Revision 3 Mechanics 16-05-16 - Hinchley Wood School

Revision 3 – Mechanics – 16-05-16 - Hinchley Wood School

Q1.The graph shows how the vertical speed of a parachutist changes with time during the first 20 s of his jump. To avoid air turbulence caused by the aircraft, he waits a short time after jumping before pulling the cord to release his parachute.

(a) Regions A, B and C of the graph show the speed before the parachute has opened. With reference to the forces acting on the parachutist, explain why the graph has this shape in the region marked

(i)A, ......

......

(ii)B, ......

......

(iii)C, ......

......

(6)

(b) Calculate the maximum deceleration of the parachutist in the region of the graph marked D, which shows how the speed changes just after the parachute has opened. Show your method clearly,

......

(2)

(c) Use the graph to find the total vertical distance fallen by the parachutist in the first 10 s of the jump. Show your method clearly.

......

(4)

(d) During his descent, the parachutist drifts sideways in the wind and hits the ground with a vertical speed of 5.0 m s–1 and a horizontal speed of 3.0 m s–1. Find

(i)the resultant speed with which he hits the ground,

......

(ii)the angle his resultant velocity makes with the vertical.

......

(2)

(Total 14 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)Figure 1 shows a skier descending the ramp of a ski jump. Figure 2 shows a graph of the distance travelled along the ramp against time, from the time the descent starts until the skier leaves the end of the ramp.

Figure 1

Figure 2

The skier of mass 80 kg (including equipment) skis down the ramp and leaves it horizontally. The skier gains 55% of the available gravitational potential energy as kinetic energy when descending the ramp.

acceleration of free fall, g = 9.8 m s–2

(i)One energy transformation which occurs as the skier skis down the ramp is from gravitational potential energy to kinetic energy of the skier. State two other energy transformations that occur as the skier skis down the ramp.

......

(2)

(ii)Use Figure 2 to show that the speed at which the skier leaves the ramp is about 23ms–1. Show your reasoning clearly.

(2)

(iii)Determine the height of the ramp.

(3)

(b) Figure 1 shows the path taken by the skier after leaving the ramp.
Assuming that there was no lift or drag due to the air during this jump, calculate:

(i)the time for which the skier was in flight;

(2)

(ii)the horizontal distance jumped by the skier before landing.

(2)

(c) On landing the skier has considerable vertical momemtum that has to be reduced to zero. The surface on which the skier lands is hard-packed snow. To reduce the force experienced by the skier, the landing surface is angled at 40° to the horizontal.

Explain briefly how angling the landing surface reduces the vertical component of the force, experienced by the skier.

......

(3)

(Total 14 marks)

Q4. A car of mass 1300 kg is stopped by a constant horizontal braking force of 6.2 kN.

(a) Show that the deceleration of the car is about 5 m s–2.

......

(3)

(b) The initial speed of the car is 27 m s–1.
Calculate the distance travelled by the car as it decelerates to rest.

......

distance travelled ...... m

(3)

(Total 6 marks)

Q5. The diagram belowshows a spacecraft that initially moves at a constant velocity of 890 m s–1 towards A.

To change course, a sideways force is produced by firing thrusters. This increases the velocity towards B from 0 to 60 m s–1 in 25 s.

(a) The spacecraft has a mass of 5.5 × 104 kg. Calculate:

(i)the acceleration of the spacecraft towards B;

Acceleration ......

(1)

(ii)the force on the spacecraft produced by the thrusters.

Force on spacecraft ......

(2)

(b) Calculate the magnitude of the resultant velocity after 25 s.

Magnitude of resultant velocity ......

(2)

Angle ......

(1)

(Total 6 marks)

Q6. A student set up the apparatus shown in the figure belowto demonstrate the principle of moments.

(a) Using the values on the figurecalculate:

(i)the magnitude of the moment about the pivot due to the tension of the spring in the spring balance;

moment due to spring tension ......

(1)

(ii)the magnitude of the moment about the pivot produced by the 2.0 N weight;

moment due to 2.0 N weight ......

(1)

(iii)the weight of the wooden bar.

weight ......

(1)

(b) (i) Calculate the magnitude of the force exerted on the bar by the pivot.

magnitude of force ......

(1)

(ii) State the direction of the force on the pivot.

......

(1)

(Total 5 marks)

Q7.To determine the force and power involved when a football is kicked, a student suspended a ball from the roof of a gymnasium by a long string as shown in Figure 1.

Figure 1

When the ball of mass 0.45 kg was kicked it rose to a maximum height of 9.0 m. The student measured the contact time between the ball and the boot as 0.12 s.

the acceleration of free fall, g = 9.8 m s–2

(a) Assume that air resistance was negligible so that all the initial kinetic energy given to the ball was converted into gravitational potential energy.

Calculate:

(i)the velocity of the ball immediately after being kicked;

(2)

(ii)the average force exerted on the ball when in contact with the boot;

(2)

(iii)the average useful power developed by the student when the ball was kicked.

(2)

(b) (i)The ball is kicked so that its initial motion is horizontal. Explain why the tension in the supporting string increases when the ball is kicked.

......

(2)

(ii)Calculate the tension in the string immediately after the ball is kicked.

(3)

Figure 2

(i)Calculate the velocity of the ball in this position.

(2)

(ii)In one test the string broke when the ball was in the position shown in Figure 2. Explain why the ball reached a lower maximum height on this occasion than it did when the string did not break.

(2)

(Total 15 marks)

M1.(a) (i)region A: uniform acceleration

(or (free-fall) acceleration = g (= 9.8(i) m s–2))

force acting on parachutist is entirely his weight

(or other forces are very small) (1)

(ii)region B: speed is still increasing

acceleration is decreasing (2)(any two)

because frictional (drag) forces become significant (at higher speeds)

(iii)region C: uniform speed (50 m s–1)

because resultant force on parachutist is zero (2)(any two)

weight balanced exactly by resistive force upwards

The Quality of Written Communication marks were awarded primarily for the quality of answers to this part

(6)

(b) deceleration is gradient of the graph (at t = 13s) (1)

(e.g. 20/1 or 40/2) = 20 m s–2(1)

(2)

(c) distance = area under graph (1)
suitable method used to determine area (e.g. counting squares) (1)
with a suitable scaling factor (e.g. area of each square = 5 m2) (1)
distance = 335 m (±15 m) (1)

(4)

(d) (i)speed = √(5.02 + 3.02) = 5.8 m s–1(1)

(ii)tanθ = gives θ = 31° (1)

(2)

[14]

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)(kinetic) energy of air / snow (pushing it out of the way)
(not internal energy)

melting / internal energy of snow / ice
or internal energy of air / skis (condone heat)

award 1 mark only for “heat and sound”
award zero marks for sound alone even if explained

(2)

(ii)clear attempt to draw tangent to graph at end of run

correct co-ordinates and manipulation for tangent
or use of 2 points on curve ≥t = 7s

(no unit penalty) (no significant figure penalty)

(2)

(iii)potential energy loss = 100 / 55 × KE

final KE = 0.5 × m × 232

equates any KE to mgh

height = 49 mallow e.c.f. from (ii)

(3)

(b) (i)s = ½ at2ors = ut + ½at2or correct numerical equivalent

4.0 s or 4.04 s

(2)

(ii)s = vt or numerical equivalent

92 m – 93 m

allowe.c.f. from (a)(ii) and (b)(i)

(2)

skier has vertical momentum after landing

change in momentum is reduced

F =

skier takes longer time to reduce (vertical) momentum

compare with time when surface is flat

F =

skier moves though longer distance (vertically) to come to rest

compare with distance to come to rest when surface is flat

F = ma

time to come to rest (vertically) is increased

acceleration is reduced

allow 1 mark for idea that “(vertical) component of normal reaction is reduced”

(3)

[14]

M4. (a)a force/1300 (condone power of ten error)

6200 ÷ 1300

4.77 (m s–2)

(b)use of suitable kinematic equation

eg distance = 272/(2 × 4.8) correct sub

76/76.4 m/72.9 from a = 5/75.9 from a = 4.8

[6]

M5. (a) (i) 2.4 m s–2

(ii)F = ma

132 000 N (ecf from (i))

(b)final speed = (8902 + 602)1/2

892 m s–1 (cao) (allow 890 m s–1 as final answer but
892 must be seen in working)

(c) tan–1 60/890 or sin–1 60/892 = 3.9° (3.86)°
or cos–1(890/892) = 3.8 (4)°
or sin–1 60/890 =3.9° (3.86)° if ecf from (b)

[6]

M6. (a) (i) 1.05 (1.1) N m (up for J)

(ii) 0.70 N m (condone 1 sf)

(iii)weight of bar = 1.59 N (1.8 if (a) (i) = 1.1)

(b) (i) 3.4 N (3.2 N if weight = 1.8 N) {ecf 5 – (a) (iii)}

(ii)upwards (not clockwise)
(allow ecf for answer consistent with weight
i.e. down if (weight +2)>7)

[5]

M7.(a) (i)mgh = ½ mv2or correct numerical substitution

13.3 m s–1

no marks for use of equation of motion for constant acceleration

allowgh = mv2 / 2 or v2 = 2gh) but not v2 = 2as

(2)

(ii)mv = Ft (or F = ma and a = v / t)

(or numerical equivalent)

48.8 to 50.0 Ne.c.f. from (i) {3.75 × (i)}

(2)

(iii)power = energy transformed / time

or power = average force × average velocity

orP = Fv leading to (i) × (ii) (664 W if (i) and (ii) are correct)

330 to 332 W e.c.f. from (i) and / or (ii) {(i) × (ii) / 2}

(2)

(b) (i)the ball accelerates toward centre (of circular path) /
the point of suspension / upwards

or the ball is changing direction upwards

centripetal force / resultant force upwards /
force towards centre of circular path

or string initially stretches producing an elastic force

(2)

(ii)T–mg = mv2 / rorF = mv2 / r (or numerical equivalent)

orF = maanda = v2 / r

centripetal / resultant accelerating force = 6.6 N

e.c.f. from (i) (0.0375 × (a)(i)2 )

tension = their centripetal force + 4.4 N (11 N)

(3)

or velocity is same as when falling 4.5 m so ½ mv2 = mg × 4.5
or KE at bottom = KE at half way + PE

allow½ mv2 = mg × 4.5

9.4 m s–1 (no marks if 9.4 m s–1 arrived at using equation of motion)

(2)

(ii)horizontal velocity is constant after string breaks

or continued movement in the horizontal direction

oridea of KE due to horizontal motion

at max height the ball still has KE so acquires less PE

ornot all KE becomes (gravitational) PE

(2)

orupward velocity = 9.4 sin 51

use of equation of motion leading to 2.72 m after the break

orafter string breaks downward force increases /
the upward force ceases to exist

there is greater vertical deceleration

[15]

E1.In part (a) most candidates were able to interpret the graph correctly and almost all understood why the parachutist reached constant terminal speed in region C of the graph. Although many also understood and stated that the acceleration in region A was constant, few stated that this was because the drag on the parachutist was negligible or much smaller than his weight. Answers were generally well expressed and a mark of four or five was most common. A minority of candidates, however, was quite incapable of using physics terms accurately and subsequently scored few marks.

Many candidates understood that the acceleration in part (b) equalled the gradient of the line in region D of the graph and arrived at a correct answer (although the unit of acceleration was often given as −1). Candidates who used a = (ʋ−u) / t very often chose points off the straight section of the graph and arrived at α value for a outside the acceptable range.

In part (c) many candidates ignored the graph and attempted to use an equation of uniform acceleration to find the distance travelled. The majority, however, made some attempt to relate distance to the area under the graph and most of these answers fell within the acceptable range.

Part (d) was most often correct, although in part (ii) many candidates inverted the tan function and found the angle to the horizontal rather than the vertical.

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.(a) (i)Most answers referred simply to 'heat and sound'. More detail was expected for full credit. Candidates could have explained where or why heating occurs.

(ii)Many candidates drew an appropriate tangent and used it correctly. Some simply used the data from the final linear portion of the graph. There were many instances of careless reading of graph scales.

There were some who applied equations of motion incorporating the 80 m fallen after leaving the ramp into their calculations.

(iii)The majority of the candidates attempted to equate P.E. to K.E. but the 55% factor was frequently applied incorrectly, yielding a wide range of answers. Some used the equation for uniformly accelerated motion which was not appropriate in this case.

(b) (i)This was often well done. The most common error was to include 23 m s–1 as an initial vertical speed.

(ii)Allowing errors carried forward many completed this successfully but incorrect answers to (i) produced ski jump distances of up to 500 in!

(c) Surprisingly few took the hint in the stem that momentum change needed to be considered. The expected response was that the skier retains some vertical momentum on impact, reducing the change in momentum and hence, from Ft = ∆(mv), reducing the force. Although such an approach was considered by the more able candidates the thinking of the majority was more muddled. Many took the view that somehow the vertical component of the force was split into a vertical and horizontal component, which consequently reduced the vertical component. Others took the view that vertical momentum was transformed into horizontal momentum.

E4. The very straight forward part (a) was well answered with generally clear solutions being provided.

Part (b) was answered more unevenly. About 60% of candidates arrived at a complete answer with the remainder making more or less progress with what was a straightforward question. Inconsistencies of sign in the treatment of deceleration and the velocities were very common and werepenalised. It is not acceptable simply to ignore the fact that an equation yields a squared quantity with a negative sign.

E5. (a) (i) This part was usually correct but the unit caused more problems than it should have done at this level.

(ii) Few candidates had problems with this part.

(b) Most were successful in this part. Those who failed usually gave (8902 + 2.42)1/2

(c) Again the majority of the candidates did this correctly. Those who failed usually determined the wrong angle and gave 86.1° as the final answer.

E6. (a) (i&ii) The structuring allowed most candidates to calculate the magnitudes of the moments correctly but many incurred a unit penalty here.

(iii) Many candidates had no appreciation of how to apply the principle of moments to this situation. A common mistake was to assume the mass to be concentrated at the end of the wooden bar.

(b) (i) The simple idea of equating the sums of the upward and downward forces to determine the force necessary to maintain the equilibrium of the bar was appreciated by relatively few candidates.

(ii) Candidates were expected to appreciate that the pivot must exert a downward force on the bar to maintain equilibrium so that there would be an upward force on the pivot.

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