Mechanics Exam Questions AS
647 minutes
638 marks
Q1.
In an experiment to measure the power output of a motor, the motor is used to lift a metal
block vertically at constant speed.
You may be awarded marks for the quality of written communication in your answers.
(a)
Describe an experiment to check whether the speed of the rising mass is constant.
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(2)
(b)
Explain how the output power of the motor is calculated, stating what measurements need
to be made.
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(2)
(Total 4 marks)
Q2.
(a) A man jumps from a plane that is travelling horizontally at a speed of 70 m s–1. If air
resistance can be ignored, determine
(i)
his horizontal velocity 2.0 s after jumping,
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(ii)
his vertical velocity 2.0 s after jumping,
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(iii)
the magnitude and direction of his resultant velocity 2.0 s after jumping.
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(5)
(b)
After 2.0 s the man opens his parachute. Air resistance is no longer negligible. Explain in
terms of Newton’s laws of motion, why
(i)
his velocity initially decreases,
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(ii)
a terminal velocity is reached.
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(4)
(Total 9 marks)
Q3.
A waiter holds a tray horizontally in one hand between fingers and thumb as shown in the
diagram.
P, Q and W are the three forces acting on the tray.
(a)
(i)
State two relationships between the forces that must be satisfied if the tray is to
remain horizontal and in equilibrium.
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(ii)
If the mass of the tray is 0.12 kg, calculate the magnitude of the force W.
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(iii)
Calculate the magnitudes of forces P and Q.
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(6)
(b)
The waiter places a glass on the tray. State and explain where the glass should be
positioned on the tray if the force, P, is to have the same value as in part (a).
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(2)
(Total 8 marks)
Q4.
A student carried out an experiment to determine the terminal speed of various ball
bearings as they fell through a viscous liquid. She did this by timing their fall between two
marks, P and Q, which were 850 mm apart on a vertical glass tube.
You may be awarded marks for the quality of written communication in your answer.
(a)
(i)
Describe the motion of a ball bearing after being released from rest at the surface.
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(ii)
In terms of the forces acting, explain why a ball bearing reaches a terminal speed
under these conditions.
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(5)
(b)
The student’s results are shown in columns A and B. Complete column C.
column A
column B
column C
column D
column E
radius of ball
bearingr / mm
time of fall /
s(through 850 mm)
terminal
speedv / mm s–
log10(r / mm)
log10(v / mm s–1)
1
1.62
32.0
0.210
1.98
21.4
0.297
2.21
17.2
0.344
2.73
11.3
0.436
3.40
7.2
0.531
4.12
4.9
0.615
(2)
(c)
The relationship between v and r is known to be of the form
v = kr ,
n
where n and k are constants.
(i)
Enter the corresponding values for log10(v / mm s–1) in column E of the table in
part (b).
(ii)
Plot a graph of log10(v / mm s–1) on the y-axis, against log10(r / mm) on the x-axis.
(Allow one sheet of graph paper)
(4)
(d)
Use your graph to determine
(i)
the constant n,
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(ii)
the constant k.
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(5)
(Total 16 marks)
Q5.
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
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(ii)
Calculate the maximum loss of potential energy of the sea water in part (i) when it is
released through the turbines.
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(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%.
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(Total 7 marks)
Q6.
(a) The graph shows the variation of tensile stress with tensile strain for two
wires X andY, having the same dimensions, but made of different materials. The materials
fracture at the points FX and FY respectively.
You may be awarded marks for the quality of written communication provided in your
answer to the following questions.
State, with a reason for each, which material, X or Y,
(i)
obeys Hooke’s law up to the point of fracture,
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(ii)
is the weaker material,
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(iii)
is ductile,
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(iv)
has the greater elastic strain energy for a given tensile stress.
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(8)
(b)
An elastic cord of unstretched length 160 mm has a cross-sectional area of 0.64 mm2. The
cord is stretched to a length of 190 mm. Assume that Hooke’s law is obeyed for this range
and that the cross-sectional area remains constant.
the Young modulus for the material of the cord = 2.0 × 107 Pa
(i)
Calculate the tension in the cord at this extension.
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(ii)
Calculate the energy stored in the cord at this extension.
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(5)
(Total 13 marks)
Q7.
The diagram shows a 250 kg iron ball being used on a demolition site. The ball is
suspended from a cable at point A, and is pulled into the position shown by a rope that is kept
horizontal. The tension in the rope is 1200 N.
(a)
In the position shown the ball is in equilibrium.
(i)
What balances the force of the rope on the ball?
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(ii)
What balances the weight of the ball?
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(2)
(b)
Determine
(i)
the magnitude of the vertical component of the tension in the cable,
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(ii)
the magnitude of the horizontal component of the tension in the cable,
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(iii)
the magnitude of the tension in the cable,
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(iv)
the angle the cable makes to the vertical.
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(6)
(Total 8 marks)
Q8.
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,
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(ii)
the apple hits the ground first.
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(Total 5 marks)
Q9.
The diagram represents an experiment that can be used to investigate stopping distances
for a moving trolley.
The trolley is placed on the raised section of the track. When released it moves down the track
and then travels along the horizontal section before colliding with the block. The trolley and
block join and move together after the collision. The distance they move is measured.
(a)
State the main energy changes taking place
(i)
as the trolley descends,
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(ii)
after the collision, as the trolley and block move together.
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(2)
(b)
Describe how the speed of the trolley, just before it collides with the block may be
measured experimentally.
You may be awarded marks for the quality of written communication in your answer.
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(3)
(c)
State and explain how the speed of the trolley, prior to impact could be varied.
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(2)
(Total 7 marks)
Q10.
While investigating projectile motion, a student used stroboscopic photography to
determine the position of a steel ball at regular intervals as it fell under gravity. With the
stroboscope flashing 20 times per second, the ball was released from rest at the top of an
inclined track, and left the foot of the track at P, as shown in the diagram below.
For each of the images on the photograph, the student calculated the horizontal distance, x, and
the vertical distance, y, covered by the ball at time t after passing P. Both distances were
measured from point P. He recorded his results for the distances x and y in the table.
(a)
image
x/cm
y/cm
t/s
1
11.6
9.3
0.05
2
22.0
21.0
0.10
3
32.4
35.0
0.15
4
44.2
51.8
0.20
5
54.8
71.0
0.25
6
66.0
92.2
0.30
(y/t)/cm s–1
Using two sets of measurements from the table, calculate the horizontal component of
velocity of the ball. Give a reason for your choice of measurements.
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(2)
(b)
The student worked out that the variables y and t in the experiment could be represented
by
= u + kt
where u and k are constants.
(i)
Complete the table above.
(ii)
Use the data in the table to plot a suitable graph to confirm the equation.
(Allow one sheet pf graph paper)
(iii)
Use your graph to find the values of u and k.
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(9)
(c)
State the physical significance of
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k ...................................................................................................................
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(2)
(d)
Calculate the magnitude of the velocity of the ball at point P.
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(2)
(Total 15 marks)
Q11.
(a)
State the principle of moments.
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(3)
(b)
(i)
Draw a labelled diagram of the apparatus you would use to verify the principle of
moments.
(ii)
Describe the procedure that would be used and state what measurements are
taken.
You may be awarded marks for the quality of written communication in your answer.
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(iii)
Explain how the results would be used to verify the principle of moments.
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(7)
(Total 10 marks)
Q12.
A constant resultant horizontal force of 1.8 × 103 N acts on a car of mass 900 kg, initially at
rest on a level road.
(a)
Calculate
(i)
the acceleration of the car,
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(ii)
the speed of the car after 8.0 s,
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(iii)
the momentum of the car after 8.0 s,
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(iv)
the distance travelled by the car in the first 8.0 s of its motion,
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(v)
the work done by the resultant horizontal force during the first 8.0 s.
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(9)
(b)
On the axes below sketch the graphs for speed, v, and distance travelled, s, against
time,t, for the first 8.0 s of the car’s motion.
(2)
(c)
In practice the resultant force on the car changes with time. Air resistance is one factor
that affects the resultant force acting on the vehicle.
You may be awarded marks for the quality of written communication in your answer.
(i)
Suggest, with a reason, how the resultant force on the car changes as its speed
increases.
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(ii)
Explain, using Newton’s laws of motion, why the vehicle has a maximum speed.
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(5)
(Total 16 marks)
Q13.
(a)
(i)
State what is meant by a scalar quantity.
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(ii)
State two examples of scalar quantities.
example 1: .........................................................................................
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example 2: ..........................................................................................
(3)
(b)
An object is acted upon by two forces at right angles to each other. One of the forces has
a magnitude of 5.0 N and the resultant force produced on the object is 9.5 N.
Determine
(i)
the magnitude of the other force,
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(ii)
the angle between the resultant force and the 5.0 N force.
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(4)
(Total 7 marks)
Q14.
1
A skydiver of mass 70 kg, jumps from a stationary balloon and reaches a speed of 45 m s–
after falling a distance of 150 m.
(a)
Calculate the skydiver’s
(i)
loss of gravitational potential energy,
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(ii)
gain in kinetic energy.
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(4)
(b)
The difference between the loss of gravitational potential energy and the gain in kinetic
energy is equal to the work done against air resistance. Use this fact to calculate
(i)
the work done against air resistance,
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(ii)
the average force due to air resistance acting on the skydiver.
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(3)
(Total 7 marks)
Q15.
(a)
Define the moment of a force.
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(2)
(b)
The diagram shows a uniform diving board of weight, W, that is fixed at A. The diving
board is supported by a cylinder at C, that exerts an upward force, P, on the board.
(i)
By considering moments about A, explain why the force P must be greater than
the weight of the board, W.
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(ii)
State and explain what would be the effect on the force P of a girl walking along the
board from A to B.
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(4)
(Total 6 marks)
Q16.
The aeroplane shown in the diagram below is travelling horizontally at 95 m s–1.
It has to drop a crate of emergency supplies.
The air resistance acting on the crate may be neglected.
(a)
(i)
The crate is released from the aircraft at point P and lands at point Q. Sketch
the path followed by the crate between P and Q as seen from the ground.
(ii)
Explain why the horizontal component of the crate’s velocity remains constant while
it is moving through the air.
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(3)
(b)
(i)
To avoid damage to the crate, the maximum vertical component of the crate’s
velocity on landing should be 32 m s–1. Show that the maximum height from
which the crate can be dropped is approximately 52 m.
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(ii)
Calculate the time taken for the crate to reach the ground if the crate is dropped
from a height of 52 m.
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(iii)
If R is a point on the ground directly below P, calculate the horizontal distance QR.
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(6)
(c)
In practice air resistance is not negligible. State and explain the effect this has on the
maximum height from which the crate can be dropped.
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(2)
(Total 11 marks)
Q17.
(a)
Define the moment of a force about a point.
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(2)
(b)
The diagram shows a trailer attached to the towbar of a stationary car. The weight of the
trailer is 1800 N and is shown acting through its centre of gravity.F is the force exerted by
the towbar on the trailer. FR is the total normal reaction force experienced by the trailer.
When stationary all forces acting on the trailer are vertical.
(i)
Explain what is meant by centre of gravity.
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(ii)
Calculate the force, F, exerted by the towbar on the trailer.
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(3)
(iii)
Calculate FR.
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(2)
(c)
The car starts to move forwards. State and explain what happens to the magnitude and
direction of force, F.
You may be awarded marks for the quality of written communication in your answer.
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(3)
(Total 10 marks)
Q18.
The graph represents the motion of two cars, A and B, as they move along a straight,
horizontal road.
(a)
Describe the motion of each car as shown on the graph.
(i)
car A: ..........................................................................……………..….
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(ii)
car B: ...........................................................................………………..
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(3)
(b)
Calculate the distance travelled by each car during the first 5.0 s.
(i)
car A: ....................................….....................................................................
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(ii)
car B: ........….......................................................................................
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(4)
(c)
At time t = 0, the two cars are level. Explain why car A is at its maximum distance ahead
of B at t = 2.5 s
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(3)
(Total 10 marks)
Q19.
A packing case is being lifted vertically at a constant speed by a cable attached to a crane.
The packing case has a mass of 640 kg.
(a)
With reference to one of Newton’s laws of motion, explain why the tension, T, in the
cable must be equal to the weight of the packing case.
You may be awarded marks for the quality of written communication in your answer.
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(3)
(b)
The packing case is lifted through a vertical height of 8.0 m in 4.5 s.
Calculate
(i)
the work done on the packing case,
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(ii)
the power output of the crane in this situation.
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(3)
(Total 6 marks)
Q20.
(a)
State the difference between vector and scalar quantities.
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(1)
(b)
State one example of a vector quantity (other than force) and one example of a scalar
quantity.
vector quantity .............................................................................................
scalar quantity ..............................................................................................
(2)
(c)
A 12.0 N force and a 8.0 N force act on a body of mass 6.5 kg at the same time.
For this body, calculate
(i)
the maximum resultant acceleration that it could experience,
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(ii)
the minimum resultant acceleration that it could experience.
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(4)
(Total 7 marks)
Q21.
A fairground ride ends with the car moving up a ramp at a slope of 30° to the horizontal as
shown in the figure below.
(a)
The car and its passengers have a total weight of 7.2 × 103 N. Show that the component
of the weight parallel to the ramp is 3.6 × 103 N.
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(1)
(b)
Calculate the deceleration of the car assuming the only force causing the car to
decelerate is that calculated in part (a).
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(2)
(c)
The car enters at the bottom of the ramp at 18 m s–1. Calculate the minimum length of the
ramp for the car to stop before it reaches the end. The length of the car should be
neglected.
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(2)
(d)
Explain why the stopping distance is, in practice, shorter than the value calculated in
part (c).
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(2)
(Total 7 marks)
Q22.
(a) Explain why a raindrop falling vertically through still air reaches a constant velocity.
You may be awarded marks for the quality of written communication in your answer.
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(4)
(b)
A raindrop falls at a constant vertical velocity of 1.8 m s–1 in still air. The mass of the
raindrop is 7.2 × 10–9 kg.
Calculate
(i)
the kinetic energy of the raindrop,
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(ii)
the work done on the raindrop as it falls through a vertical distance of 4.5 m.
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(4)
(c)
The raindrop in part (b) now falls through air in which a horizontal wind is blowing. If the
velocity of the wind is 1.4 m s–1, use a scale diagram or calculation to determine the
magnitude and direction of the resultant velocity of the raindrop.
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(3)
(Total 11 marks)
Q23.
The figure below shows a uniform steel girder being held horizontally by a crane. Two
cables are attached to the ends of the girder and the tension in each of these cables is T.
(a)
If the tension, T, in each cable is 850 N, calculate
(i)
the horizontal component of the tension in each cable,
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(ii)
the vertical component of the tension in each cable,
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(iii)
the weight of the girder.
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(4)
(b)
On the figure draw an arrow to show the line of action of the weight of the girder.
(1)
(Total 5 marks)
Q24.
The figure below shows a stationary metal block hanging from the middle of a stretched
wire which is suspended from a horizontal beam. The tension in each half of the wire is 15 N.
(a)
Calculate for the wire at A,
(i)
the resultant horizontal component of the tension forces,
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(ii)
the resultant vertical component of the tension forces.
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(3)
(b)
(i)
State the weight of the metal block.
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(ii)
Explain how you arrived at your answer, with reference to an appropriate law of
motion.
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(3)
(Total 6 marks)
Q25.
The figure below shows a supermarket trolley.
The weight of the trolley and its contents is 160 N.
(a)
Explain what is meant by centre of gravity.
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(2)
(b)
P and Q are the resultant forces that the ground exerts on the rear wheels and front
wheels respectively. Calculate the magnitude of
(i)
force P,
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(ii)
force Q.
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(3)
(c)
Calculate the minimum force that needs to be applied vertically at A to lift the front wheels
off the ground.
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(2)
(d)
State and explain, without calculation, how the minimum force that needs to be applied
vertically at A to lift the rear wheels off the ground compares to the force you calculated in
part (c).
You may be awarded marks for the quality of written communication in your answer.
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(3)
(Total 10 marks)
Q26.
The figure below shows apparatus that can be used to investigate energy changes.
The trolley and the mass are joined by an inextensible string. In an experiment to investigate
energy changes, the trolley is initially held at rest, and is then released so that the mass falls
vertically to the ground.
You may be awarded marks for the quality of written communication in your answer.
(a)
(i)
State the energy changes of the falling mass.
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(ii)
Describe the energy changes that take place in this system.
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(4)
(b)
State what measurements would need to be made to investigate the conservation of
energy.
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(2)
(c)
Describe how the measurements in part (b) would be used to investigate the conservation
of energy.
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(4)
(Total 10 marks)
Q27.
A car accelerates from rest to a speed of 26 m s–1. The table shows how the speed of the
car varies over the first 30 seconds of motion.
(a)
time/ s
0
5.0
10.0
15.0
20.0
25.0
30.0
speed/ m s–1
0
16.5
22.5
24.5
25.5
26.0
26.0
Draw a graph of speed against time on the grid provided.
(3)
(b)
Calculate the average acceleration of the car over the first 25 s.
......................................................................................................................
......................................................................................................................
(2)
(c)
Use your graph to estimate the distance travelled by the car in the first 25 s.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(d)
Using the axes below, sketch a graph to show how the resultant force acting on the car
varies over the first 30 s of motion.
(2)
(e)
Explain the shape of the graph you have sketched in part (d), with reference to the graph
you plotted in part (a).
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 11 marks)
Q28.
A supertanker of mass 4.0 × 108 kg, cruising at an initial speed of 4.5 m s–1, takes one hour
to come to rest.
(a)
Assuming that the force slowing the tanker down is constant, calculate
(i)
the deceleration of the tanker,
.............................................................................................................
.............................................................................................................
(ii)
the distance travelled by the tanker while slowing to a stop.
.............................................................................................................
.............................................................................................................
(4)
(b)
Sketch, using the axes below, a distance-time graph representing the motion of the tanker
until it stops.
(2)
(c)
Explain the shape of the graph you have sketched in part (b).
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 8 marks)
Q29.
The figure below shows an apparatus used to locate the centre of gravity of a non-uniform
metal rod.
The rod is supported horizontally by two wires, P and Q and is in equilibrium.
(a)
State two conditions that must be satisfied for the rod to be in equilibrium.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(b)
Wire Q is attached to a newtonmeter so that the force the wire exerts on the rod can be
measured. The reading on the newtonmeter is 2.0 N and the weight of the rod is 5.0 N.
Calculate
(i)
the force that wire P exerts on the rod,
.............................................................................................................
(ii)
the distance d.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(Total 5 marks)
Q30.
The figure below shows a skateboarder descending a ramp.
The skateboarder starts from rest at the top of the ramp at A and leaves the ramp
at Bhorizontally with a velocity v.
(a)
State the energy changes that take place as the skateboarder moves from A to B.
......................................................................................................................
......................................................................................................................
(2)
(b)
In going from A to B the skateboarder’s centre of gravity descends a vertical height of
1.5 m. Calculate the horizontal velocity, v, stating an assumption that you make.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(c)
Explain why the acceleration decreases as the skateboarder moves from A to B.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(d)
After leaving the ramp at B the skateboarder lands on the ground at C 0.42 s later.
Calculate for the skateboarder
(i)
the horizontal distance travelled between B and C,
.............................................................................................................
.............................................................................................................
(ii)
the vertical component of the velocity immediately before impact at C,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
the magnitude of the resultant velocity immediately before impact at C.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(Total 12 marks)
Q31.
A car accelerates from rest to a speed of 26 m s–1. The table shows how the speed of the
car varies over the first 30 seconds of motion.
(a)
time/s
0
5.0
10.0
15.0
20.0
25.0
30.0
speed/m s–1
0
16.5
22.5
24.5
25.5
26.0
26.0
Draw a graph of speed against time on the grid below.
(5)
(b)
Calculate the average acceleration of the car over the first 25 s.
(2)
(c)
Use your graph to estimate the distance travelled by the car in the first 25 s.
(2)
(d)
Using the axes below, sketch a graph to show how the resultant force acting on the car
varies over the first 30 s of motion.
(3)
(e)
Explain the shape of the graph you have sketched in part (d), with reference to the graph
you plotted in part (a).
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 14 marks)
Q32.
The diagram below shows a swimmer standing at the end of a diving board above a
swimming pool.The mass of the swimmer is 72 kg and the horizontal distance between point A
and his centre of mass is 2.4 m.
(a)
Calculate the moment of the swimmer’s weight about point A.
Gravitational field strength of the Earth, g = 9.8 N kg–1.
Moment .......................
(3)
(b)
The swimmer dives off the diving board and his centre of mass falls through 3.2 m before
he reaches the water. Calculate the swimmer’s vertical speed as he enters the water.
Neglect air resistance.
Gravitational field strength of the Earth, g = 9.8 N kg–1
Speed ...........................
(3)
(c)
The water brings the diver to rest when his centre of mass is 1.6 m below the surface of
the water. Calculate the average total upward force acting on the diver which brings his
vertical velocity to zero.
(3)
(Total 9 marks)
Q33.
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)
Q34.
(a)
(i)
State the difference between a scalar quantity and a vector quantity.
.............................................................................................................
.............................................................................................................
(1)
(ii)
State two examples of a scalar quantity and two examples of a vector quantity.
scalar quantities ..................................................................................
vector quantities …..............................................................................
(3)
(b)
The diagram below shows a ship fitted with a sail attached to a cable. The force of the
wind on the sail assists the driving force of the ship’s propellors.
The cable exerts a steady force of 2.8 kN on the ship at an angle of 35° above a
horizontal line.
(i)
Calculate the horizontal and vertical components of this force.
horizontal component of force ............. kN
vertical component of force ................. kN
(2)
(ii)
The ship is moving at a constant velocity of 8.3 m s–1 and the horizontal component
of the force of the cable on the ship acts in the direction in which the ship is moving.
Calculate the power provided by the wind to this ship, stating an appropriate unit.
Answer ..................................
(3)
(c)
The cable has a diameter of 0.014 m. Calculate the tensile stress in the cable when it
exerts a force of 2.8 kN on the ship, stating an appropriate unit.Assume the weight of the
cable is negligible.
Answer ................................
(5)
(Total 14 marks)
Q35.
(a)
(i)
State the difference between a scalar quantity and a vector quantity.
.............................................................................................................
.............................................................................................................
(1)
(ii)
State two examples of a scalar quantity and two examples of a vector quantity.
scalar quantities ..................................................................................
vector quantities …..............................................................................
(3)
(b)
The diagram below shows a ship fitted with a sail attached to a cable. The force of the
wind on the sail assists the driving force of the ship’s propellors.
The cable exerts a steady force of 2.8 kN on the ship at an angle of 35° above a
horizontal line.
(i)
Calculate the horizontal and vertical components of this force.
horizontal component of force ............. kN
vertical component of force ................. kN
(2)
(ii)
The ship is moving at a constant velocity of 8.3 m s–1 and the horizontal component
of the force of the cable on the ship acts in the direction in which the ship is moving.
Calculate the power provided by the wind to this ship, stating an appropriate unit.
Answer ..................................
(3)
(c)
The cable has a diameter of 0.014 m. Calculate the tensile stress in the cable when it
exerts a force of 2.8 kN on the ship, stating an appropriate unit.Assume the weight of the
cable is negligible.
Answer ................................
(5)
(Total 14 marks)
M1.
(a) mark out (equal) distances along height being raised (1)
measure time taken to travel each of these distances (1)
times should be equal (1)
[or use a position sensor attached to a data logger
measure distance or speeds at regular intervals
increase in distance or speeds should be constant]
max 2
(b)
find work done by motor from gain in potential energy of metal block (1)
divide work done by time to find power (1)
measurements: mass of block, height block has risen and time taken (1)
[or power = Fv
force is weight of block
velocity is velocity of block
same measurements as above]
max 2
[4]
M2.
(a)
(i)
70 m s–1 (1)
(ii)
v = 9.81 × 2.0 (1)
= 20 m s–1 (1) (19.6 m s–1)
(iii)
v = √(702 + 19.622) = 73 m s–1 (1)
direction: tan θ =
= 0.28
θ = 15.6° (1) (± 0. 1°) (to horizontal) (1)
(allow C.E. for values of v from (i) and (ii))
[or use of correct scale drawing]
5
(b)
(i)
air resistance is greater than weight (1)(hence) resultant force is
upwards (1)hence deceleration (Newton's second law) (1)
(ii)
air resistance decreases as speed decreases (1)weight equals air resistance (hence
constant speed)(hence) resultant force is zero (Newton's first law) (1)
max 4
QWC 2
[9]
M3.
(a)
(i)
resultant force acting on tray is zero [or P + W = Q] (1)
resultant torque is zero
[or correct moments equation
or anticlockwise moments = clockwise moments] (1)
(ii)
W= 0.12 × 9.81 = 1.2N (1) (1.18N)
(iii)
(taking moments about P gives)
Q × 0.1 = 0.12 × 9.81 × 0.25 (1)
Q = 2.9 N (2.94 N) (1)
P = 2.9 – 1.2 = 1.7 N (1) (or 2.94 – 1.18 = 1.76 N)
(allow C.E. for values of W and Q)
6
(b)
placed at Q (1)
no additional turning moment about Q (1)
2
[8]
M4.
(a)
(ii)
(i)
initial acceleration/increase of speed (1)
reaches a constant speed/velocity (1)
acceleration decreases to become zero (at this speed) (1)
drag/frictional forces increases with speed (1)
drag equal to weight (– upthrust) (1)
no resultant force at terminal speed
[or balanced forces or forces cancel] (1)
max 5
(b)
column C
26.6
39.7 four values correct (1)
49.4 all values correct and to 3 or 4 s.f. (1)
75.2
118
173.5
2
(c)
(i)
column E
1.421.601.69
(ii)
all values correct and to 3 or 4 s.f. (1)1.882.072.24
axes labelled and suitable scales chosen (1)at least 5 points plotted
correctly (1)acceptable line (1)
4
(d)
(i)
gradient =
= 2.0 (1)
n = gradient (= 2) (1)
(ii)
intercept on y-axis = log k (1)
intercept = 1.0 (1)
k (= 101.0) = 10 (1)
units of k: for n = 2, mm–1 s–1 (1)
max 5
[16]
M5.
(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]
M6.
(a)
(i)
X (1)stress (force)
strain (extension) for the whole length (1)
(ii)
Y (1)has lower breaking stress (or force/unit area is less) (1)
(iii)
Y (1)exhibits plastic behaviour (1)
(iv)
Y (1)for given stress, Y has greater extension[or greater area under graph] (1)
8
QWC 2
(b)
(i)
(use of E =
gives)
=
(1)
(1) for data into correct equation, (1) for correct area
= 2.4 N (1)
(allow C.E. for incorrect area conversion)
(ii)
(use of energy stored = ½Fe gives) energy =
(1)
= 36 × 10–3 J (1)
(allow C.E. for value of F from (i))
5
[13]
M7.
(a)
(ii)
(i)
horizontal component of the tension in the cable (1)
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)
use of Tvert = 2450 N then T = 2730 N)
(2773 N)(if
(allow C.E. for values from (b) (i) and (b)(ii))
(iv)
tan θ =
θ = 26° (1)
(1)
(allow C.E. for values from (b) (i) and (b)(ii))
6
[8]
M8.
(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]
M9.
(a)
(ii)
(i)
(gravitational) potential energy to kinetic energy (1)
kinetic energy to heat energy
[or work done against friction] (1)
2
(b)
e.g. when using light gatesplace piece of card on trolley of measured length (1)card
obscures light gate just before trolley strikes block (1)calculate speed from length of
card/time obscured (1)
alternative 1: measured horizontal distance (1)speed = distance/time (1)time (1)
alternative 2: measure h (1)equate potential and kinetic energy (1)v2 = gh (1)
alternative 3: data logger + sensor (1)how data processed (1)how speed found (1)
3
QWC 2
(c)
vary starting height of trolley
[or change angle] (1)
the greater the height the greater the speed of impact (1)
[or alter friction of surface (1)
greater friction, lower speed] (1)
2
[7]
M10.
(a) suitable calculation using a pair of values of x and corresponding t
to give an average of 2.2 m s–1 (± 0.05 m s–1) (1)
valid reason given (1)
(e.g. larger values are more reliable/accurate
or use of differences eliminates zero errors)
2
(b)
(i)
column D (y/t (cm s–1)
186
210
233
259
284
307
all values correct to 3 s.f. (1)
(ii)
(iii)
graph: chosen graph gives a straight line (e.g. y/t against t) (1)axes labelled
correctly (1)suitable scale chosen (1)minimum of four points correctly
plotted (1)best straight line (1)
u (= y - intercept) = 162 cm s–1 (± 4 cm s–1) (1)gradient = 495 (cm s–2) (± 25 cm s–
) (1)k = gradient (= 495 cm s–2) (1)
2
9
(c)
(i)
u : initial vertical component of velocity (1)
(ii)
k : = ½ g (1)
2
(d)
v2 = u2 + 2.22 (1)
gives v = (1.622 + 2.22)1/2 = 2.7 m s–1 (± 0.1 m s –1) (1)
2
[15]
M11.
(a) for a body in equilibrium (1)
the (sum of the) clockwise moments about a point (1)
are equal to (the sum of) the anticlockwise moments (1)
[or resultant torque about a point (1)
is zero (1)]
3
(b)
(i)
diagram to show:
pivot/fulcrum/balance point (1)
masses or appropriate objects (1)
(ii)
known masses on either side of pivot (1)
move this mass until ruler is in equilibrium/balanced (1)
measure distances (1)
repeat with other masses (1)
(iii)
(calculate) weights of masses (on left and right of pivot) (1)
product of weight and distance to pivot on either side of pivot (1)
hence should be equal (1)
max 7
QWC 2
[10]
M12.
(a)
(i)
(use of F = ma gives) 1.8 × 103 = 900 a (1)a = 2.0 m s–2 (1)
(ii)
(use of v = u + at gives) v = 2.0 × 8.0 = 16 m s–1 (1)(allow C.E. for a from (i))
(iii)
(use of p = mv gives) p = 900 × 16= 14 × 103 kg m s–1 (or N s) (1)
m s–1)(allow C.E. for v from(ii))
(iv)
(use of s = ut + ½at2 gives)
s=
(14.4 × 103 kg
(1)
= 64 m (1)
(allow C.E. for a from (i))
(v)
use of W = Fs gives) W = 1.8 × 103 × 64 (1)
= 1.2 × 105 J (1)
(1.15 × 105 J)
(allow C.E. for s from (iv))
[or Ek = ½mv2 = ½ × 900 × 162 (1)
= 1.2 × 105 J (1)
(allow C.E. for v from (ii))]
9
(b)
2
(c)
(i)
decreases (1)
air resistance increases (with speed) (1)
(ii)
eventually two forces are equal (in magnitude) (1)
resultant force is zero (1)
hence constant/terminal velocity (zero acceleration)
in accordance with Newton’s first law (1)
correct statement and application of Newton’s first
or second law (1)
max 5
QWC 2
[16]
M13.
(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]
M14.
(a)
(i)
(use of Ep = mgh gives) Ep = 70 × 9.81 × 150 (1)
= 1.0(3) × 105 J (1)
(ii)
(use of Ek = ½mv2 gives) Ek = ½ × 70 × 452 (1)
= 7.1 × 104 J (1)
(7.09 × 104 J)
4
(b)
(i)
work done (= 1.03 × 105 – 7.09 × 104) = 3.2(1) × 104 J (1)(allow C.E. for values
of Ep and Ek from (a)
(ii)
(use of work done = Fs gives)3.21 × 104 = F × 150 (1)(allow C.E. for value
of work done from (i)F = 210 N (1) (213 N)
3
[7]
M15.
(a) product of the force and the perpendicular distance (1)
reference to a point/pivot (1)
2
(b)
(i)
since W is at a greater distance from A (1)
then W must be less than P if moments are to be equal (1)
(ii)
P must increase (1)
since moment of girl’s weight increases as she moves from A to B (1)
correct statement about how P changes
(e.g. P minimum at A, maximum at B, or P increases in a
linear fashion) (1)
max 4
[6]
M16.
(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=
(ii)
(1)
(= 52.2 m)
(use of s = ½ at2 gives)
= 3.3 s (1)
52 = ½ 9.81 × t2 (1)
(3.26 s)
[or use of v = u + at gives 32 = (0) + 9.81 × t (1)
= 3.3 s (1)
(iii)
(3.26 s)]
(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]
M17.
(a)
(moment) force × perpendicular (1) distance (from the point) (1)
2
(b)
(i)
the point in a body where the resultant torque is zero
[or where the (resultant) force of gravity acts or where the weight
acts through] (1)
(ii)
F × 2.5 = 1800 × 0.35 (1)
F = 250 N (1)
(252 N)
(iii)
FR = (1800 - 252) (1)
= 1500 N (1)
(1548) N
[use of F = 250 N gives FR = 1550 N or 1600 N)
(allow C.E. for incorrect value of F from (ii))
5
(c)
force must have a horizontal component (1)F (therefore) increases in magnitude (1)and
act at an angle (to the vertical) towards the car (1)
3
QWC 1
[10]
M18.
(a)
(i)
car A: travels at constant speed (1)
(ii)
car B: accelerates for first 5 secs (or up to 18 m s–1) (1)
then travels at constant speed (1)
3
(b)
(i)
car A: distance = 5.0 × 16 (1)
= 80 m (1)
(ii)
car B: (distance = area under graph)
distance = [5.0 × ½ (18 + 14)] (1)
= 80 m (1)
4
(c)
car B is initially slower than car A (for first 2.5 s) (1)
distance apart therefore increases (1)
cars have same speed at 2.5 s(1)
after 2.5 s, car B travels faster than car A (or separation decreases) (1)
max 3
[10]
M19.
(a) resultant force on crate is zero (1)
forces must have equal magnitudes or size (1)
(but) act in opposite directions (1)
correct statement of 1st or 2nd law (1)
max 3
QWC 1
(b)
(i)
work done = F × d = 640 × 9.81 × 8.0 (1)
= 5.0(2) × 104J (1)
(ii)
(use of P =
gives)
P=
= 1.1(2) × 104W (1)
(allow C.E. for value of work done from (i))
3
[6]
M20.
(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)
(ii)
a=
= 3.1 m s–2 (1)
(3.08 m s–2)
subtraction of forces (12 – 8) (1)
a=
= 0.62 m s–2 (1)
(0.615 m s–2)
4
[7]
M21.
(a)
component (parallel to ramp) = 7.2 × 103 × sin 30 (1) (= 3.6 × 103 N)
1
(b)
mass =
a=
= 734 (kg) (1)
= 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]
M22.
(a)
weight/gravity causes raindrop to accelerate/move faster (initially) (1)resistive
forces/friction increase(s) with speed (1)resistive force (eventually) equals weight (1)[or
upward forces equal downward forces]resultant force is now zero (1)[or forces balance or
in equilibrium]no more acceleration (1)[or correct application of Newton’s Laws][if
Newton’s third law used, then may only score first two marks]
Max 4
QWC 1
(b)
(i)
Ek (= ½mv2) = ½ × 7.2 × 10–9 × 1.82 (1)
= 1.2 × 10–8 J (1) (1.17 × 10–8 J)
(ii)
work done (= mgh) = 7.2 × 10–9 × 9.81 × 4.5 (1)
= 3.2 × 10–7 J (1) (3.18 × 10–7 J)
4
(c)
vresultant = √(1.82 + 1.42) (1)
= 2.2(8) m s–1 (1)
θ = tan–1 (1.4/1.8) = 38° (1) (37.9°)
[or correct scale diagram]
3
[11]
M23.
(a)
(i)
horizontal component = 850 × cos 42 (1)
= 630 N (1) (632 N)
(ii)
vertical component = 850 × sin 42 = 570 N (1) (569 N)
(if mixed up sin and cos then CE in (ii))
(iii)
weight of girder = 2 × 570 = 1100 N (1) (1142 N)
(use of 569 N gives weight = 1138 N)
(allow C.E. for value of vertical component in (ii))
4
(b)
arrow drawn vertically downwards at centre of girder (1)
1
[5]
M24.
(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]
M25.
(a)
the point (in a body) (1)
where the weight (or gravity) of the object appears to act
[or resultant torque zero] (1)
2
(b)
(i)
P × 0.90 = 160 × 0.50 (1)
P = 89 N (88.9 N)
(ii)
Q = (160 − 89) = 71 N (1)
(allow C.E. for value of P from (i))
3
(c)
(minimum) force × 0.10 = 160 × 0.40 (1)
force = 640 N (1)
2
(d)
force is less (1)
because distance to pivot is larger (1)
smaller force gives large enough moment (1)
3
[10]
M26.
(a)
(i)
(gravitational) potential energy (1)
to kinetic energy (1)
(ii)
both trolley and mass have kinetic energy (1)
mention of thermal energy (due to friction) (1)
4
(b)
masses of trolley and falling mass (1)
distance mass falls (or trolley moves) and time taken to
fall (or speed) (1)
2
(c)
calculate loss of gravitational pot. energy of falling mass (mgh) (1)
calculate speed of trolley (as mass hits floor),
with details of speed calculation (1)
calculate kinetic energy of trolley (1)
and mass (1)
compare (loss of) potential energy with (gain of) kinetic energy (1)
Max 4
[10]
M27.
(a)
scales (1)
six points correctly plotted (1)
trendline (1)
3
(b)
average acceleration =
(1)
= 1.0(4) ms-2 (1)
(allow C.E. for incorrect values used in acceleration calculation)
2
(c)
area under graph (1)
= 510 ± 30 m (1)
2
(d)
(graph to show force starting from y-axis)
decreasing (not a straight line) (1)
to zero (at end of graph) (1)
2
(e)
(since) gradient of a velocity-time graph gives acceleration (1)
first graph shows acceleration is decreasing (1)
2
[11]
M28.
(a)
(i)
(use of
gives)
(1)
=1.25 × 10–3 ms–2 (1)
(ii)
(use of v2 = u2 +2as gives) 0=4.52 – 2 × 1.25 ×10–3 × s (1)
(1)
4
(b)
increasing curve (1)
correct curve (1)
1
(c)
gradient (slope) of graph represents speed (1)
hence graph has decreasing gradient (1)
2
[8]
M29.
(a)
resultant force zero (1)
resultant torque about any point zero (1)
2
(b)
(i)
force due to wire P = 5.0 - 2.0 = 3.0 N (1)
(ii)
(moments give) 5.0 × d = 2.0 × 0.90 (1)
d= 0.36 m (1)
3
[5]
M30.
(a)
potential energy to kinetic energy (1)mention of thermal energy and friction (1)
2
(b)
(use of ½ mv2 = mgh gives) ½ vh2 = 9.81 × 1.5 (1)vh = 5.4(2)ms–1 (1)(assumption) energy
converted to thermal energy is negligible (1)
3
(c)
component of weight down the slope causes acceleration (1)this component decreases
as skateboard moves further downthe slope (1) air resistance/friction increases (with
speed) (1)
2
(d)
(i)
distance (= 0.42 × 5.4) = 2.3m (1)
(2.27m)
(allow C.E. for value of vh from (b))
(ii)
vv = 9.8 × 0.42 (1)
= 4.1(l) m s–1 (1)
(iii)
v2 = 4.12 + 5.42 (1)
v = 6.8 m s–1 (1)
(6.78 m s–1)
(allow C.E. for value of vh from (b))
5
[12]
M31.
(a)
scales (1)(1) (one mark for each scale)
six points correctly plotted (1)(1)
(ignore 0,0 and lose one mark for each error)
trend line (1) (if misses more than two points then lose mark)
5
(b)
average acceleration = 26/25 (1)
= 1.0(4) m s–2 (1) e.c.f. from correct values used
2
(c)
area under graph (1) = 510 ± 30m (1)
2
(d)
curve decreasing (1)
to zero at end of graph (1) and starting from vertical axis within 1mm (1)
3
(e)
(since) gradient of a velocity-time graph gives acceleration (1)
(first graph shows) acceleration is decreasing (1)
or resistive force increases (with speed) (1)
so resultant force (or acceleration) decreases (1)
2
[14]
M32.
(a)
(moment) = 72 × 9.8 × 2.4 (1)
penalise 1 mark for g = 10 m s–2
= 1690 (1) Nm (1)
3
(b)
½ mv2 = mgΔh or v2 = u2 + 2gs (1)
v2 = 9.8 × 3.2 × 2 (1)
allow e.c.f. g =10 m s–2
v = 7.92 m s–1 (1)
(8.0 m s–1 with e.c.f.)
3
(c)
from Δmgh or Δ½mv2 = decelerating force × 1.6 m (1)
decelerating force = 1411N (or 1440N if g = 10 m s–2 used) (1)
total average upward force = 1411 + 706 = 2100 (2117)N (1)
3
[9]
M33.
(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)
2
(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)
4
[6]
M34.
(a)
(i)
vector has direction and a scalar does not (1)
(ii)
scalar examples; any two e.g. speed, mass, energy, time, power
vector examples; any two e.g. displacement, velocity,
acceleration, force or weight
(1)(1)(1) for 4 correct, (1)(1) for 3 correct, (1) for 2 correct
4
(b)
(i)
horizontal component (= 2.8 cos 35) = 2.3 (kN) (2293.6) (1)
vertical component (= 2.8 sin 35) = 1.6(kN) (1606.0) (1)
(ii)
power = force × velocity or 2.3 kN × 8.3 m s–1 (1) (ecf from (b) (i))
= 1.9 × 104 (19037 or 19100) (1) ecf
W (or J s–1) (1) (or 19 W (or kJ s–1))
5
(c)
(area of cross–section of cable =) π × (½ 0.014)2 (1)
= 1.5(4) × 10–4 (m2) (1)
stress (=F/A) =
(allow ecf here if attempt to calculate area) (1)
= 1.8(2) × 107 (1) ecf
Pa (or N m–2) (1)
QWC
descriptor
mark
range
goodexcellent
The candidate provides a comprehensive and coherent
description which includes all the necessary
measurements in a logical order. The description should
show awareness of the need to use a range of standard
masses. In addition, the use of the measurements is
explained clearly, including an outline of a graphical
method to find the mass of the rock sample, or calculation
using two or more standard masses and averaging. For 6
marks there must be a description of how to make
accurate measurements.
modest- The candidate’s description includes the necessary
adequate measurements using one standard mass as well as the
rock sample. The description may not be presented in a
logical order and they show little consideration in relation to
making the measurements accurately. A clear explanation
is provided of how to find the mass of the rock sample from
their measurements, including correct use of Hooke’s law
through calculations or inadequate graphical method.
poorlimited
The candidate knows the necessary measurements to be
made using a standard mass and the rock sample. The
explanation of how to find the mass of the rock sample
may be sketchy.
5-6
3-4
1-2
5
[14]
M35.
(a)
(i)
vector has direction and a scalar does not (1)
(ii)
scalar examples; any two e.g. speed, mass, energy, time, power
vector examples; any two e.g. displacement, velocity,
acceleration, force or weight
(1)(1)(1) for 4 correct, (1)(1) for 3 correct, (1) for 2 correct
4
(b)
(i)
horizontal component (= 2.8 cos 35) = 2.3 (kN) (2293.6) (1)
vertical component (= 2.8 sin 35) = 1.6(kN) (1606.0) (1)
(ii)
power = force × velocity or 2.3 kN × 8.3 m s–1 (1) (ecf from (b) (i))
= 1.9 × 104 (19037 or 19100) (1) ecf
W (or J s–1) (1) (or 19 W (or kJ s–1))
5
(c)
(area of cross–section of cable =) π × (½ 0.014)2 (1)
= 1.5(4) × 10–4 (m2) (1)
stress (=F/A) =
(allow ecf here if attempt to calculate area) (1)
= 1.8(2) × 107 (1) ecf
Pa (or N m–2) (1)
QWC
goodexcellent
descriptor
The candidate provides a comprehensive and coherent
description which includes all the necessary
measurements in a logical order. The description should
show awareness of the need to use a range of standard
masses. In addition, the use of the measurements is
explained clearly, including an outline of a graphical
method to find the mass of the rock sample, or calculation
using two or more standard masses and averaging. For 6
marks there must be a description of how to make
accurate measurements.
modest- The candidate’s description includes the necessary
adequate measurements using one standard mass as well as the
rock sample. The description may not be presented in a
logical order and they show little consideration in relation to
making the measurements accurately. A clear explanation
is provided of how to find the mass of the rock sample from
their measurements, including correct use of Hooke’s law
through calculations or inadequate graphical method.
poorlimited
The candidate knows the necessary measurements to be
made using a standard mass and the rock sample. The
explanation of how to find the mass of the rock sample
may be sketchy.
mark
range
5-6
3-4
1-2
5
[14]
E1.
Candidates found this question straightforward and were for the most part able to suggest
sensible ideas for demonstrating constant velocity and finding the power output of the motor.
Full marks were quite common. Candidates expressed themselves well and took note of the
statement regarding the quality of written communication at the start of the question.
E2.
This question also proved difficult for a number of candidates. The calculation in part (a)
was generally done well by the better candidates. Weaker candidates tended to be confused by
the idea of vector components and it was not unusual to find the inclusion of the horizontal
component of velocity in the calculation of the vertical component of velocity.
Explaining the motion of the parachutist in part (b) was not done well. Candidates in the main
did not appreciate that there was a resultant force upwards giving a decrease in velocity. The
explanation of terminal velocity has been assessed in previous papers and answers showed
that candidates are getting more confident in their explanations. They are, however, still inclined
to give expressions such as “the force of air resistance equals the terminal velocity” or “the
acceleration due to gravity equals air resistance”.
E3.
The first question on the paper was done well and candidates coped with the moments
calculation in an impressive way. They performed less confidently in part (a)(i) where the
majority of candidates only gave one of the necessary conditions for equilibrium. In part (b) the
majority of candidates knew intuitively that the drinking glass should be placed at the point Q.
Explaining why this was the case proved more of a challenge.
E4.
This data handling and graph question carried the highest marks on the paper. When
describing the motion of the ball bearing in part (a)(i), the majority of candidates were aware
that the object accelerated initially and then reached terminal speed. Few however, realised that
the acceleration became zero and some incorrectly thought that the acceleration was constant
even though they stated it had reached terminal speed.
When explaining the motion in part (ii), many candidates failed to state clearly that the drag
force increased with increase of speed. Most candidates knew that the drag force became equal
to the weight of the ball bearing but often failed to state that the two forces opposed each other
and therefore produced a zero resultant force, leading to zero acceleration at the terminal
speed.
Calculation of the data in part (b) was done very well, except that some candidates were
careless in the use of significant figures and were penalised. Other candidates gained no credit
at all through using an incorrect distance to calculate the speed.
In part (c), almost all candidates scored the mark for correct data in column E in the table.
Many also scored full marks in part (ii). Penalties were incurred through omission of labels on
the axes of the graph or failure to plot the points correctly. Surprisingly, many candidates
misplotted the datum 2.07 in column E as 2.007. Best fit lines were usually appropriately drawn.
In part (d) it was pleasing to see that many candidates used logs with confidence and
expressed the equation v = krn in the required form of y = mx + c. In part (i), full marks were
often scored with a clear gradient triangle, an appropriate calculation and a clear statement that
n is equal to the gradient. Some candidates thought that n was the intercept and k the gradient.
However, many candidates did use the intercept correctly and obtained the correct value of k. A
significant number of candidates approached the same point, and gained the same credit, by
using the value of n obtained previously, together with the co-ordinates of a point on the line in
the equation v = krn to calculate k. Very few candidates were able to give the correct unit of k.
E5.
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.
E6.
Examiners were pleased to find that part (a) was answered satisfactorily and that
candidates not only chose the correct wire but were very often able to provide the correct
reason for doing so. Many candidates gained full marks, while a large number only lost one or
two marks.
Part (i) was usually correct, although reasons such as ‘the graph is a straight line’ were not
accepted. A ‘constant gradient’ was accepted but few candidates gave this as a reason, most
giving the proportionality of the quantities involved. In part (ii) answers such as ‘Y broke before
X’ was not accepted. Examiners were looking for a reason in terms of lower breaking stress.
Answers to part (iii) were not so good and candidates who did not know the correct answer
attempted an answer in terms of the gradients of the curves or the bending of curve Y as the
tensile strain increased. Part (iv) gave the most trouble. Many candidates again tried an
explanation in terms of the gradient, but a significant number followed the correct track and
gave a reason in terms of the area under such a graph. Unfortunately the majority of these
candidates referred to the area under the whole curve, whereas it should have been the area
under the curve at a given tensile stress. Surprisingly, many candidates, even when using a
given stress, gave the area under X as being greater than that under Y.
The final calculation in part (b) did not cause too much difficulty and, provided the initial
equation for the Young modulus was correct, candidates produced a correct answer with correct
units. One common error which again arose from not reading the question thoroughly, was
using the extended length of the elastic cord as the extension. Converting the cross-sectional
area of the cord from mm2 to m2 caused some problems, but this error was carried forward after
the initial penalty had been imposed. The calculation in part (ii) was also done well by those
who knew the expression for the energy stored, or were aware that it was given in the data
sheet. Some answers, resulting from a carry forward of an incorrect force in part (i) gave
energies amounting to several million joules. This attracted no comment.
E7.
Most candidates found this question quite accessible although weaker candidates were
confused by the concept of resolving a force into two perpendicular components. A surprising
number of candidates gave incorrect responses to part (a) but were completely at ease with part
(b), scoring maximum marks. There were several approaches, correctly used, to deduce the
tension in the cable. Some of these methods were extremely complicated but eventually
generated the correct answer. This question produced the greatest number of significant figure
errors.
E8.
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.
E9.
Candidates found this question reasonably familiar and had more success, especially in part
(a), than had been the case with similar questions in the past. In part (b), more candidates than
previously seemed familiar with light gates and data loggers and were able to describe clearly
the use of these devices. A minority of candidates had obviously studied a similar question in
the January paper, a question which had required candidates to describe how the speed was
measured after a collision. These candidates then tried to answer this question in the same way
with the result that their answers were generally inappropriate.
Part (c) caused more problems than anticipated as many candidates misinterpreted the
question and explained why the speed of the trolley might vary and not how.
E10.
In part (a) many candidates failed to read the question correctly and did not realise that two
sets of measurements meant two pairs of measurements. Consequently, many candidates used
the values of x and t for the first image only and obtained an unacceptable answer. Few
candidates provided a suitably valid reason for their choice of data. Many candidates
considered a valid reason to be the constant value of the horizontal component of velocity. Only
the best candidates used the difference between the first and last values and also provided a
valid reason.
Some candidates failed to give their answers in part (b) to the same number of significant
figures as the data provided. In part (ii), most candidates were aware of the correct graph to plot
and were able to produce the necessary straight line graph. However many candidates lost a
mark through an inappropriate scale for y/t. In part (iii), the majority of candidates were able to
determine the correct values of u and k, usually providing clear indications on the graph for the
gradient calculation.
Many candidates failed to score a single mark in part (c). Some did know the physical
significance of u but almost all failed to realise the significance of k. Even the best candidates
were unable to relate the given equation in part (b) to their knowledge of projectile motion.
Very few candidates realised in part (d) that they were expected to apply their knowledge of
vectors and use appropriate values from parts (a) and (b).
E11.
This question, based on the principle of moments, differed from previous questions on this
topic, in that a descriptive question on moments had not been previously set. Candidates
generally rose to the challenge and approached the description of the experiment in parts (b)
and (c) with confidence. The procedural descriptions did tend to lack clarity but candidates were
generally able to score reasonable marks. A minority of candidates totally misunderstood the
question and described experiments which verified the conservation of momentum.
E12.
The question, as a whole, proved to be a good discriminator. The various calculations in
part (a) were quite demanding for a significant proportion of the candidates and unit errors were
quite common for all the quantities involved, including those for the more straightforward
quantities such as acceleration and speed.
Candidates were asked to sketch two graphs in part (b). This proved to be quite a difficult
exercise with the non-linear, distance vs. time, graph being the most difficult to draw correctly.
Part (c) produced the same problems that questions of this type have produced in previous
papers. Many candidates insisted on using Newton’s third law incorrectly and consequently
getting into real trouble. Statements such as “the air resistance is the reaction to the driving
force action” were common and were awarded zero marks.
E13.
This question was done well with most candidates in part (a) successfully explaining the
meaning of a scalar quantity and able to give suitable examples. Calculating the third force and
the angle in part (b) did not present too many problems and candidates across the ability range
approached this task with confidence. A few candidates did opt for a scale drawing but they
were very much in the minority.
E14.
A significant proportion of weaker candidates were not able to select the correct equations
to calculate the loss in potential energy and gain in kinetic energy, which is a little surprising
given that these have been tested frequently in the past and do not usually cause problems.
Significant figure errors were very common in this question with many candidates quoting the
changes in kinetic and potential energies to five and six significant figures respectively. Part (b)
proved to be successful with a pleasing number of candidates able to calculate the force of air
resistance from the energy differences.
E15.
This question proved to be a success with excellent discrimination and good accessibility.
In part (a) most candidates defined the moment of a force although some did lapse into woolly
explanations of the effects of moments.
Part (b) worked well, with candidates of all abilities making a good attempt at sorting out the
relationship between the forces P and W. The better candidates made good use of Physics
principles and expressed themselves cogently.
E16.
This question caused problems for all but the most able candidates. This was particularly
true with part (a) in which the path of the crate was often shown as a straight line or curving the
wrong way. Candidates also found it difficult to express themselves clearly when trying to
explain why the horizontal component of velocity remained constant.
Part (b) was better answered and frequently scored full marks. In part (c) the majority of
candidates correctly surmised that the crate could be dropped from a greater height but found it
more difficult to explain why. Very often the answer stated that air resistance slowed the crate
down, and not that it decreased its acceleration.
E17.
This question on moments provided excellent discrimination and good candidates were
given the opportunity to really show a sound grasp of forces and equilibrium. Part (a) was a
straightforward starter but weaker candidates were still unclear as to what ‘define’ means and
were inclined to describe what moments do instead of providing a formal definition.
Part (b) (i) was answered well by good candidates but caused problems for less able
candidates. The meaning of ‘centre of gravity’ was a source of confusion and a common
response was ‘where the mass of an object acts’, rather than the weight. The calculations in
parts (ii) and (iii) were demanding but a significant proportion of candidates scored full marks.
Part (c) generated some interesting responses. Most candidates realised that a horizontal force
was now involved, but failed to appreciate that the force F was still required. Many candidates
had problems trying to apply Newton’s third law and managed to convince themselves
that Fnow acted towards the trailer. Good candidates answered this question very well and
clearly relished the challenge it presented.
E18.
This question proved to be very accessible and full marks were awarded frequently. The
only real confusion arose in part (c) where weaker candidates had difficulty organising their
thoughts and expressing them in a logical way. Loose use of language was evident, with
statements such as “accelerating at a constant speed” cropping up in a significant minority of
scripts.
E19.
Responses, requiring explanations rather than calculations, to questions involving
Newton’s Laws of Motion are often poor and this was certainly the case in part (a) of this
question. The common mistake was assuming that this was an example of an application of the
third law, and answers suggesting that the tension was the reaction to the action of weight were
commonplace. Another common error was assuming that a force was needed to make the
packing case move, leading to some convoluted and ultimately incorrect explanations as to how
there was a resultant force, even though tension and weight were equal.
Candidates also had problems with the calculation in part (b), with many just multiplying
numbers together in a somewhat random manner in an attempt to arrive at the correct answer.
Significant figure errors were again commonplace in this question, although the majority of
candidates did use a value of 9.81 N kg–1 rather than 10 N kg–1 for the gravitational field strength.
E20.
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.
E21.
This question provided excellent discrimination and more able candidates analysed the
situation successfully. Less able candidates were confused with the distinction between mass
and weight and consequently calculated the acceleration as 0.50 m s–2. Calculating the stopping
distance proved more straightforward, although there was the usual confusion of signs for
acceleration and whether the given velocity of 18 m s–1 was the initial or final velocity. Most
candidates were able to identify friction as a reason for the stopping distance being shorter in
practice, but were less confident when explaining why.
E22.
Although questions of this type have not been well answered in past papers, the general
standard of answers in this paper was impressive. Candidates seem to have learnt that it is
wrong to always make use of Newton’s third law and in part (a) proved to be quite confident in
applying the concept of a resultant force and the effect it has on the motion of a body. Less able
candidates still had the tendency to make statements such as ‘the drag force becomes equal to
the acceleration’ and ‘air resistance slows the raindrop down’, but this was less of a problem
than it has been in the past. Answers were generally set out well and usually were awarded
good marks for Quality of Written Communication.
Part (b) proved to be quite discriminating and although most candidates quoted the correct
expression for kinetic energy, arithmetic errors were frequent. The most common of these were
the failure to square the speed of the raindrop and rounding errors, with candidates giving the
answer for kinetic energy as 1.1 × 10–8 J instead of 1.2 × 10–8 J. Unit errors were also quite
common in both part (b) (i) and (ii). Part (b) (ii) proved to be very discriminating with only the
best candidates being able to correctly calculate the work done on the raindrop.
Part (c) was answered well, with the only common error being failure to calculate a direction as
well as a magnitude for the resultant velocity.
E23.
This question was answered well and it was clear that candidates were skilled in resolving
vectors. Full marks were common in part (a). A minority of candidates confused the vertical and
horizontal components and consequently lost the first two marks. The arrow showing the line of
action proved a very accessible mark, although some candidates spoilt their answers by a lack
of care in their drawing. It was not clear from these drawings that the arrow pointed vertically
downwards.
E24.
Responses to this question were extremely mixed with the term resultant force once again
confusing a significant proportion of candidates. The tension was resolved correctly very
frequently, but the next step of calculating the resultant of the two horizontal and the two vertical
components was often omitted.
Part (b) produced better responses, although a significant minority failed to answer this part at
all. There were also a number of candidates who thought the weight of the block must be 30 N.
The written explanation in part (b) (ii) generated better responses than had been found in
answers to questions of this type in previous papers, but there is still much confusion about
when it is appropriate to apply Newton’s Third Law. Comments such as ‘the tension is the action
on the block, so the weight is the reaction on the block’ are still quite common. Candidates who
used Newton’s First Law were far less confused and were able to explain correctly how the
weight was determined.
E25.
Candidates found this question not very accessible. This was probably due to the trolley
having two sets of wheels in contact with the ground as opposed to only one set in the previous
paper. This meant that unless candidates were familiar with this type of problem, there was no
obvious pivot point about which to take moments. This unfamiliarity was noticeable in the
calculations for part (b), which caused difficulties for large numbers of candidates, and it was
clear that not all centres had given candidates experience of this type of question on moments.
Part (c) proved to be even more difficult and only the more able candidates were able to
calculate the force necessary to lift the front wheels off the ground. Part (d) was answered
consistently better, with even less able candidates explaining why the required force would be
less than that in part (c).
E26.
Candidates found this question far less accessible than expected. Most were able to
answer part (a) well, but found it difficult to describe the energy changes that took place in the
described experiment. There was much confusion as to exactly how energy conservation might
be demonstrated and a common mistake was to assume that gravitational potential energy was
converted to kinetic energy in either the mass or the trolley alone. There was also a tendency
for candidates to write about momentum conservation and elastic and inelastic collisions.
E27.
For the most part this question was answered well. In part (a), a significant minority of
candidates did choose inappropriate scales for the graph, but plotting errors were rare. The
calculation for the average acceleration in part (b) was done well but that for the distance
travelled, in part (c) was less so. Most candidates appreciated that they needed to estimate the
area under the graph, but many failed to do this within the allowed range.
Parts (c) and (d) caused problems to a number of candidates because they did not understand
the term resultant force. This was often confused with air resistance which, candidates correctly
assumed, increased with speed. The mark scheme did allow candidates to score marks in part
(d), even if they had misunderstood resultant force, provided their description of the effect of an
increasing resistive force was correct.
E28.
The deceleration and distance travelled by the supertanker were calculated correctly in
part (a) by a large number of candidates. The few errors that occurred were either the unit for
deceleration or an incorrect conversion of hours to seconds.
Parts (b) and (c) caused more problems and many less able candidates were unable to sketch
a correct graph, confusing distance-time with speed-time. The explanation of the shape of the
graph generated some quite vague answers with considerable confusion over what the gradient
of the graph represented.
E29.
The January 2006 examination paper contained a question on moments that was
answered very poorly. The evidence from this paper is that candidates had learned from the
earlier question and consequently answered this question very well indeed. Many scored full
marks with the only significant omission being a second condition for equilibrium.
E30.
Parts (a) and (b) of this question involved a familiar theme of energy conversion that has
been assessed before. For the most part, candidates identified the energy changes correctly.
The calculation of the horizontal velocity did however, cause problems. This was because many
candidates applied the equations of uniform acceleration which were not appropriate in this
situation as the acceleration changed.
Parts (c) and (d) proved to be challenging. Only a few of the more able candidates were able to
identify the component of the weight parallel to the slope as the cause of the acceleration. A
significant proportion of candidates concentrated on the changing frictional forces as an
explanation of decreased acceleration. Part (d) proved to be too difficult for less able candidates
who became very confused by the horizontal velocity and tried to use it in calculations for
motion in the vertical plane.
Projectile questions of a similar type have been asked before but the impression given was that
candidates found the analysis more difficult this time round. The diagram did, unfortunately,
include an incorrect label of a vertical height of 1.0 m. There was, however, no evidence that
this caused confusion and candidates were given full credit if they used this value for vertical
displacement.
E33.
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.
E34.
Part (a) was answered very successfully. A common error was to state .length. or ‘height’
rather than distance, although ‘length’ was tolerated.
On part (b) (ii), candidates knew what to do but most did not gain full marks due to confusion
over powers of ten or units for power.
Half of candidates achieved four or five marks out of five on part (c). Again there was confusion
with prefixes/powers of ten and units. It is important that candidates stated ‘Pa’ not ‘pa’.
E35.
Part (a) was answered very successfully. A common error was to state .length. or ‘height’
rather than distance, although ‘length’ was tolerated.
On part (b) (ii), candidates knew what to do but most did not gain full marks due to confusion
over powers of ten or units for power.
Half of candidates achieved four or five marks out of five on part (c). Again there was confusion
with prefixes/powers of ten and units. It is important that candidates stated ‘Pa’ not ‘pa’.