web view(ii) calculate the horizontal component of velocity as the projectile hits the ground. ......

Harris Federation - Consultants

New Document 1 Name: ________________________

Class: ________________________

Date: ________________________

Time: 48 minutes

Marks: 39 marks

Comments:


Q1.A seismometer is a device that is used to record the movement of the ground during an earthquake. A simple seismometer is shown in the diagram.

A heavy spherical ball is attached to a pivot by a rod so that the rod and ball can move in a vertical plane. The rod is suspended by a spring so that, in equilibrium, the spring is vertical and the rod is horizontal. A pen is attached to the ball. The pen draws a line on graph paper attached to a drum rotating about a vertical axis. Bolts secure the seismometer to the ground so that the frame of the seismometer moves during the earthquake.

(a) The ball is made of steel of density 8030 kg m−3 and has a diameter of 5.0 cm.

Show that the weight of the ball is approximately 5 N.

(3)


(b) The distance from the surface of the ball to the pivot is 12.0 cm, as shown in the diagram above.

Calculate the moment of the weight of the ball about the pivot when the rod is horizontal. Give an appropriate unit for your answer.

moment = ................................ unit = ........................ (3)

(c) The spring is attached at a distance of 8.0 cm from the pivot and the spring has a stiffness of 100 N m−1.

Calculate the extension of the spring when the rod is horizontal and the spring is vertical. You may assume the mass of the pen and the mass of the rod are negligible.

extension = ........................ m (3)

(d) Before an earthquake occurs, the line being drawn on the graph paper is horizontal.

Explain what happens to the line on the graph paper when an earthquake is detected and the frame of the seismometer accelerates rapidly downwards.

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(Total 11 marks)

Q2.The world record for a high dive into deep water is 54 m.

(a) Calculate the loss in gravitational potential energy (gpe) of a diver of mass 65 kg falling through 54 m.

loss in gpe = ................................... J(2)

(b) Calculate the vertical velocity of the diver the instant before he enters the water. Ignore the effects of air resistance.

velocity = ............................ ms–1

(2)

(c) Calculate the time taken for the diver to fall 54 m. Ignore the effects of air resistance.

time = ................................... s(2)

(d) Explain, with reference to energy, why the velocity of the diver is independent of his mass if air resistance is insignificant.

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(Total 9 marks)

Q3.A projectile is launched some distance above the ground at an angle of 25° above the horizontal with a vertical component of velocity of 5.0 m s−1. Figure 1 shows the flight path of the projectile. The flight takes 1.3 s.

Ignore the effects of air resistance throughout this question.

Figure 1

(a) (i) Show that the initial speed of the projectile is about 12 m s−1.

(2)

(ii) Calculate the horizontal component of velocity as the projectile hits the ground.

horizontal component of velocity = ............................... m s−1


(2)

(b) (i) Calculate the maximum height above the starting point reached by the projectile.Give your answer to an appropriate number of significant figures.

maximum height reached = ............................... m(2)

(ii) Calculate the total horizontal distance travelled by the projectile from its starting point.

horizontal distance = ............................... m(1)

(c) (i) Mark with an A on the flight path in Figure 1 the position where the speed of the projectile is greatest.

(1)

(ii) Mark with a B on the flight path in Figure 1 the position where the speed of the projectile is least.

(1)

(iii) The projectile reaches its maximum height at time tH and finishes its flight at time tF. Draw on Figure 2 a graph to show how the magnitude of the vertical component of velocity of the projectile varies with time. Numerical values are not required.

Figure 2

(2)


(Total 11 marks)

Q4.The Soyuz Spacecraft is used to transport astronauts to and from an orbiting space station. The spacecraft is made up of three sections as shown in Figure 1.

Figure 1

(a) On leaving the space station the spacecraft is given an initial horizontal thrust of 1400 N. Calculate the initial acceleration of the spacecraft during the firing of the thruster engines.

acceleration = ............................... m s–2

(2)

(b) Newton’s Third Law refers to pairs of forces.

(i) State one way in which a pair of forces referred to in Newton’s Third Law are the same.


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(ii) State one way in which a pair of forces are different.

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(c) When the spacecraft returns to the Earth’s atmosphere the orbital module and the service module are separated from the descent module. This descent module has its speed greatly reduced by drag from the atmosphere.

Figure 2 shows two of the forces acting on the descent module as it travels down through the atmosphere.

Figure 2

State one reason why the two forces shown in Figure 2 are not a pair of forces as referred to in Newton’s Third Law.

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(d) In one particular descent, the descent module has its speed reduced to 5.5 m s–1 by parachutes. The descent module also releases its empty tanks and shield to reduce its mass to 890 kg.


A final speed reduction can be carried out by using engines which operate for a maximum time of 3.5 s. When the engines are in use, the resultant upward force on the descent module is 670 N. The safe landing speed of the descent module is 3.0 m s–1.

Determine whether these engines are able to reduce the speed of the descent module to its safe value.

At these landing speeds atmospheric drag is negligible.(3)

(Total 8 marks)


M1.(a) use of V = πr3 to give V = π(2.5 × 10−2)3 ✔ = 6.5 × 10−5 m3

use of ρ = to give m = ρ V = 8030 × 6.5 × 10−5 ✔ = 0.53 kguse of W = mg to give W = 0.53 × 9.81 = 5.2 (N) ✔

the first mark is for making some attempt to calculate the volume; ignore power of ten errors.the second mark is for the correct substitution or for the calculation of massthe third mark is for going on to calculate the weightallow ce for incorrect volume or mass but 2 errors = 0/3no sf penalty but g = 10 N kg−1 loses mark

3

(b) distance of line of action of weight to pivot = (0.120 + 0.025) = 0.145 m ✔moment = force × distance = 5.2 × 0.145 = 0.75 ✔unit Nm ✔

the first mark is for identifying that the weight of the ball will act through its centre; use of 0.12 m loses this markthe second is for correctly calculating the moment; allow ce for wrong distance; condone force = 5 N (which leads to 0.725)allow suitable unit consistent with calculation, eg N cmreject ‘nm’ or ‘NM’ etc

3

(c) taking moments about the pivot clockwise moment from spring = anticlockwise moment from ballF × 0.080 = 0.75 ✔F = 9.4 N ✔

use of F = kx to give x = = 0.094 m ✔allow ce from (b)the first mark is for the use of the moment equationthe second mark is for calculating the force on the spring; condone 9.35 and 9.3the third mark is for calculating the extension; allow calculation in cmallow ce from the second mark ie use of wrong force; condone 1 sf 0.09 m if (1 sf) 5 N used in (b)

3


(d) the line / pen (initially) moves up; ignore subsequent motion ✔ (the downwards acceleration of the ball is much less than that of the frame and) the ball does not move (very far in the time taken for the frame to move down) ✔

the first mark is for stating the correct direction of the line / pen; allow ‘diagonally up’, ‘up then down’ but reject ‘up and down’the second mark is for an explanation which shows some understanding of the relative displacement of the ball and frame; this mark is consequential on the first being correct; condone ‘ball has inertia’

2[11]

M2.(a) (Ep = mgΔh)= 65 × 9.81 × 54 = 3.44 × 104 = 3.4 × 104 (J) (34433)

max 1 if g =10 used (35100 J)Correct answer gains both marks

2

(b) allow 32 (32.3) for the use of 34000allow 32.6

OR correct use of v2 = 2 g sdon’t penalise g = 10 (32.863)

2

(c) (s =1 / 2 gt 2 or other kinematics equation)

With use of g= 9.8 or 9.81 or 10 and / or various suvat equations, expect range 3.2 to 3.4 s.No penalty for using g= 10 here.

ecf from 1(b) if speed used2

(d) (all G)PE (lost) is transferred to KE


no (GP)E transferred to 'heat' / 'thermal' / internal energyOR

Must imply that all GPE is transferred to KE. E.g. accept ‘loss of GPE is gain in KE’ but not: ‘loses GPE and gains KE’.

(therefore) mass cancels.

Accept ‘m’s crossed out3

[9]

M3.(a) (i) (using sin 25° = VV/VV = VV / sin 25° )=5.0/sin25° ✔11.8 (m s-1 ) ✔(working and answer is required)

Look out for cos 65° = sin 25° in first mark.Also calculating the horizontal component using cos 25° followed by Pythagoras is a valid approach.Working backwards is not acceptable.

2

(ii) (using tan 25° = VV /VH)VH = VV /tan 25° ✔= 5 / tan 25° =11(m s-1) ✔ (10.7m s-1)Or (using cos 25° = VH / V )VH = V cos 25° ✔ = 11.8 cos 25° = 11 (m s-1) ✔ (10.7 m s-1)Or (using VH

2 + VV2 = V2)

VH2 + 52 =11.82 ✔ (0r 122)

VH = 11 (m s-1) ✔ (10.7 m s-1)Note 1/cos 65° = sin 25°and tan 25° = 1/ tan 65°Rounding means answers between 10.7 and 11 m s-1 are acceptable

2

(b) (i) (using v2 = u2 + 2as with up being positive 0 = 5.02 + 2 × -9.81 × s)s = 1.3 (m) ✔ (1.27 → 1.28 m)or (loss of KE = gain of PE½ m v2 = m g h½ 5.02 =9.81 × h)h = 1.3 (m) ✔ (1.27 → 1.28 m)


quoted to 2 sig figs ✔for the sig fig mark the answer line takes priority. If a choice of sig figs given and not in answer line – no sig fig markSig fig mark stands alone provided some working is shown

2

(ii) (using s = (u + v)t/2) or horizontal distance = speed × time)s = 11 × 1.3 = 14 (m) ✔ (using 10.7 gives the same answer)

allow CE s = (aii) × 1.3 but working must be seen1

(c) (i) A marked at the point of landing or immediately before ✔The A or its marked position must not be further to the left than the ‘c’ in the word ‘scale’

1

(ii) B marked at the maximum height of the path ✔The B must lie vertically between the ‘r’ and ‘a’ in the word ‘resistance above figure 2.

1

(d) straight line from point given down to point tH on the axis ✔straight line starting where first line stops (tH) but with opposite gradient to the first line ✔

(A measure of accuracy for the second mark) The second line must end (tH) between the height of the vertical axis and half this height.Obviously straight lines drawn by hand are acceptable.

2[11]

M4.(a) (using F = ma)a (= F/m) = 1400 / (any mass / masses taken from the table)✔ (any arrangement gains mark)= (1400/(2600 + 2900 + 1300) = 0.206 m s-2)= 0.21 (m s-2) ✔


Allow any single or combination of masses from the table for first mark.0.21 m s-2 on its own gains 2 marksbut 0.2 m s-2 on its own gains 1 mark1 sig fig is not acceptable

2

(b) (i) they have the same line of actionthey have the same magnitude (not size)the forces are of the same kind✔ any one statement

The statement must be of a general nature.The statement ‘equal magnitude + opposite direction does not make the similarity clear and so gains zero marks.If forces act against each other or cancel or act on the same body this mark is lost.

1

(ii) they are in opposite directionsthey act on different bodies✔ any one statement

The statement must be of a general nature.If forces act against each other or cancel or act on the same body this mark is lost.

1

(c) they do not have the same magnitude / sizethe forces are of different typesthey do not act on different bodiesdrag is greater than the weightthere is a resultant force (as deceleration occurs)✔ any one statement

Statements can be written negatively1

(d) actual deceleration(= F/m) = 670 / 890 = 0.75 (m s-2) ✔ (0.753 m s-2)minimum required deceleration = ∆v / ∆t = (5.5 – 3.0) / 3.5 = 0.71 (m s-2) ✔ (0.714 m s-2)therefore (compared to 0.75 m s-2) there is sufficient deceleration, yes ✔Oractual deceleration

ignore any interchange between acceleration and deceleration

(= F/m) = 670 / 890 = 0.75 (m s-2) ✔ (0.753 m s-2)


maximum required time = ∆t = ∆v/a = (5.5 – 3.0) / 0.753 = 3.3 s (3.32 s) ✔therefore (compared to 3.5 s) there is sufficient time, yes ✔Oractual deceleration(= F/m) = 670 / 890 = 0.75 (m s-2) ✔ (0.753 m s-2) (using v = u + at)v = 5.5 – 0.753 × 3.5 = 2.8 (m s-1) ✔which is a safe landing speed ✔Or(using Ft = ∆mv)670 × 3.5 = 890 × ∆v ✔∆v = 2.7landing speed = 5.5 – 2.7 = 2.8 (m s-1) ✔which is a safe landing speed ✔

3rd mark is dependent on having a valid attempt at the calculation.3rd mark can be given for wrong answer if it follows from an arithmetic error.

3[8]


E1.This question required students to apply their knowledge and understanding of physics to a simple seismometer. Although the diagram contained a lot of information, and there was a relatively long stem to the question, there was no evidence to suggest that students found the context particularly demanding.

(a) This was a multi-step calculation that most students found fairly straight forward. The common errors seen were wrong substitution of diameter (or use of a wrong formula for volume) and power of ten errors arising from calculation of volume in cm3. Students who have difficulty converting between cm3 and m3 would be better advised to work in m from the outset. Generally “show that” questions are used to provide unsuccessful students the data they would need to complete further parts of the question. Students should be reminded to provide at least 1 sf more than the “show that” value, and they should be discouraged from trying to calculate an answer backwards. Another error is forcing their answer to be near the “show that” value: many students were denied consequential error marks when, having made an error, they attempted to manipulate their answer to obtain a numerical value near 5. For example, students who used 5 cm for the radius could obtain a value of 4.2 kg for the mass. Many would then miss out the step (multiplying by g) to determine the weight, as they had already reached a value near to 5 perhaps. Many modern calculators generate results as fractions or surds. No credit is given for final answers given in such a form or with recurring notation but there is no penalty for this with intermediate results. However students should be discouraged from doing this because it makes the work less transparent and inhibits error-checking. The rounding down of intermediate results compromises the chance of full credit; any rounding down, e.g. to the same significant figures as that of the least accurate data should not be done until the final stage is complete.

(b) There were two potential errors in answers that often led students to lose at least one mark. Many students did not take the centre of mass of the ball into account, and therefore did not include the radius when calculating the distance to the pivot. Some students worked through their answers in cm, but wrote the moment unit as Nm.

(c) Many good answers were seen to this multi-step calculation and this was a good discriminator. Some students were unable to suggest much beyond picking F = 5 N and rearranging F = k∆l (given on the data sheet) to produce ∆l = 0.05 m. Spotting that this was a 3 mark question may have led some of them to realise that a more complicated calculation was needed. Others tried to calculate the extension by dividing turning moment by stiffness or by multiplying distance from the pivot by stiffness. A number of students did not attempt this question.

(d) This was a fairly demanding question that aimed to get students to think about the reason for having the heavy ball in the seismometer. Successful answers were able suggest that, in the very short length of time involved, the ball would barely move and therefore the arm holding the pen would pivot about the ball, causing the upwards line. Many incorrect answers were seen: some students were convinced there was a third law or conservation of momentum explanation while others said the spring, having become compressed, then pulled the arm up. It seemed that many felt that the downwards accelerating seismometer took the ball with it and so the line went downwards. No credit was earned for saying the pen or the arm did not move, likewise any suggestion of an ‘up and down’ motion of the pen (although ‘up then down’ could earn a mark).


E2.(a) Candidates were generally very successful on these questions.

(b) Candidates were generally very successful on these questions.

(c) Most candidates were successful. However, a few used t = s / v = 54 / 33 instead of a kinematics equation.

(d) As with previous papers, candidates lacked the depth of knowledge necessary to describe energy transfers. Perhaps some teachers assume candidates already know this from GCSE. More practice on this type of question is necessary. Too few candidates are able to make the link between potential and kinetic energy.

E3.(a) (i) The answers seen split clearly into two groups. The largest of which scored the mark because they could resolve a vector into its components. The smaller group did not score because they thought the question could be approached through the equations of motion.

(ii) Slightly fewer scored full marks compared to those in 2(a)(i) because, although just as many used a vector rather than an equation of motion approach, there was more opportunity for things to go wrong. There was more data to choose from with the velocity and the vertical component of velocity both available.

(b) (i) A majority of students performed well choosing the correct equation of motion to obtain the correct answer. There were a number of students who used the wrong equation and / or data. Also equal numbers scored one mark either by answering with a wrong answer given to two significant figures or by giving the correct answer but not using two significant figures.

(ii) Again the vast majority had no problem with this question especially when an error carried forward was allowed. Those who got it wrong normally wanted to include a non-zero acceleration and used the value of 'g'.

(c) These questions were generally answered very well.

(d) The overall correct shape was achieved by a majority of students. Some students did not realise that the lines on the graph needed to be straight because of the constant acceleration due to gravity. An equal number seemed to guess at the answer and had the maximum speed at the time corresponding to the maximum height.

E4.(a) It was only the occasional slip or showing evidence of working to one significant figure that caused marks to be lost in this relatively simple question.


(b) (i) A significant minority failed to score this mark. It was common to see students fail to read the words of the question and simply write down Newton's third law. Others did not explain themselves fully and thought, 'they are equal' was a sufficient answer. A vast majority went for the correct ‘equal magnitude’ answer.

(ii) It was again only a minority who failed to score a mark. A number of these thought that the one word answer 'direction' was sufficient rather than saying 'opposite direction'. Very few students discussed 'different bodies'.

(c) This question was tackled well by the majority who focussed on the unequal magnitudes of the forces. There were some successful references to not acting on different bodies or the forces being of different types. Students who did not score either quoted Newton III or they said the forces were 'not equal' without specifying in which way.

(d) Having the ability to organise an answer in a logical sequence was a skill that determined the students’ mark in this question. There were about half a dozen valid ways in which the conclusion that the landing speed was within a safe value could be ascertained. Most of these methods relied on a progression of calculations. Weaker responses simply put some of the data in an equation of motion almost chosen at random to get another number and produced nothing useful. Zero and full marks were seen in approximately equal numbers with a minority scoring a total of one mark for calculating the actual deceleration.

web view(ii) calculate the horizontal component of velocity as the projectile hits the ground. ......

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