Chapter 6 7 8 Review with ANSWERS - Weeblymisssugarscience.weebly.com/uploads/9/7/0/3/... · (b) Figure 1 shows a student of weight 550 N doing a “press up”. In the position shown

Drayton Manor High School

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Chapter 6, 7, 8 Old Exam Questions

Q1. The object in the diagram below is in equilibrium.

By resolving forces, calculate:

(a) the angle θ;

Angle θ ...................................................... (2)

(b) the magnitude of the force F.

Magnitude of the force F ...................................................... (1)

(Total 3 marks)

Q2.(a) State the difference between vector and scalar quantities.

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. (1)


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

Vector quantity ...............................................................................................

(1)

Scalar quantity ............................................................................................... (1)

(c) A 6.0 N force and a 4.0 N force act on a body of mass 7.0 kg at the same time. Calculate the maximum and minimum accelerations that can be experienced by the body.

Maximum acceleration........................ Minimum acceleration...................... (3)

(Total 6 marks)

Q3.Coplanar forces of 5 N, 4 N and 3 N act on an object. Which force, in N, could not possibly be the resultant of these forces?

A 0

B 4

C 12

D 16

(Total 1 mark)

Q4.A horizontal force of 1.5 kN acts on a motor car of mass 850 kg that is initially at rest.

(a) Calculate:

(i) the acceleration of the motor car; (1)


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(ii) the speed of the motor car after 15 s; (2)

(iii) the distance travelled by the motor car in the first 7.5 s of the motion; (2)

(iv) the distance travelled by the motor car in the first 15 s of the motion. (1)

(b) The diagrams below show the graph of force against time together with three incomplete sets of axes. Sketch on these axes the corresponding graphs for acceleration, speed and distance travelled for the first 15 seconds of the car’s motion.

You should include labels for the axes and any known numerical values.


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(4)

(c) In practice the resultant force exerted on the motor car will not be constant with time as suggested by the force-time graph. Air resistance is one factor that affects the resultant force acting on the vehicle.

(i) Suggest how the force-time graph will change when air resistance is taken into account. Explain your answer. You may wish to sketch a graph to illustrate your answer.

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(ii) Explain why the vehicle will eventually reach a maximum speed even though the motorist keeps the accelerator pedal fully depressed.

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(Total 15 marks)

Q5.In the system shown a light rigid beam, pivoted at X, is held in position by a string which is fixed at Y. The beam carries a load of 200 N. The load is moved towards X. Which one of the following statements is correct?

A The tension in the string increases

B The compression force in the beam increases

C The moment of the load about X increases

D The magnitude of the vertical component of the reaction at X increases


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(Total 1 mark)

Q6.A solid iron ball of mass 890 kg is used on a demolition site. It hangs from the jib of a crane suspended by a steel rope. The distance from the point of suspension to the centre of mass of the ball is 15 m.

(a) Calculate the tension in the rope when the mass hangs vertically and stationary.

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. (2)

(b) The iron ball is pulled back by a horizontal chain so that the suspension rope makes an angle of 30° with the vertical. Calculate the new tension in the suspension rope.

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(c) The ball is now released from rest and hits a brick wall just as it passes through the vertical position. It can be assumed that the ball is brought to rest by the impact with the wall in 0.2 s.


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Calculate

(i) the vertical height through which the ball falls,

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(ii) the speed of the ball just before impact,

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(iii) the average force exerted by the ball on the wall.

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. (5)

(Total 9 marks)

Q7.(a) State the conditions necessary for a body to be in equilibrium.

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(2)

(b) The boat shown in the figure below is being towed at constant velocity. The tension force (FT) in the towing rope is 2800 N. The forces resisting the motion can be assumed to be the force of the water on the keel (FK) and the force of the water on the rudder (FR).

By calculation or by scale drawing, find the size of the force, FR, needed to keep the boat in equilibrium.

(6) (Total 8 marks)

Q8.A ballbearing X of mass 2m is projected vertically upwards with speed u. A ballbearing Y of mass m is projected at 30° to the horizontal with speed 2u at the same time. Air resistance is negligible. Which of the following statements is correct?

A The horizontal component of Y's velocity is u.

B The maximum height reached by Y is half that reached by X

C X and Y reach the ground at the same time.

D X reaches the ground first.

(Total 1 mark)

Q9.(a) State the principle of moments.

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. (2)

(b) Figure 1 shows a student of weight 550 N doing a “press up”. In the position shown the body is horizontal and the forearms are vertical.

Figure 1

Assuming that each arm experiences the same force and that the forces acting on each foot are equal, calculate the compression force acting:

(i) in each of the student’s forearms; (2)

(ii) on each of the student’s feet. (1)

(c) Another student attempts the same exercise but with the forearms at an angle of 30° to the ground, as shown in Figure 2.

Figure 2

(i) The directions of some of the forces acting on the hands have been indicated. Indicate, on Figure 2, any other forces acting on the hands

(1)


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(ii) State the cause of these additional forces.

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. (1)

(iii) The reaction force at each hand is 210 N. Calculate the magnitude of the compression force in each forearm in this position.

(1) (Total 8 marks)

Q10. Figure 1 shows the speed-time graph for a swimmer performing one complete cycle of the breast stroke.

Figure 1

Figure 2


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(a) (i) Find the acceleration of the swimmer at time 0.65 s.

acceleration ...................................

(ii) Sketch, on the axes in Figure 2, a graph to show how the acceleration of the swimmer varies with time for the same time interval. You are not required to make any further calculations but your graph should show relative values.

(4)

(b) Use the graph in Figure 1 to estimate the distance travelled by the swimmer in one complete cycle of the stroke. Show your working clearly.

distance travelled .............................. m (4)

(Total 8 marks)

Q11.The diagram below shows two different rifles being fired horizontally from a height of 1.5 m above ground level. Assume the air resistance experienced by the bullets is negligible.


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(a) When rifle A is fired, the bullet has a horizontal velocity of 430 m s–1 as it leaves the rifle. Assume the ground is level.

(i) Calculate the time that the bullet is in the air before it hits the ground.

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

(ii) Calculate the horizontal distance travelled by the bullet before it hits the ground.

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

(b) Rifle B is fired and the bullet emerges with a smaller horizontal velocity than the bullet from rifle A.

Explain why the horizontal distance travelled by bullet B will be less than bullet A.


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(Total 6 marks)

Q12.A steel ball of weight W falls through oil. At a time before the ball reaches terminal velocity, the magnitude of the viscous resistance force on the ball is

A zero

B between zero and W

C equal to W

D greater than W

(Total 1 mark)

Q13.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.


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(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, ...........................................................................................................

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(ii) B, ...........................................................................................................

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(iii) C, ...........................................................................................................

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. (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,

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(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.

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(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,

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

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(Total 14 marks)

Q14.The Thrust SSC car raised the world land speed record in 1997. The mass of the car was 1.0 × 104 kg. A 12 s run by the car may be considered in two stages of constant acceleration. Stage one was from 0 to 4.0 s and stage two 4.0 s to 12 s.

(a) In stage one the car accelerates from rest to 44 m s-1 in 4.0 s. Calculate the acceleration produced and the force required to accelerate the car.

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(b) In stage two the car continued to accelerate so that it reached 280 m s–1 in a further 8.0 s. Calculate the acceleration of the car during stage two.

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(c) Calculate the distance travelled by the car from rest to reach a speed of 280 m s–1.

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. (Total 6 marks)

Q15.A lift and its passengers with a total mass of 500 kg accelerates upwards at 2 m s–2 as shown. Assume that g = 10 m s–2.

What is the tension in the cable?

A 1000 N

B 4000 N

C 5000 N


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D 6000 N

(Total 1 mark)

Q16.Two masses hang at rest from a spring, as shown in the diagram. The string separating the masses is burned through.

Which of the following gives the accelerations of the two masses as the string breaks?

acceleration of free fall = g

acceleration of 1 kg mass upwards in

m s–2

acceleration of 2 kg mass downwards

in m s–2

A 3 g 1 g

B 2 g 2 g

C 2 g 1 g

D 1 g 1 g

(Total 1 mark)

Q17. In the 1969 Moon landing, the Lunar Module separated from the Command Module above the surface of the Moon when it was travelling at a horizontal speed of 2040 m ss–1. In order to descend to the Moon’s surface the Lunar Module needed to reduce its speed using its rocket as shown in Figure 1.


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Figure 1

(a) (i) The average thrust from the rocket was 30 kN and the mass of the Lunar Module was 15100 kg. Calculate the horizontal deceleration of the Lunar Module.

answer = ............................. m s–2

(2)

(ii) Calculate the time for the Lunar Module to slow to the required horizontal velocity of 150 m s–1. Assume the mass remained constant.

answer = ................................... s (2)

(b) The rocket was then used to control the velocity of descent so that the Lunar Module descended vertically with a constant velocity as shown in Figure 2. Due to the use of fuel during the previous deceleration, the mass of the Lunar Module had fallen by 53%.

Figure 2


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acceleration due to gravity near the Moon’s surface = 1.61 m s–2

(i) Draw force vectors on Figure 2 to show the forces acting on the Lunar Module at this time. Label the vectors.

(2)

(ii) Calculate the thrust force needed to maintain a constant vertical downwards velocity.

answer = .................................. N (2)

(c) When the Lunar Module was 1.2 m from the lunar surface, the rocket was switched off. At this point the vertical velocity was 0.80 m s–1. Calculate the vertical velocity at which the Lunar Module reached the lunar surface.

answer = ............................. m s–1

(2) (Total 10 marks)


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M1. (a) 7.5 = 15 sin θ (or 15 cos θ) (i.e. attempt to resolve and equate)

C1

θ = 30° (cao) (n.b. unit accept deg or degree)

A1 2

(b) F = 15 cos 30 or 15 sin 60 (if wrong way round) = 13 N or F = (152 – 7.52)1/2

B1 1

[3]

M2.(a) vector has direction, scalar has no direction / only vector has direction

B1

(b) vector: any vector except force (accept weight) B1

scalar: any scalar B1

(c) F = ma in any form C1

maximum: 1.4 m s–2

minimum: 0.29 m s–2

A1 [6]


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M3.D [1]

M4.(a) (i) (acceleration = F / m =) 1.76 m s–2

B1 (1)

(ii) (v = u + at) = 0 + 1.76 × 15 C1

= 26.4 m s–1 allow e.c.f. from (i)

A1 (2)

(iii) (s = ut + 0.5 at2) = 0.5 × 1.76 × 7.52

C1

= 49.5 m allow e.c.f. from (i)

A1 (2)

(iv) t doubles so s quadruples = 4 × 49.5 ~ 200 m [or equivalent]

B1 (1)

(b) acceleration graph correct: same shape as F-t; 1.8 (1.76) identifiable on axis

B1

speed graph correct: straight line through origin to identifiable 26 m s–1 at 15 s

B1

distance graph correct: shape parabolic; both calculated points identifiably marked

B1


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all axes labelled with unit B1

(4)

(c) (i) total force decreases as time increases [or appropriate graph] B1

(because) speed increase leads to drag force increase B1

total thrust is sum of engine force – drag (frictional force) B1

(3)

(ii) forward thrust = friction force B1

so (total thrust = 0) and acceleration = 0 B1

(2) [15]

M5.D [1]

M6.(a) T = mg = 890 × 10 = 8900 (1) N (1) (accept alternative correct value using g = 9.81 N kg–1)

(2)

(b)

resolve vertically T cos 30° =mg (1)

T = = 10280 (1.03 × 104) N (1)


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(2)

(c) (i) vertical height fallen = l(1 – cos θ) = 15(1 – 0.866) = 2.0(1) m (1) (allow e.c.f. if h calculated wrongly)

(ii) mυ2 = mgh or reference energy (1) υ = = 6.34 m s–1 (1)

(max 1 / 3 if equations of motion used)

(iii) (1)

(allow e.c.f of υ and m as before) (5)

[9]

M7.(a) (vector) sum of forces is zero / no resultant force / no net force B1

sum of moments (about any point) zero / no resultant moment / no net moment

B1 (2)

(b) Clear attempt to resolve C1

horizontal resolution: FR cosθ = 2800 cos 20° – 3200 cos 42° C1

vertical resolution: FR sinθ = 3200 sin 42° – 2800 sin 20° C1

horizontal (253N) or vertical (1180N) resolution correct C1

correct method for resultant C1


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FR = 1200 N (1210) A1

or

scale drawing of correct approximate shape i.e. closed triangle C1

correct triangle with either labels or arrows C1

2800 and 3200 drawn to scale to nearest mm C1

angles correct to nearest° C1

FR = 1190 N to 1230 N (sf penalty for more than 2 sf) A1

acceptable scale quoted giving triangle of at least 6 cm B1

or

a closed triangle sketch C1

correct triangle with either labels or arrows C1

correct angle shown (22°) C1

cosine rule quoted correctly C1

correct substitution FR2 = 28002 + 32002 – 2(2800)(3200) cos 22°

C1

FR = 1200 N (1210) A1

(6) [8]

M8.C [1]

M9.(a) vague statement:


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e.g. clockwise moments = anticlockwise moments or recognition of the equilibrium condition precise statement: must have ‘sum of’ and equilibrium condition i.e. when in equilibrium sum of clockwise moments = sum of anticlockwise moments (about any point)

C1 or Σclockwise moments = Σanticlockwise moments or vector sum of moments = 0 or no resultant moment (or torque)

A1 (2)

(b) (i) correct moments equation (354 N seen) C1

175 N to 180 N (177 N) A1

(2)

(ii) 95 N to 100 N (98 N)

or 275 – (i)

or 550 – 354 = 196 N

(i.e. e.c.f. for those who forget about two hands and feet;

also allow reverse answers as e.c.f.) A1

(1)

(c) (i) two friction forces correctly shown at ground level

(at least one on the line) B1

(1)

(ii) friction between the hands and the floor or resistance to relative motion of hands and floor

B1 (1)

(iii) 420 N B1

(1) [8]


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M10. (a) (i) use of appropriate data from graph (1)

answer in acceptable range (to be decided) (1)

(ii) zero at 0, 0.2 0.58, 0.8 and 1 s (approx) (1)

reasonable attempt to show relative magnitudes (1) 4

(b) appreciation of area under the graph (1)

appropriate counting of squares (1)

distance per square (1)

correct answer in acceptable range (1) 4

[8]

M11.(a) (i)

Allow g=10 (0.5477)

( = 0.553) = 0.55 ( s)

0.6 gets 2 marks only if working shown. 0.6 on its own gets 1 mark.

2

(ii) (s = v t = 430 × 0.553 = 237.8 = ) 240 (m) ecf a(i)

1


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(b) their vertical motion is independent of their horizontal motion OR downward / vertical acceleration is the same for both OR acceleration due to gravity is the same for both OR vertical speed / velocity is the same for both

Allow ‘time is constant’ Don’t allow ‘similar’

(bullets A and B will be in the air) for the same time

(Horizontal acceleration is zero and thus horizontal) distance is proportional to horizontal speed OR s = ut where u is the horizontal velocity

‘velocity smaller so distance smaller’ is not sufficient

3 [6]

M12.B [1]

M13.(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


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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) tan θ = gives θ = 31° (1) (2)

[14]

M14.(a) a = = 11 ms–2 (1)

F = ma = 1.1 × 105 N (1)

(b) ∆υ = 236 m s–1

a = = 29.5 m s–2 (1)


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(c) sone = υav × t = × 4.0 = 88 m (1)

stwo = υav × t = × 8.0 (1) = 1296 (m) (1)

total distance = 1384 m (1) [6]

M15.D [1]

M16.C [1]

M17. (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)


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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]

Documents

Chapter 6 7 8 Review with ANSWERS - Weeblymisssugarscience.weebly.com/uploads/9/7/0/3/... · (b) Figure 1 shows a student of weight 550 N doing a “press up”. In the position shown