force is a maximum at t = 0, it decreases as the drag increases with speed.
It reaches zero after 9 s and remains at this value for the rest of the time /
B1
B1 / It is important to mention the drag increases as v increases.
Should see the time at which drag = driving force.
2 b / Convert 28 km h–1 to m s–1: 28´=7.78 m s–1
Max Ek =´ 7.0 ´ 10³ ´ 7.78²
= 2.12 ´ 105 J /
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A1 / Must convert km h–1 into m s–1 and square the speed after converting.
2 c / P = Fv so F =
/ C1
A1 / Marks will only be scored after substituting the values since the formula is in the data book.
2 d / Height gained/second =
Ep gained/second = 7.0 ´ 10³ ´ 9.81 ´ 0.271
Extra power required = 18.6 kW /
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A1 / In 1 s the lorry will travel (28 000/3600) m up the hill. The product of this distance multiplied by sin 2 is the height gained. Since v and KE are unchanged, the rate of increase in GPE is the increase in power required.
3 a / Work done = 45 ´ 9.81 ´ 3.2 = 1.41 ´ 10³ J / A1 / Work done is the same as the increase in GPE
3 b / Gain in GPE is same in both methods.
KE is zero at start and finish so, assuming ramp is smooth, the energy used is the same not less.
If the ramp exerts a frictional force (as is usually the case) then work must be done against this force as well as against gravitational force so more energy is used. /
B1
B1 / It is easier to pull the truck up the ramp since the tension force in the rope will be less but the distance moved by the force increases so that the work done is the same.
4 a / Upthrust reduces the tension in the thread. / B1 / The newtonmeter records the tension in the thread and as sphere is in equilibrium
T + U = W
4 b / Mass of sphere = = 0.0765 kg
F = m a gives a = = 7.85 m s–2 / C1
A1 / Must convert weight to mass.
T = 0 so resultant force is 0.6 N downwards
4 c / As sphere falls GPE is converted into thermal energy (in the sphere and the oil) rather than a gain in KE. / B1 / This is an important point to note when dealing with energy changes at terminal velocity.
5 a / GPE at B = 750 ´ 9.81 ´ 40 = 2.94 ´ 105 J / A1
5 b / Change in GPE B to D = 750 ´ 9.81 ´ (40 – 20)
= 1.47 ´ 105 J
/ C1
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A1 / Half the GPE is converted to KE as height is halved.
5 c / As carriage moves from B to D friction acts between the wheels and the track so some GPE is converted to thermal energy rather than KE.
or
Air resistance will reduce KE. / B1 / Total energy is conserved in the process but friction does work and produces thermal energy.
5 d / If the carriage had people in, the mass would increase and this would increase the value of GPE given in parta.
In partb friction is not considered and the maximum velocity would not be changed as a result of increasing mass. /
B1
B1 / Δ(m g h) = m g Δh, which is converted to KE using
m g Δh = m (vmax)2
m cancels leaving:
vmax = Ö(2 g Δh)
5 e / Useful energy output = 11 ´ 10³ ´ 60 = 6.6 ´ 105 J
Total mass lifted =
= 1.682 ´ 10³ kg
Extra mass = 1.682 ´ 10³ – 750 = 932 kg
Extra people = = 13 people / C1
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A1 / Must convert time to seconds. Assuming this is all converted to GPE
Must round down for this value since fractions of people are not possible.
6 a / Attach a set mass, m, to the motor and switch the motor on to lift the load.
Mark two points a measured distance, h, apart.
Time how long it takes for motor to lift the load
distance h.
Mechanical power output = / B1
B1
B1 / Must suggest a method for timing, e.g., stopwatch, or light gates and data logger.
6 b / Repeat steps described in parta with increasing masses (load). Measure the average wattmeter reading as the load rises between the points.
For each load calculate the efficiency using mechanical power output/electrical power input ´ 100
Plot a graph of efficiency against mass of load /
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7 a / / B1 / Could be written in standard symbols
7 b i / From parta: F = = 1.14 N /
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7 b ii / Total resistive force = 1.14 N / A1 / Since speed of car is constant driving force = total resistive force
7 b iii / / C1
A1 / It is important to rearrange this equation correctly. Remember that the input power cannot be less than the output power in any system. This often alerts you to incorrect rearranging.
7 c /
Hence max speed increases by 5.8% /
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A1 / Students often have difficulty with ratio problems. It can be solved using F = k v² provided initial values are assumed for F and v. The ratio method is much neater and can be used in many other problems so it is well worth practising.

This resource sheet may have been changed from the original 1