Mixed for Questions Unit 2 - Hinchley Wood School

Mixed FOR questions – unit 2 - Hinchley Wood School

Q1. (a) State the difference between vector and scalar quantities.

......

(1)

(b) State one example of a vector quantity (other than force) and one example of a scalar quantity.

vector quantity ......

scalar quantity ......

(2)

(i) the maximum resultant acceleration that it could experience,

......

(ii) the minimum resultant acceleration that it could experience.

......

(4)

(Total 7 marks)

Q2. The diagram below shows a dockside crane that is used to lift a container of mass 22000 kg from a cargo ship onto the quayside. The container is lifted by four identical ‘lifting’ cables attached to the top corners of the container.

(a) When the container is being raised, its centre of mass is at a horizontal distance 32 m from the nearest vertical pillar PQ of the crane’s supporting frame.

(i) Assume the tension in each of the four lifting cables is the same. Calculate the tension in each cable when the container is lifted at constant velocity.

answer ...... N

(2)

(ii) Calculate the moment of the container’s weight about the point Q on the quayside, stating an appropriate unit.

answer ......

(3)

(iii) Describe and explain one feature of the crane that prevents it from toppling over when it is lifting a container.

......

(2)

(b) Each cable has an area of cross–section of 3.8 × 10–4 m2.

(i) Calculate the tensile stress in each cable, stating an appropriate unit.

answer ......

(3)

(ii) Just before the container shown in the diagram above was raised from the ship, the length of each lifting cable was 25 m. Show that each cable extended by 17 mm when the container was raised from the ship.

Young modulus of steel = 2.1 × 1011 Pa

(2)

(Total 12 marks)

Q3. A cyclist pedals downhill on a road, as shown in the diagram below, from rest at the top of the hill and reaches a horizontal section of the road at a speed of 16 m s–1. The total mass of the cyclist and the cycle is 68 kg.

(a) (i) Calculate the total kinetic energy of the cyclist and the cycle on reaching the horizontal section of the road.

answer ...... J

(2)

(ii) The height difference between the top of the hill and the horizontal section of road is 12 m.
Calculate the loss of gravitational potential energy of the cyclist and the cycle.

answer ...... J

(2)

(iii) The work done by the cyclist when pedalling downhill is 2400 J. Account for the difference between the loss of gravitational potential energy and the gain of kinetic energy of the cyclist and the cycle.

......

(3)

(b) The cyclist stops pedalling on reaching the horizontal section of the road and slows to a standstill 160 m further along this section of the road. Assume the deceleration is uniform.

(i) Calculate the time taken by the cyclist to travel this distance.

answer...... s

(3)

(ii) Calculate the average horizontal force on the cyclist and the cycle during this time.

answer ...... N

(3)

(Total 13 marks)

Q4. The diagram below shows the path of a ball thrown horizontally from the top of a tower of height 24 m which is surrounded by level ground.

(a) Using two labelled arrows, show on the diagram above the direction of the velocity, v, and the acceleration, a, of the ball when it is at point P.

(2)

(b) (i) Calculate the time taken from when the ball is thrown to when it first hits the ground. Assume air resistance is negligible.

Answer ...... s

(2)

(ii) The ball hits the ground 27 m from the base of the tower. Calculate the speed at which the ball is thrown.

Answer ...... m s–1

(2)

(Total 6 marks)

Q5. An apple and a leaf fall from a tree at the same instant. Both apple and leaf start at the same height above the ground but the apple hits the ground first.

You may be awarded marks for the quality of written communication in your answer.

Use Newton’s laws of motion to explain why

(i) the leaf accelerates at first then reaches a terminal velocity,

......

(ii) the apple hits the ground first.

......

(Total 5 marks)

Q6. Tidal power could make a significant contribution to UK energy requirements. This question is about a tidal power station which traps sea water behind a tidal barrier at high tide and then releases the water through turbines 10.0 m below the high tide mark.

(i) Calculate the mass of sea water covering an area of 120 km2 and depth 10.0 m.

density of sea water = 1100 kg m–3

......

(ii) Calculate the maximum loss of potential energy of the sea water in part (i) when it is released through the turbines.

......

(iii) The potential energy of the sea water released through the turbines, calculated in part (ii), is lost over a period of 6.0 hours. Estimate the average power output of the power station over this time period. Assume the power station efficiency is 40%.

......

(Total 7 marks)

Q7. A digital camera was used to obtain a sequence of images of a tennis ball being struck by a tennis racket. The camera was set to take an image every 5.0 ms. The successive positions of the racket and ball are shown in the diagram below.

(a) The ball has a horizontal velocity of zero at A and reaches a constant horizontal velocity at D as it leaves the racket. The ball travels a horizontal distance of 0.68 m between Dand G.

(i) Show that the horizontal velocity of the ball between positions D and G in the diagram aboveis about 45 m s–1.

(3)

(ii) Calculate the horizontal acceleration of the ball between A and D.

answer = ...... m s–2

(1)

(b) At D, the ball was projected horizontally from a height of 2.3 m above level ground.

(i) Show that the ball would fall to the ground in about 0.7 s.

(3)

(ii) Calculate the horizontal distance that the ball will travel after it leaves the racket before hitting the ground. Assume that only gravity acts on the ball as it falls.

answer = ...... m

(2)

(iii) Explain why, in practice, the ball will not travel this far before hitting the ground.

......

(2)

(Total 11 marks)

M1. (a) vector quantities have direction (as well as magnitude)
and scalar quantities do not (1)

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

[7]

M2. (a) (i) weight of container (= mg = 22000 × 9.8(1)) = 2.16 × 105 (N) (1)

tension (= ¼ mg) = (5.39) 5.4 × 104 (N) or divide a weight by 4 (1)

(ii) moment (= force × distance) = 22000 g × 32 (1) ecf weight in (a) (i)

= 6.9 or 7.0 × 106(1) N m or correct base units (1) not J, nm, NM

(iii) the counterweight (1)

provides a (sufficiently large) anticlockwise moment (about Q)
or moment in opposite direction ( to that of the container to
prevent the crane toppling clockwise) (1)

or
left hand pillar pulls (down) (1)
and provides anticlockwise moment

or
the centre of mass of the crane(‘s frame and the counterweight)
is between the two pillars (1)

which prevents the crane toppling clockwise/to right (1)

(b) (i) (tensile) stress ecf (a) (i) (1)

= 1.4(2) × 108(1) Pa (or N m–2) (1)

(ii) extension = (1)

= and (= 1.7 × 10–2 m) = 17 (mm) (1)

[12]

M3. (a) (i) (EK = ½ mv2 =) 0.5 × 68 × 162 (1) = 8700 or 8704(J) (1)

(ii) (ΔEP = mgΔh =) 68 × 9.8(1) × 12 (1) = 8000 or 8005 (J) (1)

(iii) any three from

gain of kinetic energy loss of potential energy (1)

(because) cyclist does work (1)

energy is wasted (on the cyclist and cycle) due to air resistance
or friction or transferred to thermal/heat (1)

KE = GPE + W – energy ‘loss’ (1) (owtte)

energy wasted (= 8000 + 2400 - 8700) = 1700(J) (1)

(b) (i) (u = 16 m s–1, s = 160 m, v = 0, rearranging s = ½ (u+v) t gives)

160 = ½ × 16 × t or t = or correct alternative

(gets 2 marks) (1) = 20s (1)

(ii) acceleration a = ecf (b) (i) (1) = (–) 0.80 (m s–2)

resultant force F = ma = 68 × (–) 0.80 (1) = (–) 54 (N) (1) or 54.4
or (work done by horizontal force = loss of kinetic energy
work done = force × distance gives)

force =

ecf (a) (i) (1) = 54 (N) (1)

[13]

M4. (a) velocity vector tangential to path and drawn from the ball, arrow
in correct direction (1)

acceleration vector vertically downwards, arrow drawn and in line
with ball (1)

(b) (i)s = ½ gt2 gives t = (1) = 2.2(1) s (1)

(ii)v (= s/t) = 27/2.2(1) (1) = 12(.2 m s–1) or 12(.3) (1) (ecf from (b)(i))

(answer only gets both marks)

[6]

M5. (i) weight greater than air resistance
[or (initially only) weight/gravity acting] (1)
hence resultant force downwards or therefore acceleration (2nd law) (1)
air resistance or upward force increases with speed(1)
until air resistance equals weight or resultant force is zero(1)
leaf moves at constant velocity (1st law)
[or 1st law applied correctly] (1)

(ii) air resistance depends on shape
[or other correct statement about air resistance] (1)
air resistance less significant(1)
air resistance less, therefore greater velocity
[or average velocity greater or accelerates for longer] (1)

max 5

QWC 2

[5]

M6. (a) (i) area = 120 × 106 (m2) (1)
mass = 120 × 106 × 10 × 1100 = 1.3 × 1012 kg(1)

(ii) (use of Ep = mgh gives) ΔEp = 1.3 × 1012 × 9.8 × 5 = 6.4 × 1013 J(1)
(allow C.E. for incorrect value of mass from (i))

(iii) power (from sea water) =

[or correct use of P = Fv]
= 3000 (MW) (1)
(allow C.E. for incorrect value of ΔEp from (ii))
power output = 3000 × 0.4(1)
= 120 MW(1)
(allow C.E. for incorrect value of power)

[7]

M7. (a) (i)v = (1)

t = 0.015 (s) or 15 (ms) (1)

0.68/0.015 (1) (= 45)

(ii)= 3000 (m s–2) (3022) (1)

(b) (i)s = (ut) = gt2 or t = (1)

correct substitution seen = (1)

0.68 to 0.69 correct answer to more than one dp seen (1)

(ii) (s = vt) = 45(.3) × 0.685 or 0.7 (1)

= 30.6 to 32 (1) (m)

(iii) mention of air resistance or drag (1)

causing horizontal deceleration or ‘slowing down’ (1)

[11]

E1. Parts (a) and (b) were answered well and it was clear that candidates were confident with the distinction between scalar and vector quantities and could provide correct examples of each.

The calculation in part (c) caused more problems with a significant proportion of candidates either not combining the two forces correctly or using an incorrect version of the force equation from Newton’s second law. This was a little surprising but was probably the result of candidates trying to overcomplicate a straightforward question. Significant figure errors were common in part (c) (ii), the common answer being 0.6 m s–2 for the acceleration.

E2. A surprisingly large number of candidates divided the mass by four to get a .weight. of 5500 kg in part (a) (i). Many also forgot to divide by four in what should have been a fairly uncomplicated question.

In part (a) (ii), many candidates simply multiplied the mass of 22000 kg by 32, indicating a surprising confusion between weight and mass. For the unit mark there were many common errors such as N, NM, Nm–1, Nm–2, J, nm, kg and Nkg–1.

A very easy mark for mentioning the .counterweight. was picked up by most candidates in part (a) (iii). However, not many went on to discuss the .anticlockwise moment. that this provides.

Most picked up the first two marks to part (b) (i), some as a result of the ecf for the tension. Many candidates used wrong units; pa, PA, Nm–1, being common rather than Pa.

Those with an ecf in (b) (i) generally failed to get both marks to part (b) (ii) because they did not arrive at 17 mm. This may have given some candidates a clue that one of their previous answers was incorrect. The candidates who were successful on the first parts of the question invariably scored both marks here.

E3. Part (a) (i) was straightforward and was answered very well.

Again, part (a) (ii) proved to be a particularly accessible question and candidates performed well.

In part (a) (iii), most grasped the concept that energy was wasted but less able candidates did not realise that the cyclist did work, believing it to be a simple transfer of potential energy to kinetic energy. Many candidates perhaps did not realise that marks could be gained by performing the relevant calculation.

In part (b) (i), a surprising number of candidates used time = distance/speed = 160/16 = 10 s. They did not appreciate that the situation involved uniform deceleration and therefore a kinematics equation should be used.

Part (b) (ii) was generally done well, with most calculating the acceleration and then using F=ma. It was possible to gain full marks even with the use of an incorrect answer from the previous question.

E4. In part (a), a large number of candidates thought that the horizontal component of velocity was to be shown rather than the instantaneous velocity at a tangent to the path. Many candidates who had the right idea still did not gain marks due to a badly drawn tangential arrow. The mark for the acceleration was gained by many candidates. Of those candidates who gained no marks in part (b) (i), many simply could not rearrange s = 1/2gt^2. Those who successfully did this picked up both marks. A majority of candidates gained two marks but some believed they had to use a ‘suvat. equation and gained no marks.

E5. The application of Newton’s laws to terminal velocity has been assessed before and candidates are improving at expressing themselves when answering this type of question.

Fewer candidates now make the mistake of assuming air resistance reduces the speed as opposed to reducing the acceleration. Confusing acceleration with force is also less common and although statements such as “Air resistance equals the acceleration due to gravity” did occur in many scripts, they were certainly less frequent than previous. The commonest error which occurred was applying Newton’s third law incorrectly and identifying air resistance as an equal and opposite reaction to weight. Part (ii) produced some interesting responses and many candidates seemed unaware that gravitational acceleration is independent of mass.

E6. This question produced many high marks. The usual error was failing to calculate the area correctly in m2, usually through multiplying by 103 instead of 106. Only the very best candidates realised in part (ii) that the centre of gravity of the water dropped by 5 m and not 10 m. In part (iii) most candidates knew how to calculate the average loss of gravitational potential energy per second and how to use the efficiency correctly.

E7. In part (a) (i), most candidates quoted the equation and correctly calculated the time. The most frequent misconception was the belief that a ‘suvat’ equation should be used even though the velocity is constant.

The correct answer of 3000, or 3022, was accepted in part (a) (ii) and the majority successfully produced this value.

In part (b) (i) most select and quoted the correct equation and showed the correct substitution. Some lost the mark as they did not show the answer to more than one significant figure.

Part (b) (ii) was a very easy question for over 40% of candidates who understood that the horizontal acceleration was zero. For these students, 45.3 × 0.685 = 31 gained two marks. However, 13% did not attempt the question and another 40% misapplied a kinematics equation to the situation often using 9.81 as the acceleration.

The vast majority of candidates identified air resistance as the key factor to part (b) (iii). However, only 7% mention that horizontal deceleration is caused by air resistance.

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