4.2 mechanics and materials - scalars and vectors 2 - ms

4.2 Mechanics and Materials - Scalars and Vectors 2 – Mark schemes

Q1. (a) Calculation of energy = 12 × 7.2 × 104 = 8.64 × 105 J

Or time = 12000 / 1.5 = 8000 s ✔

Calculation of other quantity and substitution in power = useful energy / time taken ✔

Power = 110 (108 W) ✔

Or

Time = 8000 s ✔ Allow ecf for current or time

Current = charge / time = 9 A ✔

Power = VI = 108 (W) ✔ 3

(b) Attempt to use Power / velocity ✔ Allow use of 100W for P

73 N ✔ Ignore inclusion of KE in calculation If 108 used then answer is 72 N If 100 used then answer is 67 N

or

work done = F × 12000 ✔

equates to 110 × 8000 so F = 73 N ✔ allow ecf from 3.1

2

(c) Force parallel to slope = 120 × 9.81 × sin 4.5 = 92 N ✔ 1

(d) Total resistive force = ans to (c) + ans to (b) (= 165 N) ✔ Allow ecf for incorrect F

Allow 0.66 / 0.67 if 108 W or 110 W used

2

(e) Increasing the mass Reward discussion of compression of tyres

Reduces the range ✔

increases the friction on the bearings/tyres

OR More energy / power is used accelerating the user to the final speed

OR user and wheelchair have higher KE/ more energy to move ✔

Increasing the speed

Reduces the range ✔

Air resistance increases with speed ✔ Treat as independent parts If not explicit about increasing / decreasing lose the first mark in each part Within each part, second mark is dependent on the first Allow opposite answers for decreasing mass / speed

4 [12]

Q2. statement that forces up = forces down/correctly resolved vertical component of either drag or tension

C1

2600 sin 41 = W + 2200 sin 27 seen (or equivalent kN)

C1

1705.8 = W + 998.8 (condone power 10 error)

707/710 (N)

A1 3

[3]

Q3. (a) (i) air resistance/drag

B1

(normal) reaction (of the ground on the skier)

B1 2

(ii) no resultant force (in any direction)/forces in equilibrium

B1 1

(b) any closed triangle with W as a complete side

M1

closed triangle with correct lengths or angles even if P and Q are reserved

A1

correct triangle by eye

A1

force correct 490 ± 20 N

B1 4

(c) (i) appropriate force/87 ecf

C1

5.4 to 5.9 ms–2 cao

A1 2

(ii) deceleration would decrease

B1

resistance forces increase with speed/are proportional to speed2/

resultant force gets smaller as speed gets less

B1 2

[11]

Q4. (a) (quantity that has both) magnitude and direction

B1 1

(b) any vector quantity eg velocity, force, acceleration

B1 1

[2]

Q5. (a) 170 × 103 × cos 40 (or 1.3 × 105)

× 1000 (= 1.302 × 108J) 2

(b) (i) (P = Fv) = 1.3 × 105 × 7.0

= 9.1 × 105 (911.6 kW) ecf wrong force from part a

or correct alternative approach using t = s/v and P = E/t 2

(ii) (total power =) 9.1 (× 105) + 21 (× 105) ecf from bi

powers of 10 must be consistent for this mark (eg not 9.1 + 2.1)

(percentage = (9.1/9.1 + 21) × 100)

= 30 (%)

ecf for power of 10 error in first mark

ecf from bi, but must have attempted to calculate total power

eg 9.1/21 gains zero 2

(c) any two from

(surface) area (of the sail)

wind speed/strength/power/KE/force (not air resistance)

acceleration or speed of the ship 2

[8]

Q6. (a) (i) uses trigonometry (mg sin5 or mg cos85 seen)

B1

829.3 / 828.5 (N) at least 3 sf

B1 2

(ii) tension = 830 (N)

C1

E= ½ F∆L and F = k ∆L identified / or combined to E= ½ (F 2 /k)

C1

correct sub condone power 10 error

C1

13.8 (J) range 13.9 to 13.7

A1 4

(b) lower speed

B1

less extension

B1

less energy stored (in rope)

B1 3

[9]

Q7.

(a) Correct answer with no working gets 2 out of three.

OR substitution in above equation OR = 6.9767 – 5.0 = 2.0 (1.98 m s−1)

Full credit for use of gsin25 = acceleration down slope. This yields answer 3.22 m s−1

Allow 1sf answer (2). 3

(b) (i) (F = 75 × 9.81 ×) sin25 (o ) = 310 ( 311, 310.94) (N)

use of g = 10 not penalised here ‘sin25’ on its own Use of g = 10 yields 317 Allow cos65

2

(ii) W = Fs = 311 × 2.0 = 620 (622 J) ecf (2bi) × 2.0

1

(c) Idea that GPE is ultimately transferred to: internal (energy) / 'heat'/ 'thermal' (energy in the surroundings)

Allow transfer of GPE to KE and then to ‘thermal’ etc Do not allow reference to ‘sound’ on its own

Correct reference to a named resistive force: friction / drag / air resistance Don’t accept implication that a resistive force is a form of energy

All GPE becomes 'heat', etc OR no (overall) increase in KE OR reference to work done against or by a resistive force

Do not allow references to loss of body heat. Allow: ‘(GPE) not converted to KE’

3 [9]

Q8. (a) ∆h = 2.51 – 1.00 = 1.51 (m) / (s =) 1.51 m seen

M1

use of appropriate kinematics formula correctly makes t subject

M1

time = 0.555 (s) / 0.56 (s) (allow 0.55 (s))

A1 3

(b) (i) use of appropriate kinematics equation to find vertical v

C1

v = 5.4 (ms–1) (accept 5.4 to 5.9)

A1 2

(ii) any use of Pythagoras where vh = 18 or use of appropriate trig ratio where vh = 18 and angle is to horizontal

C1

velocity = 18.8 / 18.9 / 19 (ms–1)

A1

angle = 16.8 to 18.1 (°)

A1 3

[8]

Q9. (a) (i) W = 2mg cosφ ∴ m = W/(2g cosφ)✔

The question says show that, so the candidates must write down both steps.

1

(ii) Well drawn straight line of best fit.✔ The line should follow the trend of the points with an even scatter of points on either side of the line.

1

(b) (i) Triangle drawn with smallest side at least 8 cm in length. ✔ Correct readings taken from the line for the triangle ✔ Gradient in the range 0.45 to 0.49 (0.445 to 0.494) quoted to 2 or 3 significant figures ✔

The size of the triangle can be identified from readings taken from the line. The third mark is independent of the other two: error carried forward for incorrect readings (or for a poor line of best fit) which give a gradient out of range is not allowed.

3

(ii) Candidate’s answer for gradient in (b)(i) correctly multiplied by g (expected answer 4.6)✔ N ✔

No s.f. penalty. The second mark is for the unit and can be awarded if the numerical answer is incorrect.

2

(c) δx% = 0.2 and δy% = 0.5✔ δ(x/y)% = δx% + δy% = 0.2 + 0.5 = 0.7 ✔ Use of δ(x/y.)2% = 2 × δ(x/y)(%) ✔

Final answer is (±) 1.4 (%) which automatically gains all three marks Otherwise Accept only 1 s.f. for 1st and/or 2nd marks. The third mark is for the method, not the final answer

3

(d) (i) Systematic errors in measurements are errors which show a pattern or a bias or a trend ✔

Some acceptable alternatives • A systematic error is one which deviates by a fixed

amount from the true value of a measurement • An error which has the same value in all readings • A difference between the true value of a quantity and

the indicated value caused by a fault in the measuring device

• Accept a good example of systematic error. 1

(ii) y would be larger✔ because angle θ would be smaller or because friction would be opposing the increasing weight of m✔

2 [13]

Q10. (a) force is at right angles / horizontal to wall (and so no

component along the wall) ✔ 1

(b) arrow drawn upwards from point of contact with ground ✔ between ladder and vertical which when extended would pass through ‘on ladder’ part of label ✔

Must be vertical or pointing to right for first mark 1 1

(c) (use of sum of clockwise moments = sum of anticlockwise moments) EITHER F × 8.0sin 60 OR 390 × 4.0 cos60 ✔ F = 390 × 4.0 cos60/8.0sin 60= 110 (113) N ✔

Either moment correct gets first mark (does not have to be evaluated)

1 1

(c) ANY 3 (vertical reaction) force from ground increases ✔ direction of arrow of resultant force from ground changes (as ladder is ascended) ✔ friction force / horizontal component of G between ladder and ground would increase ✔ the (reaction) force from the wall increases ✔ weight (of system) / load on ladder increases ✔ as person climbs they exert a force along the ladder ✔

MAX 3 [8]

Q11. (a) (i) 1000(N) AND 6000(N) seen

Independent marks

OR

allow incorrect values seen = 6083 (N) ( = 6100) More than 2 sf seen

Allow full credit for appropriate scale drawing Ignore rounding errors in 3rd sig fig.

2

(ii) tanӨ = 1000 / 6000 or correct use of sin or cos Ө = 9.5 (9.46°) Allow range 9.4 − 10.4

Use of cos yields 10.4 Allow use of 6100 Some working required for 2 marks. Max 1 mark for correct calculation of vertical angle (range 79.6 − 80.6) some working must be seen

2

(iii) (m = W/g = ) 6500 / 9.81 ( = 662.6 kg) (a = F / m = 6083 / 662.6) = 9.2 (ms−2) (9.180)

Use of weight rather than mass gets zero Correct answer on its own gets 2 marks Penalise use of g=10 in this question part only (max 1)

2

(b) (i) = 6500 × 600 (662.6 × 9.81 × 600) = 3 900 000 (J)

Look out for W x g x h which gives 39000000 (gets zero) Correct answer on its own gets 2 marks Do not allow use of 1/2 mv2 (= 39 000)

2

(ii) (E= Pt =) 320 000 ×; 55 (= 17 600 k J ) OR P= 1(b)(i) / 55 (7.09 × 104) 3.9 / 17.6 OR 70.9 / 320 OR = 0.22(16) ecf from first line

Some valid working required for 3 marks

conversion to a percentage (= 22 %) Look out for physics error: Power / time (320/55) then use of

inverted efficiency equation yielding correct answer Do not allow percentages >= 100% for third mark

3 [11]

Q12. (a) (i) two from: velocity, acceleration, force etc

1

(ii) two from: speed, distance, mass etc 1

(b) (i) B: drag / air resistance

C: weight 2

(ii) closed triangle (of vectors)

so forces are in equilibrium / resultant force is zero / forces balance (so moving at constant velocity)

2

(c) W = 9500 sin 74

= 9100 (9132)

2 sf 3

[9]

Q13. (a) 2800/cos 20 = 3000 (2980)N (1)

B1 1

(b) 42000J (1)

B1 1

[2]

Q14. (a) (i) a quantity that has magnitude only

[or has no direction] (1)

(ii) any two: e.g. energy (1) temperature (1)

3

(b) (i)

scale (1) 5 N and 9.5 N (1) correct answer (8.1 N ± 0.2 N) (1)

[or 9.52 = 5.02 + F2 (1) F2 = 90.3 – 25 (1) F = 8.1 N (1) (8.07 N)]

(ii) cos θ =

gives θ = 58° (1) (± 2° if taken from scale diagram) 4

[7]

Q15. (a) displacement is a vector (1)

ball travels in opposite directions (1) max 1

(b) velocity is rate of change of displacement average speed is rate of change of distance velocity is a vector [or speed is a scalar) velocity changes direction

any two (1) (1) 2

(c) (i) a = (1)

= (–)140.m s–1 (1)

(allow C.E. for incorrect values of Δv)

(ii) F = 0.45 × (–) 140 = (–) 63N (1) (allow C.E for value of a)

(iii) away from wall (1) at right angles to wall (1) [or back to girl (1) (1)] [or opposite to direction of velocity at impact (1) (1)]

5 [8]

Q16.

(a) AB: (uniform) acceleration (1) BC: constant velocity / speed or zero acceleration (1) CD: negative acceleration or deceleration or decreasing speed / velocity (1) DE: stationary or zero velocity (1) EF ; (uniform) acceleration in opposite direction (1)

5

(b) area under the graph (1) 1

(c) distance is a scalar and thus is the total area under the graph [or the idea that the train travels in the opposite direction] (1) displacement is a vector and therefore the areas cancel (1)

2 [8]

Q17. (a) (i) rate of change of velocity

[or a = ] (1)

(ii) (acceleration) has (magnitude and) direction (1) 2

(b) (i) (acceleration) is the gradient (or slope) of the graph (1)

(ii) (displacement) is the area (under the graph) 2

(c)

4

[8]

Q18. (a) (i) horizontal component of the tension in the cable (1)

(ii) vertical component of the tension in the cable (1) 2

(b) (i) Tvert = 250 × 9.81 = 2500 N (1) (2452 N)

(ii) Thoriz = 1200 N

(iii) T2 = (1200)2 + (2500)2 (1) T = (1.44 × 106 + 6.25 × 106)1/2 = 2800 N (1) (2773 N) (if use of Tvert = 2450 N then T = 2730 N)

(allow C.E. for values from (b) (i) and (b)(ii))

(iv) tan θ = (1)

θ = 26° (1) (allow C.E. for values from (b) (i) and (b)(ii))

6 [8]

Q19. (a) √(1.32 + 1.02)

C1

1.6/1.64 (m s–1)

A1 2

(b) angle = tan–1 (1.0/1.3)

C1

N38°E/N37.6°E

A1 2

[4]

Q20. (a) (i)

(1)

(ii) no horizontal force acting (1) (hence) no (horizontal) acceleration (1) [or correct application of Newton’s First law]

3

(b) (i) (use of v2 = u2 + 2as gives) 322 = (0) + 2 × 9.81 × s (1)

s = (1) (= 52.2 m)

(ii) (use of s = ½ at2 gives) 52 = ½ 9.81 × t2 (1)

= 3.3 s (1) (3.26 s)

[or use of v = u + at gives 32 = (0) + 9.81 × t (1)

= 3.3 s (1) (3.26 s)]

(iii) (use of x = vt gives) × (= QR) = 95 × 3.26 (1) = 310 m (1)

(use of t = 3.3 gives x = 313.5 m) (allow C.E. for value of t from (ii)

6

(c) maximum height is greater (1) because vertical acceleration is less (1) [or longer to accelerate]

2 [11]

Q21. (a) component (parallel to ramp) = 7.2 × 103 × sin 30 (1) (= 3.6 × 103 N)

1

(b) mass = = 734 (kg) (1)

a = = 4.9(1) m s–2 (1) 2

(c) (use of v2 = u2 + 2as gives) 0 = 182 – (2 × 4.9 × s) (1) s = 33(.1) m (1) (allow C.E. for value of a from (b))

2

(d) frictional forces are acting (1) increasing resultant force [or opposing motion] (1) hence higher deceleration [or car stops quicker] (1) energy is lost as thermal energy/heat (1)

Max 2 [7]

Q22. (a) (i) (horizontal) force = zero (1)

(ii) (vertical) force = 2 × 15 sin 20 (1)

= 10(.3) N (1) 3

(b) (i) weight (of block) = 10(.3) N (1)

(allow C.E. for value from (a) (ii))

(ii) resultant force must be zero (1)

with reference to an appropriate law of motion (1) 3

[6]

Q23. (a) vector quantities have direction (as well as magnitude)

and scalar quantities do not (1) 1

(b) vector: e.g. velocity, acceleration, momentum (1) scalar: e.g. mass, temperature, energy (1)

2

(c) (i) addition of forces (12 + 8) (1)

(use of F = ma gives) a = = 3.1 m s–2 (1) (3.08 m s–2)

(ii) subtraction of forces (12 – 8) (1)

a = = 0.62 m s–2 (1) (0.615 m s–2) 4

[7]

Q24. (a) λ(=2 × 38) = 76(m)

MHz (1) 1

(b) (i) angle between cable and horizontal = (1)

T= 110 cos59° = 57N • (56.7N) (1) (allow C.E. for value of angle)

(ii) cross-sectional area (= π(2.0 × 10–3)2)

=1.3 × 10–5(m2) (1) (1.26 × 10–5(m2))

stress (1)

= 4.4 × 106Pa (1) (4.38 × 106Pa) (use of 56.7 and 1.26 gives 4.5 × 106 Pa) (allow C.E. for values of T and area)

(iii) breaking stress is 65 × stress copper is ductile copper wire could extend much more before breaking because of plastic deformation extension to breaking point unlikely

any three (1)(1)(1) 7

[8]

Q25. (a) technique one (1)

information derived from it (1)

technique two (1)

information derived from it (1) 4

(b) (i) gravitational attraction to… (1)

…centre of gravity (mass) of mountain (1)

(ii) cancellation of some systematic errors (1) 3

(c) (i) calculates volume of cone (1)

mass = density × volume seen (1)

2.2 × 1012 kg (1)

(ii) sideways force/mg = tan (0.0011) (1)

sideways force = Gmsch 0.5/(1400)2 subst seen (1)

2.4 × 1024 kg (1)

(iii) his density estimate was too low (1)

or mean density of the Earth is higher than that of the mountain (1) 7

[14]

Q26. vertical component of rope = 610 × cos (20) or 573(N)

C1

610 cos 20 seen in an equation

(vertical component of resultant) = 590 – 573 or their Fv

C1

16.8 or 17(N) cao

A1 [3]

Q27. resolving one force correctly –410 sin (65) seen

C1

doubling the force (eg 743.1 or 346.5 seen)

C1

2 sf answer (740 (N) or 370 (N))

A1 [3]

Q28. (a) 8300 × 9.81 OR = 81423 ✓

(8300 × 9.81 sin 25) = 3.4 × 104 (N) ✓ (34 411 N) ecf from first line unless g not used

msin25 gets zero Penalize use of g = 10 here only (35 077 N) Allow 9.8 in any question Correct answer only, gets both marks for all two mark questions

2

(b) (i) (Ek = ½mv2 ) = ½ × 8300 × 562 ✓ = 1.3 × 107 (J) ✓ (13 014 400) allow use of 8300 only

In general: Penalise transcription errors and rounding errors in answers

2

(ii) mgh = KE (13 014 400) for mgh allow GPE or Ep

OR 13 014 400 / 81 423 ✓ h = 160 (m) ✓ (159.8) ecf 1bi

Allow use of suvat approach 2

(c) (i) (work done) by friction \ drag \ air resistance \ resistive forces ✓ (energy converted) to internal \ thermal energy ✓

Allow ‘heat’ 2

(ii) 0.87 x (8300 x 9.81 x 140 = 9 917 000) OR V = ✓ = 49 ( = 48.88 ms-1) ✓

87% of energy for 140m or 160m only for first mark. Use of 160 (52.26) and / or incorrect or no % (52.4) gets max 1 provided working is shown Do not credit suvat approaches here

2 [10]

Q29. (a) (i) 0.416 or 0.417 and 0.495 or 0.496

1

(ii) Both plotted points to nearest mm✓ Straight line of best fit ✓

The line should be a straight line with approximately an equal number of points on either side of the line.

2

(iii) Large triangle drawn (at least 8cmx8cm) ✓

Correct values read from graph✓ Gradient value in range 0.805 to 0.837 to 2 or 3 sf✓

3

(iv) (1) For showing correct vertical component of at least one of the forces / tensions as mgcosθ or both vertical components as 2 mgcosθ Question specifically referred to resolving forces so component must include g.✓

1

(2) cosθ = = ✓

= compared to y = mx ✓

(Hence gradient is ) 2

(3) Magnitude of m correct from E.g middle gradient value gives m = 0.609 kg

✓ kg and 2 or 3 sf ✓ Allow ecf from gradient value. Sf and unit mark depends on correct calculation of m from the gradient value.

2

(v) (1) Friction at the pulleys ✓ 1

(2) Take a mean value of readings from loading and unloading ✓ 1

(b) (i) Description of technique: Use small plane mirror beneath string / use of set square / bright light source to project shadow of the strings onto the paper, and marking points on shadow to aid drawing lines ✓

Explanation: Line of sight not perpendicular from string to paper / mark on paper depends on the angle the eye is positioned at / reference to parallax error. ✓

2

Markers should measure the angle to check that no scaling error has been introduced in the photocopying of the paper.

If the angle is different, mark accordingly. Answers should be consistent with protractor precision stated by the candidate.

(ii) (1) Value of θ quoted as 30o or 31o (for a protractor with precision ±1o) OR θ = 30.0, 30.5, 31.0 (for a protractor with precision ± 0.5o) ✓

Correct computation of % uncertainty, answer quoted to 2 or 1sf ✓ Allow ecf for incorrect angle (penalised in 1st marking point).(e.g. if using a protractor with 1o precision % uncertainty will be 1/31 × 100% = 3.2% or 3% OR for candidates who measured the angle 2 θ % uncertainty = 1/62 × 100 = 1.6% or 2%. With a protractor with precision ± 0.5o the % uncertainties will be half these values) This is because the question specifically stated “as accurately as possible”. It should be clear from the candidate’s percentage uncertainty calculation whether 2θ or θ has been measured.

Extra mark for a candidate who measures the angle 2θ (rather than just the single angle θ) ✓ (This 3rd mark can also be awarded for a candidate who has measured θ on both sides of the ‘vertical line’,and taken the mean value)

3

(2) Evidence of right angled triangle drawn on to the diagram with dimensions of two sides also shown on the diagram. The minimum dimension shown must be 70mm. ✓ Correct use of cosine rule without right angled triangle is acceptable.

Angle correctly computed using sine cosine or tangent, with value quoted in the range 30.0o to 31.4o ✓

Angle quoted to 3 sf/to 0.1o

2nd mark is still available to a candidate who didn’t achieve the 1st mark.

2

(iii) Plot a graph of cos θ against 1/m AND Statement that it should give a straight line through origin. ✓

Allow graphs of 1/m against cos θ, against 1/cos θ against m, which would all be straight lines through the origin.

1 [21]

Q30. (a) (has magnitude but) no direction/has only magnitude

B1 1

(b) (i) 1N: 3.9 or 4 cm/1 cm: 0.25 or 0.26 N

B1

1

(ii) completes parallelogram correctly/or ‘nose to tail’

M1

measures/draw correct diagonal

M1

2.1 ± 0.1 N

A1 3

[5]

Q31. (a) (i) velocity is constant (1)

no acceleration (1)

(ii) 1.5 sin 50 = D cos 55 (1)

2.0 kN (1) 4

(b) (i) 1.15 kN (1)

(ii) total resistance to motion = 1200 + 1150 N (1)

use of power = Fv (1)

20 (1)

kW (1) 5

(c) boat now has resultant force of 1200 N acting on it (1)

boat will accelerate (until resistance of water = 2350 N) (1) 2

[11]

Q32.

(a) (i) (1)

= 7.5 × 103 (V m–1) (1) 2

(ii) force F (= EQ) = 7500 × 0.17 × 10–6 (1) (= 1.28 × 10–3 N) 1

(b) (i) correct labelled arrows placed on diagram to show the three forces acting;

• electric force F (or 1.3 mN) horizontally to left (1)

• W (or mg) vertically down and

• tension T upwards along the thread (1) 2

(ii) F = T sinθ and mg = T cosθ give F = mg tanθ (1) (or by triangle or parallelogram methods)

tanθ = (1)

gives θ = 15(.2) (°) (1) 3

[8]

Q33. (a) Velocity and speed correct ✓

Distance and displacement correct ✓

velocity speed distance displacement

vector ✓ ✓

scalar ✓ ✓

2

(b) (i) v2 = u2 + 2as

v = ✓ v = ✓

= (−)3.9 (m s−1) ✓two or more sig fig needed (− 3.87337 m s−1) 1st mark for equation rearranged to make v the subject (note sq’ root may be implied by a later calculation) penalise the use of g = 10 m s2 only on this question 2nd mark for substituting numbers into any valid equation 3rd mark for answer Alt’ approach is gainKE = lossPE missing out u gives zero marks answer only gains one mark [Note it is possible to achieve the correct answer by a wrong calculation]

3

(ii) velocity / ms−1

first line descends from X to the dotted line at tA or up to one division sooner ✓ (allow line to curve)

first line is straight and descends from X to v = −4 (m s−1) ✓(allow tolerance one division) second line has same gradient as the first, straight and descends to v = 1(m s−1) ✓(tolerance ½ division) a steep line may join the two straight lines but its width must be less than 2 divisions

3

(c) s = ut + 1 / 2at2

t = OR correct substitution seen into either equation t = ✓

= 0.49 (s) ✓ (0.4946 s)

working must be shown for the first mark but not the subsequent marks

v = s / t = 5.0 / 0.49 = 10 (m s−1) ✓ (10.2 m s−1) (allow CE from their time)

[note it is possible to achieve the correct answer by a wrong calculation]

3 [11]

Q34.

(a) (i) (1) = (–)2.0 (= 1.99 m s–2) (1) 2

(ii) (v = u + at) or substitution (1)

= = 950 (s) (1) ecf from (i) 2

(b) (i)

opposing vertical arrows of roughly equal length or labelled weight/mg/gravity/W and thrust/reaction/R/F/TF/engine force/rocket force/motor force/motive force/driving force (1)

correctly labelled + arrows vertical + not more than 2 mm apart + roughly central + weight arrow originates within rectangular section and thrust originates within rectangular section or on jet outlet (1)

2

(ii) new mass = 15100 × 0.47 = 7097 (kg) (1)

(F = mg = 7097 × 16(1)) = 11000 (= 11426 N) (1) 2

(c) (v2 = u2 + 2as v = correct u, a and s clearly identified (1)

= 2.1 (= 2.122 m s–1) (1) 2

[10]

Q35.

(a) Substitution of data in

3.1 × 10–3 (m) ✔ 2 marks can be awarded if 4mm used to show T>500 N provided an explanation is provided, otherwise award zero.

2

(b) (500 = Tcos 65)

T = 1200 N ✔ 1

(c) Wind produces a wave / disturbance that travels along the wire ✔

Wave is reflected at each end / waves travel in opposite

directions✔

(Incident and reflected) waves interfere / superpose ✔

Only certain frequencies since fixed ends have to be nodes. ✔

4

(d) Mass per m of the wire = 0.14(2) kg ✔ 1

(e) Use of to find fundamental

Third harmonic = 7.4 (Hz) ✔ The second mark is for multiplying the fundamental frequency by 3 – allow ecf

2

(f) Diagram showing three approximately equally spaced loops Condone single line

1

(g) Copper may be stretched beyond elastic limit / may deform plastically ✔

Permenant deformation / Does not return to original length ✔ Allow 'will remain longer than original' or 'will be permenantly deformed'

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