web viewa serious omission in many answers was that of the word ‘kinetic ... frequencies and...

Unit 4 – Mixed questions – 02-06-14 - Hinchley Wood School

Q1. Deep space probes often carry modules which may be ejected from them by an explosion. A space probe of total mass 500 kg is travelling in a straight line through free space at 160 m s–1 when it ejects a capsule of mass 150 kg explosively, releasing energy. Immediately after the explosion the probe, now of mass 350 kg, continues to travel in the original straight line but travels at 240 m s–1, as shown in the figure below.

(a) Discuss how the principles of conservation of momentum and conservation of energy apply in this instance.

The quality of your written communication will be assessed in this question.

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(b) (i) Calculate the magnitude of the velocity of the capsule immediately after the explosion and state its direction of movement.

magnitude of velocity = ....................................... m s–1

direction of movement ............................................................(3)

(ii) Determine the total amount of energy given to the probe and capsule by the explosion.

answer = ....................................... J(4)

(Total 13 marks)

Q2. The graph shows the variation with time, t, of the force, F, acting on a body.


What physical quantity does the area X represent?

A the displacement of the body

B the acceleration of the body

C the change in momentum of the body

D the change in kinetic energy of the body(Total 1 mark)

Q3. Which row, A to D, in the table correctly shows the quantities conserved in an inelastic collision?

mass momentum kinetic energy total energy

A conserved not conserved conserved conserved

B not conserved conserved conserved not conserved

C conserved conserved conserved conserved

D conserved conserved not conserved conserved

(Total 1 mark)

Q4. What is the angular speed of a point on the Earth’s equator?

A 7.3 × 10–5 rad s–1


B 4.2 × 10–3 rad s–1

C 2.6 × 10–1 rad s–1

D 15 rad s–1

(Total 1 mark)

Q5.

A force, F, varies with time, t, as shown by the graph and is applied to a body initially at rest on a smooth surface. What is the momentum of the body after 5.0 s?

A zero.

B 12.5 N s.

C 25 N s.

D 50 N s.(Total 1 mark)

Q6. A particle of mass m oscillates with amplitude A at frequency f. What is the maximum kinetic energy of the particle?

A π2 mf 2A2

B π2 mf 2A2

C 2 π2 mf 2A2


D 4 π2 mf 2A2

(Total 1 mark)

Q7. (a) A spring, which hangs from a fixed support, extends by 40 mm when a mass of 0.25 kg is suspended from it.

(i) Calculate the spring constant of the spring.

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(ii) An additional mass of 0.44 kg is then placed on the spring and the system is set into vertical oscillation. Show that the oscillation frequency is 1.5 Hz.

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(b) With both masses still in place, the spring is now suspended from a horizontal support rod that can be made to oscillate vertically, as shown in the figure below, with amplitude 30 mm at several different frequencies.


Describe fully, with reference to amplitude, frequency and phase, the motion of the masses suspended from the spring in each of the following cases.

(i) The support rod oscillates at a frequency of 0.2 Hz.

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(ii) The support rod oscillates at a frequency of 1.5 Hz.

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(iii) The support rod oscillates at a frequency of 10 Hz.

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(Total 10 marks)


Q8. A particle, whose equilibrium position is at Q, is set into oscillation by being displaced to P, 50 mm from Q, and then released from rest. Its subsequent motion is simple harmonic, but subject to damping. On the first swing, the particle comes to rest momentarily at R, 45 mm from Q.

During this first swing, the greatest value of the acceleration of the particle is when it is at

A P.

B Q.

C R.

D P and R.(Total 1 mark)

Q9. (a) Define the gravitational potential at a point in a gravitational field.

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(b) The figure below, which is not drawn to scale, shows the region between the Earth (E) and the Moon (M).

(i) The gravitational potential at the Earth’s surface is –62.6 MJ kg–1. Point X shown in the figure above is on the line of centres between the Earth and the


Moon. At X the resultant gravitational field is zero, and the gravitational potential is –1.3 MJ kg–1.

Calculate the minimum amount of energy that would be required to move a Moon probe of mass 1.2 × 104 kg from the surface of the Earth to point X. Express your answer to an appropriate number of significant figures.

answer = .................................. J(3)

(ii) Explain why, once the probe is beyond X, no further energy would have to be supplied in order for it to reach the surface of the Moon.

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(iii) In the vicinity of the Earth’s orbit the gravitational potential due to the Sun’s mass is –885 MJ kg–1. With reference to the variation in gravitational potential with distance, explain why the gravitational potential due to the Sun’s mass need not be considered when carrying out the calculation in part (b)(i).

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(c) The amount of energy required to move a manned spacecraft from the Earth to the Moon is much greater than that required to return it to the Earth. By reference to the forces involved, to gravitational field strength and gravitational potential, and to the point X, explain why this is so.

The quality of your written communication will be assessed in your answer.

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(Total 14 marks)

Q10.

An α particle travels towards a gold nucleus and at P reverses its direction.Which one of the following statements is incorrect?

A The electric potential energy of the α particle is a maximum at P.

B The kinetic energy of the α particle is a minimum at P.

C The total energy of the α particle is zero.

D The total energy of the α particle has a constant positive value.(Total 1 mark)


Q11. A small charged sphere of mass 2.1 × 10–4 kg, suspended from a thread of insulating material, was placed between two vertical parallel plates 60 mm apart. When a potential difference of 4200 V was applied to the plates, the sphere moved until the thread made an angle of 6.0° to the vertical, as shown in the diagram below.

(a) Show that the electrostatic force F on the sphere is given by

F = mg tan 6.0°

where m is the mass of the sphere.

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(b) Calculate

(i) the electric field strength between the plates,

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(ii) the charge on the sphere.

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(3)(Total 6 marks)

M1. (a) The candidate’s writing should be legible and the spelling,punctuation and grammar should be sufficiently accurate for themeaning to be clear.

The candidate’s answer will be assessed holistically. The answer will beassigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised, logical andcoherent, using appropriate specialist vocabulary correctly. The form andstyle of writing is appropriate to answer the question.

The candidate states that momentum is conserved, supported by reasoningto explain why the conditions required for momentum conservation aresatisfied in this case.

The candidate also gives a statement that total energy is conserved, givingdetailed consideration of the energy conversions which take place,described in the correct sequence, when there is an explosion on a bodythat is already moving.

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well organised andnot fully coherent. There is less use of specialist vocabulary, or specialistvocabulary may be used incorrectly. The form and style of writing is lessappropriate.

The candidate states that momentum is conserved, but the reasoning ismuch more limited.

and/or

There is a statement that (total) energy is conserved, with basicunderstanding that some energy is released by the explosion.

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised and may notbe relevant or coherent. There is little correct use of specialist vocabulary.The form and style of writing may be only partly appropriate.

The candidate indicates that either momentum or energy is conserved, or


that both are conserved. There are very limited attempts to explain eitherof them.

The explanation expected in a competent answer should include acoherent selection of the following points concerning the physicalprinciples involved and their consequences in this case.

Momentum

• momentum is conserved because there are no external forcesacting on the overall system (probe plus capsule) – or because it’sfree space

• they are moving in free space and are therefore so far from largemasses that gravitational forces are negligible

• during the explosion, there are equal and opposite forces actingbetween the probe and the capsule

• these are internal forces that act within the overall system

• because momentum has to be conserved, and it is a vector, thecapsule must move along the original line of movement after theexplosion

Energy

• total energy is always conserved in any physical process becauseenergy can be neither created nor destroyed

• however, energy may be converted from one form to another

• the probe is already moving and has kinetic energy

• in the explosion, some chemical energy is converted into kineticenergy (and some energy is lost in heating the surroundings)

• the system of probe and capsule has more kinetic energy than theprobe had originally, because some kinetic energy is released bythe explosion

max 6

(b) (i) conservation of momentum gives (500 × 160)= 150 v + (350 × 240) (1)from which v = (−)26(.7) (m s−1) (1)

direction: opposite horizontal direction to larger fragment[or to the left, or backwards] (1)

Unit 4 – Mixed questions – 02-06-14 - Hinchley Wood School 3

(ii) initial Ek = ½ × 500 × 1602 (1) (= 6.40 × 106 J)

final Ek = (½ × 350 × 2402) + (½ × 150 × 26.72) (1) (= 1.01 × 107 J)

energy released by explosion = final Ek − initial Ek (1)

= 3.7 × 106 (J) (1)4

[13]

M2. C[1]

M3. D[1]

M4. A[1]


M5. C[1]

M6. C[1]

M7. (a) (i) mg = ke (1)

= 61(.3) N m−1 (1)

(1) (= 0.667 s)

(ii) (1)(= 1.50 Hz)4

(b) (i) forced vibrations (at 0.2 Hz) (1)amplitude less than resonance (≈ 30 mm) (1)(almost) in phase with driver (1)

(ii) resonance [or oscillates at 1.5 Hz] (1)amplitude very large (> 30 mm) (1)oscillations may appear violent (1)phase difference is 90º (1)


(iii) forced vibrations (at 10 Hz) (1)small amplitude (1)out of phase with driver [or phase lag of(almost) π on driver] (1)

Max 6[10]

M8. A[1]

M9. (a) work done (or energy required) per unit mass

in moving a mass from infinity to the point 2

(b) (i) ΔV (= – 1.3 – (–62.6)) = 61.3 (MJ kg–1)

energy required (= mΔV) = 1.2 × 104 × 61.3 × 106

= 7.4 × 1011 (J) to 2SF only 3

(ii) beyond X, gravitational potential decreases as Moon is approached [or gravitational field (or force) of Moon will now attract the probe]

1

(iii) distance from Earth to Sun » distance from Earth to Moon

change in Vsun (or in gsun) over Earth to Moon distance is negligible

value of Vsun (or gsun) is not (significantly) changed by relative positions of E+M

Unit 4 – Mixed questions – 02-06-14 - Hinchley Wood School 2

(c) The candidate’s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear.

The candidate’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marksThe information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question.The candidate discusses the forces of attraction due to the Earth and due to the Moon, appreciates that they act in opposite directions, and that the former is generally much greater than the latter.The candidate discusses the resultant gravitational field between E and M, understands that there is a ‘neutral’ point at which the resultant field strength is zero and that this point is much closer to M than E. It is recognised that this point has to be passed for the journey to be completed in either direction.There is a discussion of gravitational potential, in which it is pointed out that the resultant potential rises to a maximum at the neutral point. There is a reference to the much greater amount of work that has to be done on the spacecraft to reach this point from E than from M.

Intermediate Level (Modest to adequate): 3 or 4 marksThe information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate.

The candidate discusses the forces of attraction due to the Earth and the Moon, and appreciates either that they act in opposite directions, or that the former is much greater than the latter. There is a relevant discussion of field strength or potential. The significance of the neutral point may not be appreciated. The candidate is likely to make some reference to the work that has to be done on the spacecraft.

Low Level (Poor to limited): 1 or 2 marksThe information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.The candidate has some understanding of the forces that act during the journey but makes very limited references to the significance of the variation of the gravitational field. Discussions of gravitational potential and/or work done are likely to be superficial and may be absent.

The explanation expected in a competent answer should include a coherent selection of the following points concerning the physical principles involved and their consequences in this case.


Gravitational forces

The spacecraft experiences gravitational attractions to both the Earth and the Moon during its journey.

These forces pull in opposite directions on the spacecraft.

Because E is much more massive than M, for most of the outward journey the force towards E is greater than that towards M.

Only in the later stages of the outward journey is the resultant force directed towards M.

On the return journey the resultant force is predominantly towards E.

Gravitational field strength

During the outward journey E’s gravitational field becomes weaker and M’s becomes stronger.

The resultant field is the vector sum of those due to E and M separately.

A point (X) is reached at which these two component fields are equal and opposite, giving zero resultant.

X is much closer to M than E.

Once X has been passed, the spacecraft will be attracted to M by M’s gravitational field.

On the return journey the spacecraft will ‘fall’ to E once it is beyond X.

Gravitational potential

The gravitational potential due E increases (i.e. becomes less negative) as the spacecraft moves away from E.

The resultant gravitational potential is the (scalar) sum of those due to E and M separately.

At X the gravitational potential reaches a maximum value before decreasing as M is approached.

In order to reach M on the outward journey, the spacecraft has to be given at least enough energy to reach X, and vice-versa for the return.


Much more work is needed to move the spacecraft from E to X than from M to X, since a larger force has to be overcome over a larger distance.

6[14]

M10. C[1]

M11. (a) mg = T cos 6 (1)F = T sin 6 (1)hence F = mg tan 6 (1)[or correct use of triangle: (1) for sides correct, (1) for 6°, (1) for tan 6 = F/mg

or FΔx = mg Δh , tan θ = tan 6° = 3

(b) (i) (use of E = gives) E = = 7.0 × 104 V m–1 (1)

(ii) (use of Q = gives) Q =

= 3.1 × 10–9 C

(allow C.E. for value of E from (i))3

[6]


E1. It was evident from their attempts at part (a) that during their courses many candidates had considered the application of conservation of momentum to events involving an explosion. It was less clear that they had ever considered an explosion that takes place in a moving object, or considered how conservation of energy applies in an explosion. Consequently, part (a) of the question proved to be difficult, not least because it was unfamiliar territory for so many. Part (b), which was formulaic and involved much less original thinking, brought much more success for the majority.

In part (a) only a very small proportion of the candidates were able to produce answers that were well organised, coherent, detailed and contained correct physics to merit a ‘high level’ mark of five or six. More answers fell into the ‘intermediate level’ (three or four marks) and even more into the ‘low level’ (one or two marks). A major failing in most answers was to overlook the question’s requirement to address the two conservation laws ‘in this instance’. For a high level answer, it was necessary to consider an explosion on a moving space vehicle travelling in a straight line in deep space. All of the italicised section is significant. The system has momentum before exploding (unlike a straightforward recoil example); this momentum has to be conserved because there are no external forces in deep space. Hence the probe speeds up and the capsule must be ejected along the original line of movement (although it may not be possible to tell that this is ‘backwards’ until the calculation has been done). Forces between probe and capsule during the explosion are equal and opposite, but they are internal forces for the system. When considering momentum, it was common for candidates to conclude that ‘momentum must be conserved because momentum is always conserved’.

In the explosion, chemical energy is converted into kinetic energy; this increases the total kinetic energy of the system, which is shared between probe and capsule. Examiners saw many very weak answers that showed total confusion – such as momentum being converted into energy, mass being converted into energy, or energy not being conserved. A serious omission in many answers was that of the word ‘kinetic’ before ‘energy’, whilst many answers referred to the event as an ‘inelastic collision’. There was seldom any reference to conservation of the total energy of the system taken as a whole.

Most candidates recovered from their poor attempts at part (a) to gain all three marks for the calculation in part (b) (i). There were also many awards of full marks in part (b) (ii), where the main mistake was to calculate only the kinetic energy of the system (probe + capsule) after the explosion, and to regard this as the answer to the question. Apparently, the candidates who did this had not realised that the system had an initial kinetic energy.


E2. This question, on factual knowledge of the impulse – momentum relationship, was an easy starter with a facility of 85%.

E3. This was a straightforward test of candidates’ knowledge. It required candidates to decide whether or not mass, momentum, kinetic energy and total energy would be conserved in an inelastic collision. 85% of the candidates appreciated that everything except kinetic energy would be conserved. Incorrect responses were fairly evenly spread around the other three distractors.

E4. This question proved to be the easiest question, with a facility of 87%. Application of ω = 2π/T with T equal to the period of Earth’s rotation readily gave the correct answer.

E6. Application of ½mv2, together with vmax = 2πfA, readily gave the correct response for 70% of the candidates in this question; this was a much higher percentage than that achieved when the question was pre-tested. The most common wrong response was

distractor D, no doubt chosen by those who overlooked the factor of½ .

E7. Good progress was generally made in part (a), but the unit of the spring constant was not always correct and often omitted. Clear and concise answers were common, usually allowing all four marks to be awarded. The most common difficulty occurred where the candidate thought that k = (m/e) instead of (mg/e); these candidates were then unable to show that the frequency was 1.5 Hz.

Part (b) caused great difficulty for a majority of candidates, many of whom seemed to have little or no detailed knowledge concerning forced vibrations and resonance. Phase


relationships proved to be particularly demanding, although the mark scheme was adopted and made it possible to score all six marks without referring to phase at all. Responses were often confusing, making it difficult for examiners to decide whether the frequencies and amplitudes referred to were those of the support rod, the spring, or (as the question intended) the masses. More candidates ought to have realised that phase could only be correctly described by comparing the oscillation of the masses (the driven system) with that of the support rod (the driver). They should also know that phase is measured by an angle, not a wavelength. There were many references to the frequencies and amplitudes of waves, and even to interference. Perhaps the rather simple demonstration that formed the basis of this question should receive greater prominence when teaching the characteristics of vibrations.

E8. This question tested candidates’ understanding of the acceleration of a particle moving with simple harmonic motion. Over half of the candidates gave the correct response, but one in five of them thought that the acceleration was greatest at zero displacement.

E9. In the definition of gravitational potential (part (a)) the common failings were the omission of per unit mass and a wrong direction of displacement – from a point in the field to infinity instead of in the opposite direction. The calculation in part (b)(i) was usually well rewarded. The number of significant figures to be quoted in the final answer should have been limited to the smallest number of significant figures provided in the data, which was 2 in this case. Not all students realised that, in order to reach the Moon, the minimum increase in gravitational potential required was the difference between the values at X and at the Earth’s surface, some thinking that the probe had to be given an increase of 62.6 MJ kg–1 (which would be enough to remove it to infinity).

Part (b)(ii) had a high proportion of correct answers, usually given in terms of field strength or gravitational force rather than gravitational potential. A common error here was to state that the Earth’s field had no effect beyond X. Very few answers to part (b)(iii) scored both marks and most received no credit at all. The simplistic, incorrect answer – that V is proportional to 1/r implied that the potential would be negligible at large r – was given by many. The question had informed students that near the Earth the value of gravitational potential due to the Sun is –885 MJ kg–1, which is certainly not negligible! The important facts in this part were that the Earth-Moon distance is much less than the Earth-Sun distance, so the change in V due to the Sun is negligible over the distance moved by the probe.

To merit a mark of 5 or 6 (a high level answer) in part (c) the students were expected to make detailed correct references to all of the four factors mentioned in the question: forces, field strengths, potentials and the significance of point X. Few were able to do this, and so the majority of answers fell to the intermediate or low level. The understanding of gravitational potential was particularly weak, compounded by misinterpretation of the


negative sign in a scalar quantity. It was commonly stated that the gravitational potential of the Earth is greater than the gravitational potential of the Moon; in fact the values are –62.6 and –3.9 MJ kg–1 respectively, so it appears that a large proportion of A level students did not understand that 0 is greater than a negative number. Most answers concentrated on the Earth’s gravitational field being larger than the Moon’s because of their different masses. Many answers reiterated the question, without explaining the reasons satisfactorily, and in a coherent, sequenced, well organised way. Students should have been able to explain why X is closer to the Moon than to Earth, that work has to be done on the probe only as far as X, and that the larger distance from the Earth to X, and the average larger force required to get there, imply that more work has to be done on the probe when travelling to the Moon than when returning to Earth. Loose use of terminology was a frequent detractor when making these answers: “the Earth’s larger potential pulls the probe back to Earth”, etc. References to escape speed, which is not very relevant in the context of this question, were fairly common.

E10. The statement that is incorrect was to be chosen in this question, about the energy of an α particle during a head-on encounter with a gold nucleus. The facility of this question was 62%, the most common incorrect choice being distractor D (19%).

E11. It surprised the examiners that only a minority of candidates gained full marks in part (a). Successful solutions were usually based on the triangle of forces. Only the best candidates resolved the tension into components and equated the components to the weight and the electrostatic force respectively. Many candidates incorrectly resolved the weight into components parallel and perpendicular to the thread.

The majority of candidates obtained the correct value of the electric field strength in part (b) and were able to make good progress in part (ii). Candidates who equated g to 10 N kg–1 or rounded off incorrectly at the end were penalised. Other candidates attempted inappropriate solutions involving Coulomb’s law and did not realise that the force = qE. A small minority of candidates attempted incorrectly to relate the gain of gravitational potential energy to an electrostatic energy formula such as ½QV.

web viewa serious omission in many answers was that of the word ‘kinetic ... frequencies and...

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