3.4 Mechanics and materials — revision question pack

10 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.4. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.4.1.1 · Scalars and vectors

Explanation

Scalars have magnitude only; vectors have magnitude and direction. Speed, distance and mass are scalars, whereas velocity, displacement, acceleration, force and weight are vectors.
Vectors may be added by a scale drawing or, for two perpendicular vectors, by Pythagoras and trigonometry.
A vector may also be resolved into two perpendicular components; for a force $F$ at angle $\theta$ to the horizontal, these are $F\cos\theta$ horizontally and $F\sin\theta$ vertically.
For two or three coplanar forces acting at a point, equilibrium means zero resultant force, so the object is at rest or moves with constant velocity.
Examiners expect both a correct magnitude and an unambiguous direction.

A vector resolved into perpendicular horizontal and vertical components.

Worked example

A velocity has components $6.0\,\text{m s}^{-1}$ east and $8.0\,\text{m s}^{-1}$ north. Determine its magnitude and direction.

1.Use perpendicular components: $v=\sqrt{6.0^2+8.0^2}=10.0\,\text{m s}^{-1}$ .
2.Measure the direction from east: $\theta=\tan^{-1}(8.0/6.0)=53.1^{\circ}$ .
3.State the complete vector direction as north of east.

Answer: $10.0\,\text{m s}^{-1}$ at $53.1^{\circ}$ north of east.

Common mistakes

Don't use $F\sin\theta$ for the component adjacent to the stated angle.
Don't treat equilibrium as meaning stationary, excluding constant-velocity motion.
Don't give an angle without saying which compass direction or axis it is measured from.

Exam tip

For a calculation, show the two perpendicular components before quoting the resultant magnitude and direction.

Tier 1 · Easy

A boat has velocity components $3.0\,\text{m s}^{-1}$ east and $4.0\,\text{m s}^{-1}$ north. Determine the magnitude and direction of its velocity.
[2 marks]
Total for this question: 2
State which two of distance, displacement, speed and velocity are vector quantities.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A $12.0\,\text{kg}$ crate rests on a smooth plane inclined at $30.0^{\circ}$ to the horizontal. Resolve its weight into components parallel and perpendicular to the plane.
[3 marks]
Total for this question: 3
A walker travels $180\,\text{m}$ east and then $240\,\text{m}$ north in a total time of $300\,\text{s}$ . Determine the magnitude and direction of the resultant displacement and the magnitude of the average velocity.
[3 marks]
Total for this question: 3
A ring is in equilibrium under three coplanar forces. One force is $50.0\,\text{N}$ due north, a second force has magnitude $F$ and acts at $35.0^{\circ}$ south of east, and the third force has magnitude $P$ and acts due west. Determine $F$ and $P$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

Two forces act at a point. One is $180\,\text{N}$ due east. The other is $120\,\text{N}$ at $110^{\circ}$ anticlockwise from east. Determine the magnitude and direction of the single force that would produce equilibrium.
[5 marks]
Total for this question: 5
An aircraft must have a ground velocity of $75.0\,\text{m s}^{-1}$ due north while a wind blows at $20.0\,\text{m s}^{-1}$ due east. Determine the required velocity of the aircraft relative to the air and the distance it travels relative to the air in $12.0\,\text{min}$ .
[5 marks]
Total for this question: 5
A survey vessel travels $3.20\,\text{km}$ at $40.0^{\circ}$ east of north and then $4.50\,\text{km}$ due east. It must finish at a point $2.00\,\text{km}$ due north of its starting point by travelling at constant velocity for a further $15.0\,\text{min}$ . Determine the required velocity for the final part of the journey.
[5 marks]
Total for this question: 5
A force of $260\,\text{N}$ acts at $28.0^{\circ}$ north of east. A second force of unknown magnitude acts due west. Their resultant acts at $75.0^{\circ}$ north of east. Determine the second force, the magnitude of the resultant and the single force that produces equilibrium.
[5 marks]
Total for this question: 5
A load is supported in equilibrium by two cables. The left cable is at $38.0^{\circ}$ above the horizontal and can sustain at most $480\,\text{N}$ . The right cable is at $57.0^{\circ}$ above the horizontal. Determine the greatest mass that can be supported and the tension in the right cable.
[5 marks]
Total for this question: 5

3.4.1.2 · Moments

Explanation

The moment of a force about a point is $M=Fd_{\perp}$ , where $d_{\perp}$ is the perpendicular distance from the point to the force’s line of action.
For rotational equilibrium, the total clockwise moment equals the total anticlockwise moment; full equilibrium also requires zero resultant force.
A couple consists of equal and opposite coplanar forces on different lines of action, producing a moment equal to one force multiplied by the perpendicular separation of the lines.
The weight of a uniform regular solid acts through its centre of mass at its geometric centre.
Examiners expect the chosen pivot, sense of each moment and perpendicular distances to be clear, followed by a separate force balance if an unknown reaction is required.

A downward force on a horizontal beam, showing its perpendicular distance from the pivot.

Worked example

A uniform $4.0\,\text{m}$ beam of weight $200\,\text{N}$ is hinged at one end and supported vertically at the other. Determine the support force.

1.The beam’s weight acts at its centre, $2.0\,\text{m}$ from the hinge.
2.Take moments about the hinge: $R(4.0)=200(2.0)$ .
3.Evaluate the support force: $R=100\,\text{N}$ .

Answer: The vertical support force is $100\,\text{N}$ upward.

Common mistakes

Don't use the distance along a sloping beam instead of the perpendicular distance to the force line.
Don't omit the beam’s own weight when its mass or weight is given.
Don't balance moments and fail to balance forces when a hinge reaction is also required.

Exam tip

Taking moments about a point through an unknown reaction removes that reaction from the moment equation.

Tier 1 · Easy

A $25\,\text{N}$ force acts perpendicular to a spanner $0.40\,\text{m}$ from its pivot. Calculate the moment of the force about the pivot.
[1 mark]
Total for this question: 1
State the principle of moments for an object in equilibrium.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A uniform horizontal beam of length $3.0\,\text{m}$ and weight $180\,\text{N}$ is hinged at its left end and supported vertically at its right end. Determine the vertical support force and the vertical force exerted by the hinge.
[3 marks]
Total for this question: 3
Two equal and opposite parallel forces of $45.0\,\text{N}$ act on a steering wheel. Their lines of action are $0.320\,\text{m}$ apart. Determine the moment of the couple and state the resultant force.
[3 marks]
Total for this question: 3
A light horizontal beam is pivoted at one end. A $300\,\text{N}$ load acts vertically downward $0.800\,\text{m}$ from the pivot. A force $F$ is applied $1.20\,\text{m}$ from the pivot at $50.0^{\circ}$ to the beam. Determine $F$ when the beam is in rotational equilibrium.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A uniform horizontal beam is $5.0\,\text{m}$ long and weighs $240\,\text{N}$ . It is hinged to a wall at one end. A $360\,\text{N}$ load hangs $3.8\,\text{m}$ from the hinge, and a cable attached to the free end makes $35^{\circ}$ above the beam. Determine the cable tension and the magnitude and direction of the force exerted by the hinge on the beam.
[6 marks]
Total for this question: 6
A vehicle of weight $12.0\,\text{kN}$ has a wheelbase of $2.80\,\text{m}$ and its centre of mass is $1.10\,\text{m}$ behind the front axle. A $2.00\,\text{kN}$ load acts directly above the rear axle. Determine the vertical force at each axle and the position of the combined centre of mass behind the front axle.
[5 marks]
Total for this question: 5
A sign extends horizontally from a wall and has weight $240\,\text{N}$ acting $0.420\,\text{m}$ from the wall. It is fixed by an upper and a lower bolt whose centres are $0.180\,\text{m}$ apart vertically. The horizontal components of the bolt forces form a couple. The lower bolt exerts only a horizontal force, while the upper bolt also supplies the whole upward force. Determine the force exerted by each bolt on the sign.
[5 marks]
Total for this question: 5
A non-uniform $3.60\,\text{m}$ beam of weight $280\,\text{N}$ is supported at points $0.400\,\text{m}$ and $3.10\,\text{m}$ from its left end. A $160\,\text{N}$ load acts $2.70\,\text{m}$ from the left end. The support forces are $140\,\text{N}$ and $300\,\text{N}$ respectively. Determine the position of the centre of mass. The load and both supports are then removed; the beam is pivoted $0.400\,\text{m}$ from its left end and held horizontal by a vertical cable attached $3.40\,\text{m}$ from the left end. Determine the cable tension and the vertical force exerted by the pivot.
[6 marks]
Total for this question: 6
A uniform horizontal beam of length $4.20\,\text{m}$ and weight $310\,\text{N}$ is hinged at its left end. An upward force of $180\,\text{N}$ acts $3.40\,\text{m}$ from the hinge, and an anticlockwise couple of moment $96.0\,\text{N m}$ acts on the beam. A $240\,\text{N}$ load is added. Determine where the load must act for equilibrium and find the vertical hinge force.
[5 marks]
Total for this question: 5

3.4.1.3 · Motion along a straight line

Explanation

Velocity is the rate of change of displacement, $v=\Delta s/\Delta t$ , and acceleration is the rate of change of velocity, $a=\Delta v/\Delta t$ . An instantaneous value is obtained from the tangent gradient, whereas an average value uses the change over an interval.
On a displacement–time graph the gradient is velocity; on a velocity–time graph the gradient is acceleration and signed area is displacement; on an acceleration–time graph signed area is change in velocity.
The equations of uniform acceleration apply only when acceleration is constant.
For free fall near Earth, acceleration has magnitude $g=9.81\,\text{m s}^{-2}$ , with its sign set by the chosen positive direction.
Required practical work may determine $g$ from an appropriate graph and evaluate random and systematic errors.

A velocity–time graph whose gradient gives acceleration and whose signed area gives displacement.

Worked example

A car accelerates uniformly from $4.0\,\text{m s}^{-1}$ to $16.0\,\text{m s}^{-1}$ in $6.0\,\text{s}$ . Determine its acceleration and displacement.

1.Calculate the gradient: $a=(16.0-4.0)/6.0=2.0\,\text{m s}^{-2}$ .
2.For uniform acceleration, the mean velocity is $(4.0+16.0)/2=10.0\,\text{m s}^{-1}$ .
3.Use the area under the velocity–time graph: $s=10.0\times6.0=60\,\text{m}$ .

Answer: $a=2.0\,\text{m s}^{-2}$ and $s=60\,\text{m}$ .

Common mistakes

Don't use the gradient of a velocity–time graph as displacement.
Don't treat the area below the time axis as positive when finding displacement.
Don't apply a uniform-acceleration equation across a curved graph section.

Exam tip

For graph questions, state whether a gradient or an area is being used before evaluating it.

Tier 1 · Easy

An object moves at constant velocity $7.5\,\text{m s}^{-1}$ for $4.0\,\text{s}$ . Calculate its displacement.
[1 mark]
Total for this question: 1
State the physical quantity represented by the gradient of a displacement-time graph.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A car accelerates uniformly from $5.0\,\text{m s}^{-1}$ to $17.0\,\text{m s}^{-1}$ in $6.0\,\text{s}$ . Determine its acceleration and displacement during this interval.
[3 marks]
Total for this question: 3
A cyclist travelling at $14.0\,\text{m s}^{-1}$ has a reaction time of $0.700\,\text{s}$ and then brakes uniformly at $3.50\,\text{m s}^{-2}$ . Determine the total stopping distance.
[3 marks]
Total for this question: 3
An object has initial velocity $-3.0\,\text{m s}^{-1}$ . Its acceleration is $+2.0\,\text{m s}^{-2}$ for the next $4.0\,\text{s}$ and then $-1.0\,\text{m s}^{-2}$ for a further $6.0\,\text{s}$ . Determine its final velocity and every time at which it is instantaneously at rest.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A velocity-time graph consists of three straight sections: velocity rises from $0$ to $18\,\text{m s}^{-1}$ in $5.0\,\text{s}$ , remains at $18\,\text{m s}^{-1}$ for $4.0\,\text{s}$ , then falls uniformly to $-6.0\,\text{m s}^{-1}$ in $3.0\,\text{s}$ . Determine the final acceleration, total displacement and total distance travelled.
[5 marks]
Total for this question: 5
A train starts from rest, accelerates uniformly at $2.40\,\text{m s}^{-2}$ until it reaches $18.0\,\text{m s}^{-1}$ , and then continues at constant speed. Determine the time taken to travel $240\,\text{m}$ and its average speed over this distance.
[5 marks]
Total for this question: 5
Car A passes point P at a constant speed of $12.0\,\text{m s}^{-1}$ . Four seconds later, car B starts from rest at P and accelerates at $3.00\,\text{m s}^{-2}$ until reaching $24.0\,\text{m s}^{-1}$ , after which it maintains that speed. Determine when and where B catches A.
[5 marks]
Total for this question: 5
An experiment records the speed $v$ of an object after it has fallen through distance $h$ . A graph of $v^2$ against $h$ is a straight line with gradient $19.42\,\text{m s}^{-2}$ and vertical intercept $1.44\,\text{m}^2\,\text{s}^{-2}$ . Assuming constant acceleration, determine $g$ and explain what the intercept implies about the release.
[4 marks]
Total for this question: 4
A package is released from a balloon rising vertically at $12.5\,\text{m s}^{-1}$ . It reaches the ground $5.80\,\text{s}$ later. Neglect air resistance. Determine the balloon's height at release, the package's greatest height above the ground, its velocity on impact and the total distance travelled by the package.
[6 marks]
Total for this question: 6

3.4.1.4 · Projectile motion

Explanation

In a uniform gravitational field with air resistance neglected, horizontal and vertical motions are independent. Horizontal velocity remains constant, while vertical acceleration is $g$ downward.
Resolve the launch velocity into perpendicular components, apply the uniform-acceleration equations separately, and use the common time to connect the two directions. At maximum height the vertical velocity is zero, but the horizontal velocity is not.
Friction, lift and drag require qualitative treatment: air resistance increases with speed, reduces range and makes a projectile trajectory asymmetric. Terminal speed occurs when drag balances weight, making the resultant force and acceleration zero.
For a vehicle, maximum speed occurs when the driving force is balanced by resistive forces.
Examiners expect the no-air-resistance model to be stated before calculations and drag effects to be described through forces and acceleration.

A projectile trajectory with the launch velocity resolved horizontally and vertically and gravity acting downward.

Worked example

A ball leaves a horizontal platform at $15.0\,\text{m s}^{-1}$ and falls $19.6\,\text{m}$ . Air resistance is negligible. Determine the flight time and horizontal range.

1.Use vertical motion with $u_y=0$ : $19.6=\tfrac12(9.81)t^2$ .
2.Evaluate the common time: $t=\sqrt{2(19.6)/9.81}=2.00\,\text{s}$ .
3.Use constant horizontal velocity: $x=15.0\times2.00=30.0\,\text{m}$ .

Answer: The flight time is $2.00\,\text{s}$ and the range is $30.0\,\text{m}$ .

Common mistakes

Don't use $g$ as a horizontal acceleration when air resistance is neglected.
Don't set the total velocity to zero at maximum height instead of only the vertical component.
Don't use different flight times for the horizontal and vertical calculations.

Exam tip

Resolve the launch velocity first, then write separate horizontal and vertical equations linked by the same time.

Tier 1 · Easy

Air resistance is neglected. State the horizontal and vertical accelerations of a projectile after it has been released.
[2 marks]
Total for this question: 2
A water droplet leaves a horizontal nozzle. Calculate its vertical velocity after $0.800\,\text{s}$ when air resistance is negligible.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A ball leaves a horizontal platform at $12.0\,\text{m s}^{-1}$ . The platform is $20.0\,\text{m}$ above level ground. Air resistance is neglected. Determine the time to reach the ground and the horizontal range.
[3 marks]
Total for this question: 3
A football is kicked at $18.0\,\text{m s}^{-1}$ and $42.0^{\circ}$ above the horizontal. Determine its velocity after $1.00\,\text{s}$ , neglecting air resistance.
[3 marks]
Total for this question: 3
A projectile reaches maximum height $1.20\,\text{s}$ after launch. During this time it travels $21.6\,\text{m}$ horizontally. Determine its launch speed and angle above the horizontal, neglecting air resistance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A projectile is launched from level ground at $22.0\,\text{m s}^{-1}$ and $35.0^{\circ}$ above the horizontal. It lands at the launch height. Air resistance is neglected. Determine its time of flight, horizontal range and maximum height.
[6 marks]
Total for this question: 6
A flare is observed $30.0\,\text{m}$ horizontally from its launch point and $12.0\,\text{m}$ above it after $1.50\,\text{s}$ . Determine the flare's initial velocity and its velocity when observed, neglecting air resistance.
[5 marks]
Total for this question: 5
A projectile is launched at $24.0\,\text{m s}^{-1}$ from the origin and must pass through a point $30.0\,\text{m}$ horizontally away and $5.00\,\text{m}$ higher. Determine both possible launch angles above the horizontal, neglecting air resistance. You may use $1+\tan^2\theta=\sec^2\theta$ .
[5 marks]
Total for this question: 5
A projectile is launched from the edge of a $45.0\,\text{m}$ cliff at $19.0\,\text{m s}^{-1}$ and $28.0^{\circ}$ above the horizontal. It lands on level ground below. Neglect air resistance. Determine the flight time, horizontal distance travelled, and the magnitude and direction of its velocity immediately before impact.
[6 marks]
Total for this question: 6
A projectile launched from the origin passes through $(18.0\,\text{m},11.0\,\text{m})$ and later through $(42.0\,\text{m},7.00\,\text{m})$ . Neglect air resistance. Determine the horizontal and vertical components of its launch velocity, its launch speed and angle, and its maximum height above the origin.
[6 marks]
Total for this question: 6

3.4.1.5 · Newton's laws of motion

Explanation

Newton’s first law states that an object remains at rest or moves with constant velocity unless acted on by a resultant external force.
For constant mass, the second law is $\sum F=ma$ ; a free-body diagram should show only forces acting on the chosen object.
Newton’s third-law pairs are equal and opposite forces of the same interaction acting on different objects, so they do not cancel on one object’s diagram.
Weight is $mg$ , while normal contact force, tension and drag must be obtained from the situation rather than assumed equal to weight.
Examiners expect a declared positive direction, a signed resultant-force equation and a third-law pair identified by naming both interacting objects.

A free-body diagram for a pulled block, showing normal contact force, weight, pull and friction.

Worked example

A $5.0\,\text{kg}$ block is pulled horizontally by $24\,\text{N}$ while friction is $9.0\,\text{N}$ . Determine its acceleration.

1.Choose the pull direction as positive.
2.Find the resultant force: $F=24-9.0=15\,\text{N}$ .
3.Apply $F=ma$ : $a=15/5.0=3.0\,\text{m s}^{-2}$ .

Answer: $3.0\,\text{m s}^{-2}$ in the direction of the pull.

Common mistakes

Don't use the applied force rather than the resultant force in $F=ma$ .
Don't place a third-law partner on the same free-body diagram even though it acts on another object.
Don't conclude that zero resultant force means zero velocity.

Exam tip

A free-body diagram and one signed equation for the chosen direction usually secure the method before any arithmetic.

Tier 1 · Easy

A resultant force of $12\,\text{N}$ acts on a $4.0\,\text{kg}$ object. Calculate its acceleration.
[1 mark]
Total for this question: 1
State Newton's first law of motion.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A lift of total mass $750\,\text{kg}$ accelerates vertically upward at $1.20\,\text{m s}^{-2}$ . Determine the tension in its supporting cable.
[3 marks]
Total for this question: 3
A $1650\,\text{kg}$ car accelerates at $0.850\,\text{m s}^{-2}$ while experiencing a total resistive force of $420\,\text{N}$ . Determine the driving force.
[3 marks]
Total for this question: 3
A $10.0\,\text{kg}$ sled is pulled across horizontal ground by a $50.0\,\text{N}$ force at $30.0^{\circ}$ above the horizontal. A constant frictional force of $20.0\,\text{N}$ opposes the motion. Determine the acceleration of the sled.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Blocks A and B, of masses $4.0\,\text{kg}$ and $6.0\,\text{kg}$ , are joined by a light horizontal cord. A horizontal force of $38\,\text{N}$ pulls B. Friction on A is $5.0\,\text{N}$ and friction on B is $7.0\,\text{N}$ , both opposing motion. Determine the acceleration and cord tension. Identify the Newton's-third-law partner of the cord's pull on A.
[5 marks]
Total for this question: 5
Masses of $3.00\,\text{kg}$ and $5.00\,\text{kg}$ hang on opposite sides of a light inextensible string over a light, smooth pulley. Determine the acceleration, string tension and magnitude of the force exerted by the support on the pulley after release.
[5 marks]
Total for this question: 5
A $12.0\,\text{kg}$ block moves up a plane inclined at $25.0^{\circ}$ to the horizontal. A horizontal force of $85.0\,\text{N}$ acts towards the raised end of the plane, and a frictional force of $18.0\,\text{N}$ acts down the plane. Take $g=9.81\,\text{N kg}^{-1}$ . Determine the block's acceleration and the normal contact force.
[5 marks]
Total for this question: 5
A falling instrument has mass $0.850\,\text{kg}$ . At one instant its downward acceleration is $2.60\,\text{m s}^{-2}$ ; the upthrust is $0.00950\,\text{N}$ and air resistance acts upward. Determine the air resistance at this instant and its value when the instrument later reaches terminal speed.
[4 marks]
Total for this question: 4
For a trolley on a horizontal track, a graph of acceleration $a$ against applied force $F$ has gradient $0.625\,\text{kg}^{-1}$ and vertical intercept $-1.50\,\text{m s}^{-2}$ . Assume the resistive force is constant. Determine the trolley's mass, the resistive force and the applied force needed for $a=3.80\,\text{m s}^{-2}$ . Predict the new gradient and intercept if $0.400\,\text{kg}$ is added; assume the resistive force is unchanged when the mass is added.
[6 marks]
Total for this question: 6

3.4.1.6 · Momentum

Explanation

Linear momentum is the vector $p=mv$ . Total momentum is conserved in an isolated system, so the signed total before a one-dimensional collision or explosion equals the signed total afterwards.
Force is the rate of change of momentum, $F=\Delta(mv)/\Delta t$ , and impulse is the change in momentum. For a constant force, impulse is $F\Delta t$ ; for a varying force, it is the signed area under the force–time graph.
Elastic collisions conserve both momentum and kinetic energy, whereas inelastic collisions conserve momentum but not kinetic energy.
Examiners expect a positive direction, signed velocities and a clear distinction between momentum conservation and kinetic-energy conservation.
Crumple zones and protective packaging increase contact time, reducing mean impact force for the same momentum change; ethical transport design applies this principle to occupant safety.

A force–time pulse whose area represents impulse and hence change in momentum.

Worked example

A $0.20\,\text{kg}$ ball travels at $12\,\text{m s}^{-1}$ east and rebounds at $8.0\,\text{m s}^{-1}$ west. Contact lasts $0.050\,\text{s}$ . Determine the mean force on the ball.

1.Take east as positive, so $u=+12\,\text{m s}^{-1}$ and $v=-8.0\,\text{m s}^{-1}$ .
2.Calculate the momentum change: $\Delta p=0.20(-8.0-12)=-4.0\,\text{kg m s}^{-1}$ .
3.Use $F=\Delta p/\Delta t=-4.0/0.050=-80\,\text{N}$ .

Answer: The mean force is $80\,\text{N}$ west.

Common mistakes

Don't substitute rebound speed as positive after choosing the incident direction as positive.
Don't claim that kinetic energy must be conserved in every collision.
Don't use the peak force instead of the force–time area to find impulse.

Exam tip

Write the positive direction beside the momentum equation before substituting any velocity.

Tier 1 · Easy

A $0.80\,\text{kg}$ trolley moving at $3.0\,\text{m s}^{-1}$ collides with a stationary $0.40\,\text{kg}$ trolley. They join, so the collision is perfectly inelastic. Calculate their common velocity.
[2 marks]
Total for this question: 2
A constant force of $240\,\text{N}$ acts for $15.0\,\text{ms}$ . Calculate the impulse.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $0.40\,\text{kg}$ trolley moving at $5.0\,\text{m s}^{-1}$ collides with a stationary $0.60\,\text{kg}$ trolley. Afterwards their velocities are $-1.0\,\text{m s}^{-1}$ and $4.0\,\text{m s}^{-1}$ respectively. Show that the collision is elastic.
[4 marks]
Total for this question: 4
A stationary $2.40\,\text{kg}$ launcher fires a $0.120\,\text{kg}$ projectile horizontally at $180\,\text{m s}^{-1}$ . Determine the launcher's recoil velocity.
[3 marks]
Total for this question: 3
A $0.0200\,\text{kg}$ bullet travelling at $300\,\text{m s}^{-1}$ embeds in a stationary $0.980\,\text{kg}$ wooden block. Determine their common velocity and the percentage of the bullet's initial kinetic energy that is dissipated.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $1200\,\text{kg}$ car moving east at $18\,\text{m s}^{-1}$ collides head-on with an $800\,\text{kg}$ car moving west at $6.0\,\text{m s}^{-1}$ . They lock together, so the collision is perfectly inelastic. Determine their velocity, the kinetic energy dissipated, and the mean force on the $1200\,\text{kg}$ car if contact lasts $0.18\,\text{s}$ .
[6 marks]
Total for this question: 6
A $0.200\,\text{kg}$ ball approaches a wall at $12.0\,\text{m s}^{-1}$ . The opposing force-time graph during contact is triangular, with duration $18.0\,\text{ms}$ and peak force $400\,\text{N}$ . Determine the ball's rebound velocity and the change in its kinetic energy.
[5 marks]
Total for this question: 5
A $12.0\,\text{kg}$ object moving at $4.00\,\text{m s}^{-1}$ explodes into three fragments. A $3.00\,\text{kg}$ fragment moves at $14.0\,\text{m s}^{-1}$ in the original direction and a $4.00\,\text{kg}$ fragment moves at $2.00\,\text{m s}^{-1}$ in the opposite direction. Determine the velocity of the third fragment and the energy released in the explosion.
[5 marks]
Total for this question: 5
A $0.145\,\text{kg}$ ball strikes a horizontal floor at $8.60\,\text{m s}^{-1}$ and rebounds vertically at $6.20\,\text{m s}^{-1}$ . Contact lasts $9.50\,\text{ms}$ . Determine the mean contact force exerted by the floor and the decrease in the ball's kinetic energy.
[5 marks]
Total for this question: 5
A $1.80\times10^4\,\text{kg}$ rail vehicle moving at $7.20\,\text{m s}^{-1}$ couples with a $1.20\times10^4\,\text{kg}$ vehicle moving towards it at $1.50\,\text{m s}^{-1}$ . Immediately afterwards, braking force decreases linearly from $48.0\,\text{kN}$ to $12.0\,\text{kN}$ over time $t$ , then remains at $12.0\,\text{kN}$ for $6.00\,\text{s}$ until the vehicles stop. Determine their common velocity, the kinetic energy dissipated in coupling and $t$ .
[6 marks]
Total for this question: 6

3.4.1.7 · Work, energy and power

Explanation

Work is energy transferred by a force. For a constant force, $W=Fs\cos\theta$ , where $\theta$ is the angle between force and displacement.
For a variable force, work is the area under the force–displacement graph. Power is the rate of doing work or transferring energy, $P=\Delta W/\Delta t$ ; when a force is parallel to steady velocity, $P=Fv$ .
Efficiency is useful output energy divided by total input energy, or useful output power divided by total input power, and may be quoted as a decimal or percentage.
A motor-lifting experiment compares measured useful output power with electrical input power and evaluates random and systematic errors.
Examiners expect the parallel force component, correct graph area and like quantities in an efficiency ratio.

A variable force–displacement graph with the area under the curve representing work done.

Worked example

A motor lifts a $30\,\text{kg}$ load through $5.0\,\text{m}$ in $4.0\,\text{s}$ while taking $450\,\text{W}$ input power. Determine its useful power and efficiency.

1.Find useful energy: $\Delta E_p=mgh=30(9.81)(5.0)=1471.5\,\text{J}$ .
2.Find useful power: $P=1471.5/4.0=368\,\text{W}$ .
3.Calculate efficiency: $368/450=0.818$ , or $81.8\%$ .

Answer: The useful power is $368\,\text{W}$ and the efficiency is $81.8\%$ .

Common mistakes

Don't use the whole force in $W=Fs$ when the force is not parallel to the displacement.
Don't use one endpoint force times the full displacement for a varying force.
Don't divide an output energy by an input power when calculating efficiency.

Exam tip

For a graph-based work question, identify the required geometric areas and include their units before summing them.

Tier 1 · Easy

A constant force of $75\,\text{N}$ acts in the direction of motion through $12\,\text{m}$ . Calculate the work done.
[1 mark]
Total for this question: 1
State the SI units of work and power.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A motor raises a $25\,\text{kg}$ load vertically through $8.0\,\text{m}$ in $6.0\,\text{s}$ . Its electrical input power is $420\,\text{W}$ . Determine its useful output power and efficiency.
[3 marks]
Total for this question: 3
The power dissipated by a brake rises linearly from zero to $12.0\,\text{kW}$ in $5.00\,\text{s}$ and then remains at $12.0\,\text{kW}$ for $4.00\,\text{s}$ . Determine the energy dissipated.
[3 marks]
Total for this question: 3
A worker pulls a crate through $22.0\,\text{m}$ using a constant force of $140\,\text{N}$ at $28.0^{\circ}$ above the direction of motion. The movement takes $18.0\,\text{s}$ . Determine the work done by the pulling force and its average power.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A force acting on a $2.0\,\text{kg}$ trolley increases linearly from $10\,\text{N}$ to $50\,\text{N}$ over the first $4.0\,\text{m}$ , then remains at $50\,\text{N}$ for another $2.0\,\text{m}$ . The trolley starts from rest, resistive forces are negligible, and the motion lasts $3.0\,\text{s}$ . Determine the work done, final speed and average power.
[5 marks]
Total for this question: 5
A $900\,\text{kg}$ car climbs a $6.00^{\circ}$ slope at a constant $20.0\,\text{m s}^{-1}$ against a $500\,\text{N}$ resistive force. The engine converts $25.0\%$ of its input energy into work done by the driving force. Determine the driving force and the input energy supplied in $120\,\text{s}$ .
[5 marks]
Total for this question: 5
A cyclist and bicycle have a combined mass of $70.0\,\text{kg}$ . Their speed increases from $5.00\,\text{m s}^{-1}$ to $15.0\,\text{m s}^{-1}$ over a distance of $120\,\text{m}$ in $12.0\,\text{s}$ . A constant resistive force of $30.0\,\text{N}$ acts throughout. The cyclist's body is $22.0\%$ efficient at converting input energy into mechanical work. Determine the average input power.
[5 marks]
Total for this question: 5
A motor lifts a $2.40\,\text{kg}$ mass through $1.85\,\text{m}$ in $6.42\,\text{s}$ . Its supply readings are $11.8\,\text{V}$ and $0.920\,\text{A}$ . Determine the useful output power, the electrical energy supplied during the lift and the energy transferred to the surroundings.
[5 marks]
Total for this question: 5
A $1280\,\text{kg}$ car's engine delivers constant useful power of $72.0\,\text{kW}$ while the resistive force is constant at $540\,\text{N}$ . Determine the car's acceleration at $12.0\,\text{m s}^{-1}$ and at $30.0\,\text{m s}^{-1}$ . Explain the change and calculate the theoretical maximum speed under these assumptions.
[6 marks]
Total for this question: 6

3.4.1.8 · Conservation of energy

Explanation

The principle of conservation of energy states that energy cannot be created or destroyed; it is transferred between stores while total energy remains constant in a closed system.
Near Earth, a change in gravitational potential energy is $\Delta E_p=mg\Delta h$ and kinetic energy is $E_k=\tfrac12mv^2$ .
With negligible resistance, a decrease in gravitational potential energy may equal an increase in kinetic energy.
If resistive forces act, some mechanical energy is transferred to internal energy of the object and surroundings, so it is not lost.
Examiners expect a complete energy balance that includes any initial kinetic energy and work done against resistance, together with explicit assumptions such as negligible air resistance.

Energy transferred from the gravitational potential store to kinetic and, when resistance acts, internal stores.

Worked example

A $60\,\text{kg}$ cyclist descends $12\,\text{m}$ from rest. Resistive forces transfer $2.0\,\text{kJ}$ to internal energy. Determine the final speed.

1.Calculate the gravitational energy decrease: $mg\Delta h=60(9.81)(12)=7063\,\text{J}$ .
2.Subtract the resistive transfer: $E_k=7063-2000=5063\,\text{J}$ .
3.Use $\tfrac12mv^2=5063$ , giving $v=\sqrt{2(5063)/60}=13.0\,\text{m s}^{-1}$ .

Answer: The final speed is $13.0\,\text{m s}^{-1}$ .

Common mistakes

Don't describe energy transferred by resistance as destroyed or lost.
Don't use the final height instead of the change in height in $mg\Delta h$ .
Don't equate gravitational and kinetic energy without accounting for stated resistive work.

Exam tip

Write a single before-and-after energy balance and include every stated transfer before rearranging.

Tier 1 · Easy

A stone is released from rest and falls through $1.80\,\text{m}$ . Air resistance is neglected. Use energy conservation to calculate its speed.
[2 marks]
Total for this question: 2
State the principle of conservation of energy.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A roller-coaster car starts from rest $12.0\,\text{m}$ above a reference level. Determine its speed when it is $3.0\,\text{m}$ above that level. Resistive forces are neglected.
[3 marks]
Total for this question: 3
A $0.500\,\text{kg}$ ball is dropped from $2.00\,\text{m}$ and rebounds to $1.20\,\text{m}$ . Air resistance is negligible. Determine the energy transferred to internal energy during the impact.
[3 marks]
Total for this question: 3
A $2.00\,\text{kg}$ object is projected vertically upward at $12.0\,\text{m s}^{-1}$ . A constant resistive force of $5.00\,\text{N}$ acts downward throughout the ascent. Determine the maximum height reached above the launch point.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $70\,\text{kg}$ skier initially travels at $4.0\,\text{m s}^{-1}$ and descends through a vertical height of $25\,\text{m}$ . During the descent, $8.0\,\text{kJ}$ is transferred to internal energy by resistive forces. Determine the skier's final speed.
[5 marks]
Total for this question: 5
A $900\,\text{kg}$ car freewheels with the engine disengaged while descending through $15.0\,\text{m}$ and travelling $180\,\text{m}$ along a road. Its speed increases from $12.0\,\text{m s}^{-1}$ to $18.0\,\text{m s}^{-1}$ . Determine the energy transferred by resistive forces and the mean resistive force.
[5 marks]
Total for this question: 5
A $50.0\,\text{kg}$ sledge starts from rest $12.0\,\text{m}$ vertically above the bottom of a track. It travels $20.0\,\text{m}$ down to the bottom and then climbs a straight slope at $30.0^{\circ}$ to the horizontal. A constant resistive force of $40.0\,\text{N}$ acts throughout. Determine the greatest vertical height reached on the second slope.
[5 marks]
Total for this question: 5
A $0.350\,\text{kg}$ pendulum bob is released from rest $0.480\,\text{m}$ above its lowest point. It passes the lowest point at $2.75\,\text{m s}^{-1}$ . During the subsequent rise, a further $0.420\,\text{J}$ is transferred to internal energy. Determine the energy transferred before the bob first reaches the lowest point and the height reached on the other side.
[5 marks]
Total for this question: 5
A $1.80\,\text{kg}$ object is projected vertically upward at $14.0\,\text{m s}^{-1}$ . A constant resistive force of $3.20\,\text{N}$ acts during the ascent and $2.40\,\text{N}$ during the descent. Determine the maximum height, the speed when the object returns to its launch level and the total energy transferred to internal energy.
[6 marks]
Total for this question: 6

3.4.2.1 · Bulk properties of solids

Explanation

Density is $\rho=m/V$ . Within the limit of proportionality, Hooke’s law gives $F=k\Delta L$ , where $k$ is stiffness.
Elastic strain energy is the area under a force–extension graph and equals $\tfrac12F\Delta L$ only for a straight line through the origin. Tensile stress is $F/A$ and tensile strain is $\Delta L/L$ ; breaking stress is the stress at fracture.
Below the elastic limit a material returns to its original dimensions after unloading, while plastic deformation leaves permanent extension. Brittle materials fracture with little plastic deformation.
Examiners expect force–extension and stress–strain graphs to show the limit of proportionality, elastic limit, energy area and fracture behaviour, and may link spring energy to kinetic or gravitational energy.
Ethical transport design weighs deformation energy and occupant protection against material and environmental costs.

A simple ductile-material stress–strain curve showing linear elastic behaviour, plastic deformation and fracture.

Worked example

A spring of stiffness $250\,\text{N m}^{-1}$ is extended by $80\,\text{mm}$ within its limit of proportionality. Determine the force and elastic strain energy.

1.Convert the extension: $\Delta L=0.080\,\text{m}$ .
2.Apply Hooke’s law: $F=k\Delta L=250(0.080)=20\,\text{N}$ .
3.Use the triangular graph area: $E=\tfrac12F\Delta L=\tfrac12(20)(0.080)=0.80\,\text{J}$ .

Answer: The force is $20\,\text{N}$ and the stored energy is $0.80\,\text{J}$ .

Common mistakes

Don't use millimetres directly in $F=k\Delta L$ when stiffness is in $\text{N m}^{-1}$ .
Don't treat the limit of proportionality and elastic limit as identical definitions.
Don't calculate breaking stress from breaking force without dividing by cross-sectional area.

Exam tip

When a graph is supplied, use its area for energy unless a linear Hooke’s-law section is explicitly established.

Tier 1 · Easy

A solid block has mass $2.16\,\text{kg}$ and volume $8.00\times10^{-4}\,\text{m}^3$ . Calculate its density.
[1 mark]
Total for this question: 1
State how elastic deformation differs from plastic deformation after a force is removed.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A spring of stiffness $320\,\text{N m}^{-1}$ is compressed by $75\,\text{mm}$ . It launches a $0.150\,\text{kg}$ cart horizontally. Determine the elastic strain energy and the cart's speed if all this energy becomes kinetic energy.
[3 marks]
Total for this question: 3
An aluminium block measures $0.100\,\text{m}\times0.0800\,\text{m}\times0.0500\,\text{m}$ and has mass $0.918\,\text{kg}$ . Aluminium has density $2700\,\text{kg m}^{-3}$ . Determine the volume of a cavity inside the block.
[3 marks]
Total for this question: 3
A material's stress-strain graph is straight from the origin to point P and then curves. When the material is unloaded from point Q, beyond P, it returns to zero strain. Any increase in stress beyond Q causes a permanent strain after unloading. Identify P and Q, and state the graph evidence for P.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A metal wire's force-extension graph is straight from the origin to its elastic limit at $720\,\text{N}$ and $0.015\,\text{m}$ . The force then rises linearly to $900\,\text{N}$ at fracture, where the extension is $0.027\,\text{m}$ . Determine the work done in stretching the wire to fracture and explain why unloading after the elastic limit would not return all this energy.
[5 marks]
Total for this question: 5
A brittle wire of diameter $1.20\,\text{mm}$ fractures at $800\,\text{N}$ . Calculate its breaking stress. Describe two features expected on its stress-strain graph.
[5 marks]
Total for this question: 5
A compressed spring launches a $0.200\,\text{kg}$ block vertically and then loses contact with it. The spring's force increases linearly from zero to $40.0\,\text{N}$ over the first $0.100\,\text{m}$ of compression, then linearly to $60.0\,\text{N}$ at $0.150\,\text{m}$ . All the work done in compressing the spring is stored elastically. The spring has negligible mass and returns all its stored energy to the block. Determine the block's speed after it has risen $1.50\,\text{m}$ from its compressed starting position. Resistive forces are negligible.
[5 marks]
Total for this question: 5
A wire is loaded along a force–extension path from $(0,0)$ to $(12.0\,\text{mm},300\,\text{N})$ and then to $(26.0\,\text{mm},420\,\text{N})$ , with straight sections between the points. It is unloaded along a straight line to zero force at $4.00\,\text{mm}$ extension. Determine the work done during loading, the energy recovered and the energy transferred to internal energy.
[6 marks]
Total for this question: 6
Material A has a stress–strain graph that rises linearly to $280\,\text{MPa}$ at strain $0.0040$ , then remains at $250\,\text{MPa}$ until fracture at strain $0.240$ . Material B rises linearly to fracture at $620\,\text{MPa}$ and strain $0.0060$ . Estimate the energy absorbed per unit volume by each material and evaluate which is more suitable for the deforming part of a crumple zone.
[6 marks]
Total for this question: 6

3.4.2.2 · The Young modulus

Explanation

Tensile stress is $\sigma=F/A$ in pascals and tensile strain is $\varepsilon=\Delta L/L$ , which is dimensionless. Within the linear elastic region, Young modulus is $E=\sigma/\varepsilon=FL/(A\Delta L)$ and is the gradient of a stress–strain graph.
A simple determination uses a long wire of measured original length and diameter, adds known loads below the limit of proportionality and measures each extension.
The diameter should be measured with a micrometer at several positions and perpendicular orientations because area depends on diameter squared.
Examiners expect SI conversions, a valid graph-based method, and practical controls such as a reference wire or correction for support movement.
Required practical 4 assesses this measurement and its uncertainties.

A simple Young-modulus arrangement using adjacent reference and test wires suspended from the same support.

Worked example

A wire has length $2.0\,\text{m}$ , area $4.0\times10^{-7}\,\text{m}^2$ and extends $1.5\,\text{mm}$ under a $30\,\text{N}$ load. Determine its Young modulus.

1.Convert extension: $\Delta L=1.5\times10^{-3}\,\text{m}$ .
2.Substitute into $E=FL/(A\Delta L)$ .
3. $E=30(2.0)/[(4.0\times10^{-7})(1.5\times10^{-3})]=1.0\times10^{11}\,\text{Pa}$ .

Answer: $E=1.0\times10^{11}\,\text{Pa}$ , or $100\,\text{GPa}$ .

Common mistakes

Don't use extension instead of strain, omitting division by the original length.
Don't convert $\text{mm}$ to metres but leave $\text{mm}^2$ unconverted for area.
Don't take a graph gradient beyond the linear elastic region.

Exam tip

For a practical-method question, state how length, extension and diameter are measured and keep all loads below the limit of proportionality.

Tier 1 · Easy

A tensile force of $120\,\text{N}$ acts on a wire of cross-sectional area $2.0\,\text{mm}^2$ . Calculate the tensile stress in pascals.
[2 marks]
Total for this question: 2
State the unit of the Young modulus and explain why tensile strain has no unit.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A wire is $2.00\,\text{m}$ long and $0.800\,\text{mm}$ in diameter. A tensile force of $55.3\,\text{N}$ produces an extension of $2.00\,\text{mm}$ . Determine the tensile stress, tensile strain and Young modulus.
[4 marks]
Total for this question: 4
A rod has length $1.80\,\text{m}$ , cross-sectional area $3.00\,\text{mm}^2$ and Young modulus $2.00\times10^{11}\,\text{Pa}$ . Determine its extension under a $500\,\text{N}$ tensile force and the resulting tensile strain.
[3 marks]
Total for this question: 3
Wire A and wire B are made from the same material. Wire A is $1.20\,\text{m}$ long, has diameter $0.500\,\text{mm}$ and extends by $1.80\,\text{mm}$ under a $60.0\,\text{N}$ force. Wire B is $0.800\,\text{m}$ long, has diameter $0.800\,\text{mm}$ and is loaded by $100\,\text{N}$ . Determine the extension of wire B.
[4 marks]
Total for this question: 4

Tier 3 · Hard

In a Young-modulus experiment, a wire has original length $2.50\,\text{m}$ and mean diameter $0.420\,\text{mm}$ . Below the limit of proportionality, a force-extension graph has gradient $6.20\times10^3\,\text{N m}^{-1}$ . Starting from the definitions of stress and strain, determine the Young modulus. Explain why the diameter should be measured at several positions and in two perpendicular directions.
[6 marks]
Total for this question: 6
A steel wire and a brass wire are joined in series and carry a $120\,\text{N}$ tensile force. The steel wire has length $1.20\,\text{m}$ , area $0.800\,\text{mm}^2$ and Young modulus $2.00\times10^{11}\,\text{Pa}$ . The brass wire has length $0.800\,\text{m}$ , area $1.20\,\text{mm}^2$ and Young modulus $1.00\times10^{11}\,\text{Pa}$ . Determine their total extension and the effective stiffness of the joined wires.
[5 marks]
Total for this question: 5
In a Young-modulus experiment, a $2.00\,\text{m}$ test wire of diameter $0.700\,\text{mm}$ is beside a reference wire carrying an unchanged load. When an additional $80.0\,\text{N}$ is applied to the test wire, its indicator reading increases by $2.46\,\text{mm}$ and the reference-wire reading increases by $0.320\,\text{mm}$ because the support moves. Determine the corrected Young modulus and the percentage by which it would be underestimated if the reference reading were ignored.
[5 marks]
Total for this question: 5
A steel wire and a copper wire of equal length $1.50\,\text{m}$ are joined in parallel and support a total tensile force of $80.0\,\text{N}$ . The steel wire has diameter $0.600\,\text{mm}$ and Young modulus $2.00\times10^{11}\,\text{Pa}$ . The copper wire has diameter $0.900\,\text{mm}$ and Young modulus $1.20\times10^{11}\,\text{Pa}$ . Determine the common extension and the force carried by each wire.
[6 marks]
Total for this question: 6
A wire has length $(1.800\pm0.002)\,\text{m}$ and diameter $(0.460\pm0.006)\,\text{mm}$ . The gradient of its force–extension graph is $(4.85\pm0.08)\,\text{kN m}^{-1}$ . Determine the Young modulus and estimate its maximum uncertainty. Identify the measurement that contributes most to the percentage uncertainty.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.4.1.1 · Scalars and vectors

Tier 1 · Easy

Mark scheme for 3.4.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$5.0\,\text{m s}^{-1}$ at $53^{\circ}$ north of east	The magnitude is $v=\sqrt{3.0^2+4.0^2}=5.0\,\text{m s}^{-1}$ . The direction from east is $\theta=\tan^{-1}(4.0/3.0)=53.1^{\circ}$ , so it is $53^{\circ}$ north of east.	2
02.1	Displacement and velocity	A vector has magnitude and direction. Displacement and velocity include direction, whereas distance and speed are scalars.	1

Tier 2 · Standard

Mark scheme for 3.4.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$58.9\,\text{N}$ down the plane and $102\,\text{N}$ into the plane	The weight is $W=mg=12.0\times9.81=117.72\,\text{N}$ . The parallel component is $W\sin30.0^{\circ}=58.9\,\text{N}$ down the plane. The perpendicular component is $W\cos30.0^{\circ}=102\,\text{N}$ into the plane.	3
02.1	$300\,\text{m}$ at $53.1^{\circ}$ north of east; average velocity $1.00\,\text{m s}^{-1}$	The two displacements are perpendicular, so $R=\sqrt{180^2+240^2}=300\,\text{m}$ . Its direction is $\tan^{-1}(240/180)=53.1^{\circ}$ north of east. The magnitude of the average velocity is resultant displacement divided by total time: $300/300=1.00\,\text{m s}^{-1}$ .	3
03.1	$F=87.2\,\text{N}$ and $P=71.4\,\text{N}$	Vertical equilibrium requires $F\sin35.0^{\circ}=50.0$ , so $F=50.0/\sin35.0^{\circ}=87.2\,\text{N}$ . Horizontal equilibrium then requires the westward force to balance the eastward component: $P=F\cos35.0^{\circ}=71.4\,\text{N}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$179\,\text{N}$ at $39.0^{\circ}$ south of west	Resolve the second force: $F_x=120\cos110^{\circ}=-41.0\,\text{N}$ and $F_y=120\sin110^{\circ}=112.8\,\text{N}$ . The resultant of the given forces is $(180-41.0,112.8)=(139.0,112.8)\,\text{N}$ . Its magnitude is $\sqrt{139.0^2+112.8^2}=179\,\text{N}$ and its angle is $\tan^{-1}(112.8/139.0)=39.0^{\circ}$ north of east. The single force that produces equilibrium is equal and opposite, so it is $179\,\text{N}$ at $39.0^{\circ}$ south of west.	5
02.1	$77.6\,\text{m s}^{-1}$ at $14.9^{\circ}$ west of north; $55.9\,\text{km}$	Ground velocity equals velocity relative to the air plus wind velocity. Therefore the required air-relative components are $20.0\,\text{m s}^{-1}$ west and $75.0\,\text{m s}^{-1}$ north. Its magnitude is $\sqrt{75.0^2+20.0^2}=77.6\,\text{m s}^{-1}$ and its direction is $\tan^{-1}(20.0/75.0)=14.9^{\circ}$ west of north. In $12.0\,\text{min}=720\,\text{s}$ , the air-relative distance is $77.6209(720)=5.59\times10^4\,\text{m}=55.9\,\text{km}$ .	5
03.1	$7.30\,\text{m s}^{-1}$ at $3.94^{\circ}$ south of west	After the first two parts, the east displacement is $3.20\sin40.0^{\circ}+4.50=6.5569\,\text{km}$ and the north displacement is $3.20\cos40.0^{\circ}=2.4513\,\text{km}$ . The final displacement must therefore be $6.5569\,\text{km}$ west and $0.4513\,\text{km}$ south. Its magnitude is $\sqrt{6.5569^2+0.4513^2}=6.5724\,\text{km}$ and its direction is $\tan^{-1}(0.4513/6.5569)=3.94^{\circ}$ south of west. Over $900\,\text{s}$ , the required speed is $6572.4/900=7.30\,\text{m s}^{-1}$ .	5
04.1	$197\,\text{N}$ west; resultant $126\,\text{N}$ at $75.0^{\circ}$ north of east; the single force that produces equilibrium is $126\,\text{N}$ in the opposite direction, at $15.0^{\circ}$ west of south	The north component is fixed at $R_y=260\sin28.0^{\circ}=122.063\,\text{N}$ . Since $R_y/R_x=\tan75.0^{\circ}$ , $R_x=122.063/\tan75.0^{\circ}=32.7066\,\text{N}$ . The first force has east component $260\cos28.0^{\circ}=229.566\,\text{N}$ , so the westward force is $229.566-32.7066=196.860\,\text{N}=197\,\text{N}$ . The resultant is $R=122.063/\sin75.0^{\circ}=126.369\,\text{N}=126\,\text{N}$ . The single force that produces equilibrium has this magnitude and acts oppositely, at $15.0^{\circ}$ west of south.	5
05.1	$89.5\,\text{kg}$ and $694\,\text{N}$	At the limiting load the left tension is $T_L=480\,\text{N}$ . Horizontal equilibrium gives $T_L\cos38.0^{\circ}=T_R\cos57.0^{\circ}$ , so $T_R=480\cos38.0^{\circ}/\cos57.0^{\circ}=694.488\,\text{N}$ . Vertical equilibrium gives $mg=480\sin38.0^{\circ}+694.488\sin57.0^{\circ}=877.964\,\text{N}$ . Thus $m=877.964/9.81=89.4968\,\text{kg}=89.5\,\text{kg}$ .	5

3.4.1.2 · Moments

Tier 1 · Easy

Mark scheme for 3.4.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$10\,\text{N m}$	The perpendicular distance is $0.40\,\text{m}$ , so the moment is $Fd=25\times0.40=10\,\text{N m}$ .	1
02.1	The total clockwise moment about any point equals the total anticlockwise moment about that point.	Equilibrium requires zero resultant moment, so clockwise and anticlockwise turning effects balance about the chosen point.	2

Tier 2 · Standard

Mark scheme for 3.4.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Support force $90\,\text{N}$ upward; hinge force $90\,\text{N}$ upward	The beam's weight acts $1.5\,\text{m}$ from the hinge. Taking moments about the hinge gives $R(3.0)=180(1.5)$ , so $R=90\,\text{N}$ . Vertical equilibrium then gives $H+90-180=0$ , so the hinge force is also $H=90\,\text{N}$ upward.	3
02.1	Moment $14.4\,\text{N m}$ ; resultant force zero	The moment of a couple is one force multiplied by the perpendicular separation of the lines of action: $M=Fd=45.0(0.320)=14.4\,\text{N m}$ . The two forces are equal and opposite, so their vector sum and hence the resultant force are zero.	3
03.1	$261\,\text{N}$	Only the component $F\sin50.0^{\circ}$ perpendicular to the beam produces a moment. Equating moments about the pivot gives $(F\sin50.0^{\circ})(1.20)=300(0.800)$ . Hence $F=300(0.800)/[1.20\sin50.0^{\circ}]=261\,\text{N}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Cable tension $686\,\text{N}$ ; hinge force $599\,\text{N}$ at $20.2^{\circ}$ above the horizontal, away from the wall	Only the cable's vertical component produces a moment about the hinge. Thus $(T\sin35^{\circ})(5.0)=240(2.5)+360(3.8)=1968\,\text{N m}$ . Hence $T\sin35^{\circ}=393.6\,\text{N}$ and $T=686\,\text{N}$ . The cable pulls horizontally towards the wall, so equilibrium requires the hinge force $H_x=T\cos35^{\circ}=562\,\text{N}$ away from the wall. Vertically, $H_y+393.6-240-360=0$ , giving $H_y=206.4\,\text{N}$ upward. Therefore $H=\sqrt{562^2+206.4^2}=599\,\text{N}$ and $\theta=\tan^{-1}(206.4/562)=20.2^{\circ}$ above the horizontal.	6
02.1	Front axle $7.29\,\text{kN}$ ; rear axle $6.71\,\text{kN}$ ; combined centre of mass $1.34\,\text{m}$ behind the front axle	Let the rear reaction be $R$ . Taking moments about the front axle gives $R(2.80)=12.0(1.10)+2.00(2.80)=18.8\,\text{kN m}$ , so $R=6.71\,\text{kN}$ . Vertical equilibrium gives the front reaction $=12.0+2.00-6.71=7.29\,\text{kN}$ . If the combined centre of mass is $x$ behind the front axle, then $(12.0+2.00)x=18.8$ , giving $x=1.34\,\text{m}$ .	5
03.1	Lower bolt: $560\,\text{N}$ horizontally away from the wall; upper bolt: $609\,\text{N}$ at $23.2^{\circ}$ above the horizontal towards the wall	The weight produces a moment $240(0.420)=100.8\,\text{N m}$ . If each horizontal bolt-force component is $C$ , the balancing couple has moment $C(0.180)$ , so $C=100.8/0.180=560\,\text{N}$ . The lower bolt therefore pushes horizontally away from the wall and the upper bolt has an equal horizontal component towards the wall. Vertical equilibrium requires an upward component of $240\,\text{N}$ at the upper bolt. Its resultant is $\sqrt{560^2+240^2}=609\,\text{N}$ at $\tan^{-1}(240/560)=23.2^{\circ}$ above the horizontal.	5
04.1	Centre of mass $1.98\,\text{m}$ from the left end; cable tension $147\,\text{N}$ ; pivot force $133\,\text{N}$ upward	Take moments about the original left support. If the centre of mass is at $x$ , $300(3.10-0.400)=280(x-0.400)+160(2.70-0.400)$ . Hence $810=280(x-0.400)+368$ , so $x=1.97857\,\text{m}=1.98\,\text{m}$ . With the load and supports removed, take moments about the pivot: $T(3.40-0.400)=280(1.97857-0.400)$ , so $T=147.333\,\text{N}=147\,\text{N}$ . Vertical equilibrium gives the pivot force $R=280-147.333=132.667\,\text{N}=133\,\text{N}$ upward.	6
05.1	The load acts $0.238\,\text{m}$ from the hinge; the vertical hinge force is $370\,\text{N}$ upward	Take anticlockwise moments as positive about the hinge. The couple contributes its full moment independently of position: $180(3.40)+96.0-310(2.10)-240x=0$ . Thus $x=[612+96.0-651]/240=0.2375\,\text{m}=0.238\,\text{m}$ . Vertical equilibrium gives $R+180-310-240=0$ , so $R=370\,\text{N}$ upward.	5

3.4.1.3 · Motion along a straight line

Tier 1 · Easy

Mark scheme for 3.4.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$30\,\text{m}$	The displacement is the area under the velocity-time graph: $s=vt=7.5\times4.0=30\,\text{m}$ .	1
02.1	Velocity	The gradient is change in displacement divided by change in time, which is velocity. Its sign gives the direction of motion.	1

Tier 2 · Standard

Mark scheme for 3.4.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.0\,\text{m s}^{-2}$ and $66\,\text{m}$	The acceleration is $a=(v-u)/t=(17.0-5.0)/6.0=2.0\,\text{m s}^{-2}$ . With uniform acceleration, the mean velocity is $(u+v)/2=11.0\,\text{m s}^{-1}$ , so $s=11.0\times6.0=66\,\text{m}$ .	3
02.1	$37.8\,\text{m}$	The reaction distance is $ut=14.0(0.700)=9.80\,\text{m}$ . During braking, $v^2=u^2+2as$ gives $0=14.0^2-2(3.50)s$ , so $s=28.0\,\text{m}$ . Total stopping distance is $9.80+28.0=37.8\,\text{m}$ .	3
03.1	$-1.0\,\text{m s}^{-1}$ ; at $1.5\,\text{s}$ and $9.0\,\text{s}$ after the start	During the first stage, $v=-3.0+2.0t$ , so $v=0$ at $t=1.5\,\text{s}$ and $v=+5.0\,\text{m s}^{-1}$ at $t=4.0\,\text{s}$ . During the second stage, $v=5.0-1.0(t-4.0)$ , which is zero at $t=9.0\,\text{s}$ and equals $-1.0\,\text{m s}^{-1}$ at $t=10.0\,\text{s}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$-8.0\,\text{m s}^{-2}$ ; displacement $135\,\text{m}$ ; distance $140\,\text{m}$	For the final section, $a=(-6-18)/3.0=-8.0\,\text{m s}^{-2}$ . Signed area gives displacement: first triangle $=\tfrac12(5.0)(18)=45\,\text{m}$ , rectangle $=(4.0)(18)=72\,\text{m}$ , and final trapezium $=\tfrac12(18-6)(3.0)=18\,\text{m}$ . Total displacement is $45+72+18=135\,\text{m}$ . In the last section the velocity reaches zero after $18/8=2.25\,\text{s}$ . Its distance is $\tfrac12(2.25)(18)+\tfrac12(0.75)(6)=22.5\,\text{m}$ . Total distance is $45+72+22.5=139.5\,\text{m}$ , or $140\,\text{m}$ to two significant figures.	5
02.1	$17.1\,\text{s}$ and $14.0\,\text{m s}^{-1}$	The acceleration time is $t_1=v/a=18.0/2.40=7.50\,\text{s}$ . The distance covered while accelerating is $s_1=(u+v)t_1/2=(0+18.0)(7.50)/2=67.5\,\text{m}$ . The remaining $172.5\,\text{m}$ takes $t_2=172.5/18.0=9.583\,\text{s}$ . Thus the total time is $17.083\,\text{s}=17.1\,\text{s}$ . The average speed is $240/17.083=14.0\,\text{m s}^{-1}$ to three significant figures.	5
03.1	$16.0\,\text{s}$ after A passes P, at $192\,\text{m}$ from P	B accelerates for $24.0/3.00=8.00\,\text{s}$ , so it reaches maximum speed $12.0\,\text{s}$ after A passes P. By then B has travelled $\tfrac12(3.00)(8.00)^2=96.0\,\text{m}$ , while A has travelled $12.0(12.0)=144\,\text{m}$ . The gap is $48.0\,\text{m}$ and the relative speed is $24.0-12.0=12.0\,\text{m s}^{-1}$ , so B needs a further $4.00\,\text{s}$ . The catch occurs at $t=16.0\,\text{s}$ and $x=12.0(16.0)=192\,\text{m}$ .	5
04.1	$g=9.71\,\text{m s}^{-2}$ ; the object had initial speed $1.20\,\text{m s}^{-1}$ rather than being released from rest	For constant acceleration, $v^2=u^2+2gh$ . Therefore the graph gradient is $2g$ , giving $g=19.42/2=9.71\,\text{m s}^{-2}$ . The intercept is $u^2=1.44\,\text{m}^2\,\text{s}^{-2}$ , so $u=\sqrt{1.44}=1.20\,\text{m s}^{-1}$ . A non-zero intercept therefore indicates that the object was already moving at the stated zero of distance.	4
05.1	$92.5\,\text{m}$ ; $100\,\text{m}$ ; $44.4\,\text{m s}^{-1}$ downward; $108\,\text{m}$	Take upward as positive. From ground displacement $-h=ut-\tfrac12gt^2$ , $h=\tfrac12(9.81)(5.80)^2-12.5(5.80)=92.5042\,\text{m}$ . The further rise is $u^2/(2g)=12.5^2/[2(9.81)]=7.96381\,\text{m}$ , so the greatest height is $100.468\,\text{m}$ . At impact, $v=u-gt=12.5-9.81(5.80)=-44.398\,\text{m s}^{-1}$ , or $44.4\,\text{m s}^{-1}$ downward. Total distance is $7.96381+100.468=108.432\,\text{m}=108\,\text{m}$ .	6

3.4.1.4 · Projectile motion

Tier 1 · Easy

Mark scheme for 3.4.1.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Horizontal acceleration $0$ ; vertical acceleration $9.81\,\text{m s}^{-2}$ downward.	With no air resistance, gravity is the only force. It has no horizontal component, so horizontal acceleration is zero, and it produces vertical acceleration $g=9.81\,\text{m s}^{-2}$ downward.	2
02.1	$7.85\,\text{m s}^{-1}$ downward	Its initial vertical velocity is zero. Using $v_y=u_y+gt$ , $v_y=0+9.81(0.800)=7.848\,\text{m s}^{-1}$ , which is $7.85\,\text{m s}^{-1}$ downward to three significant figures.	2

Tier 2 · Standard

Mark scheme for 3.4.1.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.02\,\text{s}$ and $24.2\,\text{m}$	Vertically, $u_y=0$ and $s=20.0\,\text{m}$ downward, so $s=\tfrac12gt^2$ gives $t=\sqrt{2(20.0)/9.81}=2.02\,\text{s}$ . Horizontal speed is constant, so the range is $x=u_xt=12.0\times2.02=24.2\,\text{m}$ .	3
02.1	$13.6\,\text{m s}^{-1}$ at $9.48^{\circ}$ above the horizontal	The horizontal component remains $v_x=18.0\cos42.0^{\circ}=13.38\,\text{m s}^{-1}$ . Vertically, $v_y=18.0\sin42.0^{\circ}-9.81(1.00)=2.23\,\text{m s}^{-1}$ . The speed is $\sqrt{13.38^2+2.23^2}=13.6\,\text{m s}^{-1}$ and the direction is $\tan^{-1}(2.23/13.38)=9.48^{\circ}$ above horizontal.	3
03.1	$21.5\,\text{m s}^{-1}$ at $33.2^{\circ}$ above the horizontal	Horizontal speed is constant, so $u_x=21.6/1.20=18.0\,\text{m s}^{-1}$ . At maximum height $v_y=0$ , giving $u_y=gt=9.81(1.20)=11.772\,\text{m s}^{-1}$ . Therefore $u=\sqrt{18.0^2+11.772^2}=21.5\,\text{m s}^{-1}$ and $\theta=\tan^{-1}(11.772/18.0)=33.2^{\circ}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.57\,\text{s}$ ; $46.4\,\text{m}$ ; $8.12\,\text{m}$	Resolve the initial velocity: $u_x=22.0\cos35.0^{\circ}=18.02\,\text{m s}^{-1}$ and $u_y=22.0\sin35.0^{\circ}=12.62\,\text{m s}^{-1}$ . Returning to the launch height gives $t=2u_y/g=2(12.62)/9.81=2.573\,\text{s}$ . The range is $u_xt=18.02\times2.573=46.4\,\text{m}$ . At maximum height $v_y=0$ , so $0=u_y^2-2gh$ and $h=u_y^2/(2g)=12.62^2/[2(9.81)]=8.12\,\text{m}$ .	6
02.1	Initially $25.2\,\text{m s}^{-1}$ at $37.5^{\circ}$ above the horizontal; when observed $20.0\,\text{m s}^{-1}$ at $1.84^{\circ}$ above the horizontal	Horizontally, $u_x=x/t=30.0/1.50=20.0\,\text{m s}^{-1}$ . Vertically, $y=u_yt-\tfrac12gt^2$ , so $u_y=[12.0+\tfrac12(9.81)(1.50)^2]/1.50=15.3575\,\text{m s}^{-1}$ . The initial speed is $\sqrt{20.0^2+15.3575^2}=25.2\,\text{m s}^{-1}$ and the launch direction is $37.5^{\circ}$ above horizontal. When observed, $v_x=20.0\,\text{m s}^{-1}$ and $v_y=15.3575-9.81(1.50)=0.6425\,\text{m s}^{-1}$ . Hence its velocity is $20.0\,\text{m s}^{-1}$ at $\tan^{-1}(0.6425/20.0)=1.84^{\circ}$ above the horizontal.	5
03.1	$25.7^{\circ}$ or $73.8^{\circ}$	Using $x=u\cos\theta\,t$ gives $t=x/(u\cos\theta)$ . Substitute this into $y=u\sin\theta\,t-\tfrac12gt^2$ and let $q=\tan\theta$ : $y=xq-[gx^2/(2u^2)](1+q^2)$ . Here $gx^2/(2u^2)=7.664$ , so $5.00=30.0q-7.664(1+q^2)$ , or $7.664q^2-30.0q+12.664=0$ . Solving gives $q=0.4813$ or $3.433$ , hence $\theta=25.7^{\circ}$ or $73.8^{\circ}$ .	5
04.1	$4.07\,\text{s}$ ; $68.3\,\text{m}$ ; $35.3\,\text{m s}^{-1}$ at $61.6^{\circ}$ below the horizontal	The launch components are $u_x=19.0\cos28.0^{\circ}=16.7760\,\text{m s}^{-1}$ and $u_y=19.0\sin28.0^{\circ}=8.91996\,\text{m s}^{-1}$ . Vertically, $-45.0=8.91996t-4.905t^2$ , whose positive root is $t=4.07172\,\text{s}$ . Thus $x=u_xt=16.7760(4.07172)=68.3072\,\text{m}$ . The final vertical velocity is $v_y=8.91996-9.81(4.07172)=-31.0236\,\text{m s}^{-1}$ . Therefore $v=\sqrt{16.7760^2+31.0236^2}=35.2690\,\text{m s}^{-1}$ and the direction is $\tan^{-1}(31.0236/16.7760)=61.5977^{\circ}$ below horizontal.	6
05.1	$u_x=16.3\,\text{m s}^{-1}$ ; $u_y=15.4\,\text{m s}^{-1}$ ; $22.4\,\text{m s}^{-1}$ at $43.4^{\circ}$ ; maximum height $12.0\,\text{m}$	Eliminating time from $x=u_xt$ and $y=u_yt-\tfrac12gt^2$ gives $y=Ax-Bx^2$ , where $A=u_y/u_x$ and $B=g/(2u_x^2)$ . The two points give $11.0=18.0A-324B$ and $7.00=42.0A-1764B$ . Solving gives $B=0.0185185\,\text{m}^{-1}$ and $A=0.944444$ . Hence $u_x=\sqrt{9.81/(2B)}=16.2748\,\text{m s}^{-1}$ and $u_y=Au_x=15.3707\,\text{m s}^{-1}$ . The speed is $22.3859\,\text{m s}^{-1}$ and $\theta=\tan^{-1}A=43.3634^{\circ}$ . The maximum height is $u_y^2/(2g)=12.0417\,\text{m}$ .	6

3.4.1.5 · Newton's laws of motion

Tier 1 · Easy

Mark scheme for 3.4.1.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.0\,\text{m s}^{-2}$	Newton's second law gives $a=F/m=12/4.0=3.0\,\text{m s}^{-2}$ .	1
02.1	An object remains at rest or moves with constant velocity unless acted on by a resultant external force.	Zero resultant external force means zero acceleration, so velocity does not change. Rest is the special case of constant zero velocity.	2

Tier 2 · Standard

Mark scheme for 3.4.1.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$8.26\,\text{kN}$	Taking upward as positive, $T-mg=ma$ . Therefore $T=m(g+a)=750(9.81+1.20)=8257.5\,\text{N}=8.26\,\text{kN}$ .	3
02.1	$1.82\,\text{kN}$	The resultant force is $ma=1650(0.850)=1402.5\,\text{N}$ . Since driving force minus resistance equals the resultant, $F_{\rm drive}=1402.5+420=1822.5\,\text{N}=1.82\,\text{kN}$ to three significant figures.	3
03.1	$2.33\,\text{m s}^{-2}$	The horizontal component of the pull is $50.0\cos30.0^{\circ}=43.3\,\text{N}$ . The resultant horizontal force is $43.3-20.0=23.3\,\text{N}$ . Newton's second law gives $a=F/m=23.3/10.0=2.33\,\text{m s}^{-2}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.6\,\text{m s}^{-2}$ ; $15.4\,\text{N}$ ; A exerts an equal and opposite pull on the cord.	For the two-block system, the external resultant is $38-5.0-7.0=26\,\text{N}$ and the mass is $10.0\,\text{kg}$ , so $a=26/10.0=2.6\,\text{m s}^{-2}$ . For A alone, $T-5.0=4.0(2.6)$ , hence $T=15.4\,\text{N}$ . The third-law partner of the cord's force on A is the force of A on the cord; it is not another force acting on A.	5
02.1	$2.45\,\text{m s}^{-2}$ ; $36.8\,\text{N}$ ; support force $73.6\,\text{N}$	For the two-mass system, the driving force is $(5.00-3.00)g$ and the total mass is $8.00\,\text{kg}$ . Thus $a=2.00(9.81)/8.00=2.4525\,\text{m s}^{-2}=2.45\,\text{m s}^{-2}$ . For the rising $3.00\,\text{kg}$ mass, $T-3.00g=3.00a$ , so $T=3.00(9.81+2.4525)=36.7875\,\text{N}=36.8\,\text{N}$ . The two vertical string sections each pull down on the light pulley with tension $T$ , so its support exerts an equal upward force $2T=73.6\,\text{N}$ .	5
03.1	$0.774\,\text{m s}^{-2}$ up the plane and $143\,\text{N}$	The horizontal force has components $85.0\cos25.0^{\circ}$ up the plane and $85.0\sin25.0^{\circ}$ into it. Along the plane, the resultant is $85.0\cos25.0^{\circ}-12.0(9.81)\sin25.0^{\circ}-18.0=9.286\,\text{N}$ . Hence $a=9.286/12.0=0.774\,\text{m s}^{-2}$ up the plane. Perpendicular to the plane there is no acceleration, so $R=12.0(9.81)\cos25.0^{\circ}+85.0\sin25.0^{\circ}=142.6\,\text{N}=143\,\text{N}$ .	5
04.1	$6.12\,\text{N}$ and $8.33\,\text{N}$	Take downward as positive. At the stated instant, $mg-U-D=ma$ , so $D=0.850(9.81)-0.00950-0.850(2.60)=6.119\,\text{N}=6.12\,\text{N}$ . At terminal speed $a=0$ , hence $D_{\rm terminal}=mg-U=0.850(9.81)-0.00950=8.329\,\text{N}=8.33\,\text{N}$ . The larger drag balances the remaining downward force.	4
05.1	$1.60\,\text{kg}$ ; $2.40\,\text{N}$ ; $8.48\,\text{N}$ ; new gradient $0.500\,\text{kg}^{-1}$ and intercept $-1.20\,\text{m s}^{-2}$	From $F-R=ma$ , $a=(1/m)F-R/m$ . Thus $m=1/0.625=1.60\,\text{kg}$ and $R=m(1.50)=2.40\,\text{N}$ . For $a=3.80\,\text{m s}^{-2}$ , $F=ma+R=1.60(3.80)+2.40=8.48\,\text{N}$ . Adding $0.400\,\text{kg}$ gives $m'=2.00\,\text{kg}$ , so the new gradient is $1/m'=0.500\,\text{kg}^{-1}$ . If resistance is unchanged, the intercept is $-R/m'=-2.40/2.00=-1.20\,\text{m s}^{-2}$ .	6

3.4.1.6 · Momentum

Tier 1 · Easy

Mark scheme for 3.4.1.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.0\,\text{m s}^{-1}$ in the original direction	Momentum is conserved: initial momentum $=0.80(3.0)=2.4\,\text{kg m s}^{-1}$ . The joined mass is $1.20\,\text{kg}$ , so $v=2.4/1.20=2.0\,\text{m s}^{-1}$ .	2
02.1	$3.60\,\text{N s}$	Convert the time: $15.0\,\text{ms}=0.0150\,\text{s}$ . The impulse is $F\Delta t=240(0.0150)=3.60\,\text{N s}$ .	2

Tier 2 · Standard

Mark scheme for 3.4.1.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Both momentum and kinetic energy are conserved, so the collision is elastic.	Initial momentum is $0.40(5.0)=2.0\,\text{kg m s}^{-1}$ . Final momentum is $0.40(-1.0)+0.60(4.0)=-0.40+2.40=2.0\,\text{kg m s}^{-1}$ . Initial kinetic energy is $\tfrac12(0.40)(5.0)^2=5.0\,\text{J}$ . Final kinetic energy is $\tfrac12(0.40)(1.0)^2+\tfrac12(0.60)(4.0)^2=0.20+4.80=5.0\,\text{J}$ . Since both totals are unchanged, the collision is elastic.	4
02.1	$9.00\,\text{m s}^{-1}$ opposite to the projectile	Initial momentum is zero. Taking the projectile direction as positive, $0=(0.120)(180)+(2.40)v$ . Hence $v=-21.6/2.40=-9.00\,\text{m s}^{-1}$ , so the launcher recoils at $9.00\,\text{m s}^{-1}$ in the opposite direction.	3
03.1	$6.00\,\text{m s}^{-1}$ and $98.0\%$	Momentum conservation gives $0.0200(300)=(0.0200+0.980)v$ , so $v=6.00\,\text{m s}^{-1}$ . The bullet's initial kinetic energy is $\tfrac12(0.0200)(300)^2=900\,\text{J}$ . The final kinetic energy is $\tfrac12(1.000)(6.00)^2=18.0\,\text{J}$ , so $882\,\text{J}$ is dissipated. The percentage dissipated is $(882/900)100=98.0\%$ .	4

Tier 3 · Hard

Mark scheme for 3.4.1.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$8.4\,\text{m s}^{-1}$ east; $1.38\times10^5\,\text{J}$ ; $6.4\times10^4\,\text{N}$ west	Take east as positive. Conservation of momentum gives $(1200)(18)+(800)(-6.0)=(2000)v$ , so $v=16800/2000=8.4\,\text{m s}^{-1}$ east. Initial kinetic energy is $\tfrac12(1200)(18)^2+\tfrac12(800)(6.0)^2=208800\,\text{J}$ . Final kinetic energy is $\tfrac12(2000)(8.4)^2=70560\,\text{J}$ , so $138240\,\text{J}=1.38\times10^5\,\text{J}$ is dissipated. For the heavier car, $\Delta p=1200(8.4-18)=-11520\,\text{kg m s}^{-1}$ . Thus $F_{\rm mean}=\Delta p/\Delta t=-11520/0.18=-6.4\times10^4\,\text{N}$ , meaning west.	6
02.1	$6.00\,\text{m s}^{-1}$ away from the wall; kinetic energy decreases by $10.8\,\text{J}$	Take motion towards the wall as positive. The opposing impulse is $J=-\tfrac12(0.0180)(400)=-3.60\,\text{N s}$ . Initial momentum is $0.200(12.0)=2.40\,\text{kg m s}^{-1}$ , so final momentum is $2.40-3.60=-1.20\,\text{kg m s}^{-1}$ and $v=-1.20/0.200=-6.00\,\text{m s}^{-1}$ . Initial kinetic energy is $14.4\,\text{J}$ and final kinetic energy is $3.60\,\text{J}$ , a decrease of $10.8\,\text{J}$ .	5
03.1	$2.80\,\text{m s}^{-1}$ in the original direction and $226\,\text{J}$	Take the original direction as positive. The third fragment has mass $12.0-3.00-4.00=5.00\,\text{kg}$ . Momentum conservation gives $12.0(4.00)=3.00(14.0)+4.00(-2.00)+5.00v$ , so $v=2.80\,\text{m s}^{-1}$ . Initial kinetic energy is $\tfrac12(12.0)(4.00)^2=96.0\,\text{J}$ . Final kinetic energy is $\tfrac12(3.00)(14.0)^2+\tfrac12(4.00)(2.00)^2+\tfrac12(5.00)(2.80)^2=321.6\,\text{J}$ . The increase is $321.6-96.0=225.6\,\text{J}=226\,\text{J}$ .	5
04.1	$227\,\text{N}$ upward; kinetic energy decreases by $2.58\,\text{J}$	Take upward as positive. The momentum change is $\Delta p=0.145[6.20-(-8.60)]=2.146\,\text{kg m s}^{-1}$ . Hence the mean resultant force is $\Delta p/\Delta t=2.146/(9.50\times10^{-3})=225.895\,\text{N}$ upward. If $R$ is the floor's contact force, $R-mg=225.895$ , so $R=225.895+0.145(9.81)=227.317\,\text{N}=227\,\text{N}$ . The kinetic-energy decrease is $\tfrac12(0.145)(8.60^2-6.20^2)=2.5752\,\text{J}=2.58\,\text{J}$ .	5
05.1	$3.72\,\text{m s}^{-1}$ ; $2.72\times10^5\,\text{J}$ ; $1.32\,\text{s}$	Take the first vehicle's direction as positive. Momentum conservation gives $v=[(1.80\times10^4)(7.20)+(1.20\times10^4)(-1.50)]/(3.00\times10^4)=3.72\,\text{m s}^{-1}$ . Initial kinetic energy is $\tfrac12(1.80\times10^4)(7.20)^2+\tfrac12(1.20\times10^4)(1.50)^2=480060\,\text{J}$ . After coupling it is $\tfrac12(3.00\times10^4)(3.72)^2=207576\,\text{J}$ , so $272484\,\text{J}=2.72\times10^5\,\text{J}$ is dissipated. The braking impulse must equal $(3.00\times10^4)(3.72)=111600\,\text{N s}$ . The force–time area is $[(48.0+12.0)/2]\times10^3t+(12.0\times10^3)(6.00)$ , so $30000t+72000=111600$ and $t=1.32\,\text{s}$ .	6

3.4.1.7 · Work, energy and power

Tier 1 · Easy

Mark scheme for 3.4.1.7 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$900\,\text{J}$	Because force and displacement are parallel, $W=Fs=75\times12=900\,\text{J}$ .	1
02.1	Work is measured in joules; power is measured in watts.	One joule is one newton metre of work. One watt is a transfer rate of one joule per second.	2

Tier 2 · Standard

Mark scheme for 3.4.1.7 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$327\,\text{W}$ and $77.9\%$	The useful energy transferred is $mgh=25(9.81)(8.0)=1962\,\text{J}$ . Useful power is $P=1962/6.0=327\,\text{W}$ . Efficiency $=327/420=0.7786$ , or $77.9\%$ .	3
02.1	$78.0\,\text{kJ}$	Energy is the area under the power-time graph. The triangular section gives $\tfrac12(5.00)(12.0)=30.0\,\text{kJ}$ and the rectangular section gives $(4.00)(12.0)=48.0\,\text{kJ}$ . The total is $78.0\,\text{kJ}$ .	3
03.1	$2.72\,\text{kJ}$ and $151\,\text{W}$	The force component along the displacement is $140\cos28.0^{\circ}$ . Hence $W=Fs\cos\theta=140(22.0)\cos28.0^{\circ}=2719\,\text{J}=2.72\,\text{kJ}$ . The average power is $P=W/t=2719/18.0=151\,\text{W}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.7 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$220\,\text{J}$ ; $14.8\,\text{m s}^{-1}$ ; $73.3\,\text{W}$	Work is the area under the force-displacement graph. Over the first $4.0\,\text{m}$ , $W_1=\tfrac12(10+50)(4.0)=120\,\text{J}$ . Over the next $2.0\,\text{m}$ , $W_2=50(2.0)=100\,\text{J}$ , so total $W=220\,\text{J}$ . With negligible resistance, $W=\Delta E_k=\tfrac12mv^2$ , hence $v=\sqrt{2(220)/2.0}=14.8\,\text{m s}^{-1}$ . Average power is $W/t=220/3.0=73.3\,\text{W}$ .	5
02.1	$1.42\,\text{kN}$ and $13.7\,\text{MJ}$	At constant speed, the driving force is $F=mg\sin6.00^{\circ}+500=900(9.81)\sin6.00^{\circ}+500=1422.88\,\text{N}=1.42\,\text{kN}$ . In $120\,\text{s}$ the distance is $s=vt=20.0(120)=2400\,\text{m}$ . The work done by the driving force is $Fs=1422.88(2400)=3.41492\times10^6\,\text{J}=3.41\times10^6\,\text{J}$ . Since efficiency is useful output divided by input, the input energy is $(3.41492\times10^6)/0.250=1.36597\times10^7\,\text{J}=13.7\,\text{MJ}$ to three significant figures.	5
03.1	$4.02\,\text{kW}$	The kinetic-energy increase is $\tfrac12m(v^2-u^2)=\tfrac12(70.0)(15.0^2-5.00^2)=7000\,\text{J}$ . Work against resistance is $Fs=30.0(120)=3600\,\text{J}$ . The total mechanical work is therefore $10600\,\text{J}$ , giving average mechanical power $10600/12.0=883\,\text{W}$ . Since efficiency is output divided by input, the average input power is $883/0.220=4015\,\text{W}=4.02\,\text{kW}$ .	5
04.1	$6.78\,\text{W}$ ; $69.7\,\text{J}$ ; $26.1\,\text{J}$	The useful output power is $P_{\rm out}=mgh/t=2.40(9.81)(1.85)/6.42=6.78449\,\text{W}=6.78\,\text{W}$ . The electrical energy supplied is $E_{\rm in}=VIt=11.8(0.920)(6.42)=69.6955\,\text{J}=69.7\,\text{J}$ . The useful energy gained is $mgh=43.5564\,\text{J}$ , so the energy transferred to the surroundings is $69.6955-43.5564=26.1391\,\text{J}=26.1\,\text{J}$ .	5
05.1	$4.27\,\text{m s}^{-2}$ ; $1.45\,\text{m s}^{-2}$ ; maximum speed $133\,\text{m s}^{-1}$	Since $P=F_{\rm drive}v$ , at $12.0\,\text{m s}^{-1}$ the driving force is $72000/12.0=6000\,\text{N}$ . Thus $a=(6000-540)/1280=4.26563\,\text{m s}^{-2}=4.27\,\text{m s}^{-2}$ . At $30.0\,\text{m s}^{-1}$ , $F_{\rm drive}=72000/30.0=2400\,\text{N}$ and $a=(2400-540)/1280=1.45313\,\text{m s}^{-2}=1.45\,\text{m s}^{-2}$ . Constant power means the driving force falls as speed rises. At maximum speed the driving force equals resistance, so $v_{\max}=P/F=72000/540=133.333\,\text{m s}^{-1}=133\,\text{m s}^{-1}$ .	6

3.4.1.8 · Conservation of energy

Tier 1 · Easy

Mark scheme for 3.4.1.8 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$5.94\,\text{m s}^{-1}$	The loss of gravitational potential energy becomes kinetic energy: $mgh=\tfrac12mv^2$ . Mass cancels, so $v=\sqrt{2gh}=\sqrt{2(9.81)(1.80)}=5.94\,\text{m s}^{-1}$ .	2
02.1	Energy cannot be created or destroyed; the total energy of a closed system remains constant.	Energy may be transferred between stores, but in a closed system the sum across all stores does not change.	2

Tier 2 · Standard

Mark scheme for 3.4.1.8 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$13.3\,\text{m s}^{-1}$	The vertical drop is $12.0-3.0=9.0\,\text{m}$ . With no resistive work, $mg(9.0)=\tfrac12mv^2$ . Thus $v=\sqrt{2(9.81)(9.0)}=13.3\,\text{m s}^{-1}$ .	3
02.1	$3.92\,\text{J}$	The mechanical energy before impact is $mg(2.00)$ and that after impact is $mg(1.20)$ . The difference transferred to internal energy is $mg(2.00-1.20)=0.500(9.81)(0.800)=3.924\,\text{J}=3.92\,\text{J}$ to three significant figures.	3
03.1	$5.85\,\text{m}$	The initial kinetic energy is $\tfrac12mv^2=\tfrac12(2.00)(12.0)^2=144\,\text{J}$ . At maximum height this has become gravitational potential energy and work against resistance: $144=mgh+Fh=[2.00(9.81)+5.00]h$ . Therefore $h=144/24.62=5.85\,\text{m}$ .	3

Tier 3 · Hard

Mark scheme for 3.4.1.8 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$16.7\,\text{m s}^{-1}$	Initial kinetic energy is $\tfrac12(70)(4.0)^2=560\,\text{J}$ . The gravitational energy decrease is $mgh=70(9.81)(25)=17167.5\,\text{J}$ . Subtracting the $8000\,\text{J}$ resistive transfer gives final kinetic energy $560+17167.5-8000=9727.5\,\text{J}$ . Hence $v=\sqrt{2E_k/m}=\sqrt{2(9727.5)/70}=16.7\,\text{m s}^{-1}$ .	5
02.1	$51.4\,\text{kJ}$ and $286\,\text{N}$	The gravitational energy decrease is $mgh=900(9.81)(15.0)=132435\,\text{J}$ . The kinetic energy increase is $\tfrac12m(v^2-u^2)=\tfrac12(900)(18.0^2-12.0^2)=81000\,\text{J}$ . The energy transferred by resistance is $132435-81000=51435\,\text{J}=51.4\,\text{kJ}$ . Hence the mean resistive force is $F=W/s=51435/180=285.75\,\text{N}=286\,\text{N}$ to three significant figures.	5
03.1	$8.91\,\text{m}$ above the bottom of the track	Let the final vertical height be $h$ . The distance travelled up the $30.0^{\circ}$ slope is $h/\sin30.0^{\circ}=2h$ . Conservation of energy including resistive transfer gives $mg(12.0)=40.0(20.0+2h)+mgh$ . Thus $50.0(9.81)(12.0)-800=[50.0(9.81)+80.0]h$ , so $h=5086/570.5=8.91\,\text{m}$ .	5
04.1	$0.325\,\text{J}$ and $0.263\,\text{m}$	Initially the gravitational energy relative to the lowest point is $mgh=0.350(9.81)(0.480)=1.64808\,\text{J}$ . At the lowest point the kinetic energy is $\tfrac12(0.350)(2.75)^2=1.32344\,\text{J}$ , so $1.64808-1.32344=0.324643\,\text{J}=0.325\,\text{J}$ has been transferred. After a further $0.420\,\text{J}$ transfer, the available energy is $1.32344-0.420=0.903438\,\text{J}$ . Hence $h=0.903438/[0.350(9.81)]=0.263124\,\text{m}=0.263\,\text{m}$ .	5
05.1	$8.46\,\text{m}$ ; $12.0\,\text{m s}^{-1}$ ; $47.4\,\text{J}$	On ascent, the initial kinetic energy does work against weight and resistance: $\tfrac12(1.80)(14.0)^2=[1.80(9.81)+3.20]h$ . Thus $h=176.4/20.858=8.45719\,\text{m}$ . On descent, the kinetic energy at launch level is $[1.80(9.81)-2.40]h=15.258(8.45719)=129.040\,\text{J}$ . Hence $v=\sqrt{2(129.040)/1.80}=11.9740\,\text{m s}^{-1}=12.0\,\text{m s}^{-1}$ . The total transfer by resistance is $(3.20+2.40)h=5.60(8.45719)=47.3602\,\text{J}=47.4\,\text{J}$ .	6

3.4.2.1 · Bulk properties of solids

Tier 1 · Easy

Mark scheme for 3.4.2.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.70\times10^3\,\text{kg m}^{-3}$	Use $\rho=m/V=2.16/(8.00\times10^{-4})=2.70\times10^3\,\text{kg m}^{-3}$ .	1
02.1	Elastic deformation is fully reversed; plastic deformation leaves a permanent change in dimensions.	An elastically deformed object returns to its original dimensions when unloaded. Plastic deformation remains after unloading.	2

Tier 2 · Standard

Mark scheme for 3.4.2.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.900\,\text{J}$ and $3.46\,\text{m s}^{-1}$	Convert the compression: $x=0.075\,\text{m}$ . The elastic strain energy is $E=\tfrac12kx^2=\tfrac12(320)(0.075)^2=0.900\,\text{J}$ . With $E=\tfrac12mv^2$ , $v=\sqrt{2E/m}=\sqrt{1.800/0.150}=3.46\,\text{m s}^{-1}$ .	3
02.1	$6.00\times10^{-5}\,\text{m}^3$	The external volume is $0.100(0.0800)(0.0500)=4.00\times10^{-4}\,\text{m}^3$ . The aluminium volume is $V=m/\rho=0.918/2700=3.40\times10^{-4}\,\text{m}^3$ . Therefore the cavity volume is $4.00\times10^{-4}-3.40\times10^{-4}=6.00\times10^{-5}\,\text{m}^3$ .	3
03.1	P is the limit of proportionality (accept: the point where the graph stops being straight, so Hooke's law fails); Q is the elastic limit. Evidence for P: the graph is straight from the origin up to P and curved beyond it.	At P the graph stops being straight, so stress is no longer proportional to strain and P is the limit of proportionality. Q is the greatest stress for which unloading removes all strain; any greater stress leaves permanent deformation. Therefore Q is the elastic limit.	3

Tier 3 · Hard

Mark scheme for 3.4.2.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$15.1\,\text{J}$ ; plastic deformation leaves permanent extension and transfers some energy to internal energy.	Work is the area under the graph. Before the elastic limit, $W_1=\tfrac12(720)(0.015)=5.40\,\text{J}$ . From $0.015$ to $0.027\,\text{m}$ , the trapezium area is $W_2=\tfrac12(720+900)(0.027-0.015)=9.72\,\text{J}$ . Total work is $15.12\,\text{J}=15.1\,\text{J}$ . Beyond the elastic limit the wire deforms plastically, so it retains a permanent extension and some work is dissipated as internal energy rather than being recoverable elastic energy.	5
02.1	$7.07\times10^8\,\text{Pa}$ ; an almost linear elastic region followed by fracture with little or no plastic deformation	The radius is $0.600\,\text{mm}=6.00\times10^{-4}\,\text{m}$ , so $A=\pi r^2=1.131\times10^{-6}\,\text{m}^2$ . Breaking stress is $800/A=7.07\times10^8\,\text{Pa}$ . A brittle material shows an almost linear elastic region and then fractures with little or no plastic region.	5
03.1	$3.95\,\text{m s}^{-1}$	The spring energy is the area under the force-compression graph. The first triangle gives $\tfrac12(40.0)(0.100)=2.00\,\text{J}$ and the remaining trapezium gives $\tfrac12(40.0+60.0)(0.150-0.100)=2.50\,\text{J}$ , for a total of $4.50\,\text{J}$ . After rising $1.50\,\text{m}$ , the gravitational energy increase is $mgh=0.200(9.81)(1.50)=2.943\,\text{J}$ . Thus $\tfrac12mv^2=4.50-2.943=1.557\,\text{J}$ and $v=\sqrt{2(1.557)/0.200}=3.95\,\text{m s}^{-1}$ .	5
04.1	$6.84\,\text{J}$ ; $4.62\,\text{J}$ ; $2.22\,\text{J}$	Loading work is the area under both straight sections: $W_{\rm load}=\tfrac12(300)(0.0120)+\tfrac12(300+420)(0.0260-0.0120)=1.80+5.04=6.84\,\text{J}$ . The unloading line encloses a triangle above the permanent extension, so $W_{\rm recovered}=\tfrac12(420)(0.0260-0.00400)=4.62\,\text{J}$ . The difference is transferred to internal energy: $6.84-4.62=2.22\,\text{J}$ .	6
05.1	A absorbs about $5.96\times10^7\,\text{J m}^{-3}$ and B about $1.86\times10^6\,\text{J m}^{-3}$ ; A is more suitable for controlled energy absorption because its large plastic strain absorbs far more energy before fracture, whereas B is comparatively brittle	The identity $\sigma\varepsilon=(F/A)(\Delta L/L)=F\Delta L/(AL)$ links the work done per unit volume to the area under a stress–strain graph. Energy absorbed per unit volume is therefore the area under the graph. For A, the estimate is $\tfrac12(280\times10^6)(0.0040)+(250\times10^6)(0.240-0.0040)=5.956\times10^7\,\text{J m}^{-3}$ . For B it is $\tfrac12(620\times10^6)(0.0060)=1.86\times10^6\,\text{J m}^{-3}$ . A therefore absorbs about 32 times as much energy per unit volume through substantial plastic deformation. That controlled deformation is desirable in a crumple zone, subject to separate checks on mass, geometry, cost and the strength required for the protected passenger cell.	6

3.4.2.2 · The Young modulus

Tier 1 · Easy

Mark scheme for 3.4.2.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.0\times10^7\,\text{Pa}$	The area is $2.0\,\text{mm}^2=2.0\times10^{-6}\,\text{m}^2$ . Therefore $\sigma=F/A=120/(2.0\times10^{-6})=6.0\times10^7\,\text{Pa}$ , equal to $60\,\text{MPa}$ .	2
02.1	Young modulus is measured in pascals; strain is a ratio of two lengths, so its units cancel.	Young modulus is stress divided by strain, so it has the stress unit pascal. Tensile strain is $\Delta L/L$ and both numerator and denominator are lengths in the same unit.	2

Tier 2 · Standard

Mark scheme for 3.4.2.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.10\times10^8\,\text{Pa}$ ; $1.00\times10^{-3}$ ; $1.10\times10^{11}\,\text{Pa}$	The radius is $0.400\,\text{mm}=4.00\times10^{-4}\,\text{m}$ , so $A=\pi r^2=5.03\times10^{-7}\,\text{m}^2$ . Stress is $\sigma=F/A=55.3/(5.03\times10^{-7})=1.10\times10^8\,\text{Pa}$ . Strain is $\varepsilon=\Delta L/L=(2.00\times10^{-3})/2.00=1.00\times10^{-3}$ . Hence $E=\sigma/\varepsilon=(1.10\times10^8)/(1.00\times10^{-3})=1.10\times10^{11}\,\text{Pa}=110\,\text{GPa}$ .	4
02.1	$1.50\,\text{mm}$ and $8.33\times10^{-4}$	Use $E=FL/(A\Delta L)$ with $A=3.00\times10^{-6}\,\text{m}^2$ . Therefore $\Delta L=FL/(AE)=500(1.80)/[(3.00\times10^{-6})(2.00\times10^{11})]=1.50\times10^{-3}\,\text{m}=1.50\,\text{mm}$ . The tensile strain is $\Delta L/L=(1.50\times10^{-3})/1.80=8.33\times10^{-4}$ .	3
03.1	$0.781\,\text{mm}$	For a wire, $\Delta L=FL/(AE)$ , and the common Young modulus cancels in a ratio. Since $A\propto d^2$ , $\Delta L_B/\Delta L_A=(F_BL_Bd_A^2)/(F_AL_Ad_B^2)$ . Therefore $\Delta L_B=1.80[100(0.800)(0.500)^2]/[60.0(1.20)(0.800)^2]=0.78125\,\text{mm}=0.781\,\text{mm}$ .	4

Tier 3 · Hard

Mark scheme for 3.4.2.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.12\times10^{11}\,\text{Pa}$ ; repeated perpendicular diameter readings account for variation and non-circularity and reduce uncertainty in area.	From $E=\sigma/\varepsilon$ , with $\sigma=F/A$ and $\varepsilon=\Delta L/L$ , $E=(F/A)/(\Delta L/L)=(F/\Delta L)(L/A)$ . The radius is $0.210\,\text{mm}=2.10\times10^{-4}\,\text{m}$ , so $A=\pi r^2=1.385\times10^{-7}\,\text{m}^2$ . Since $F/\Delta L$ is the graph gradient, $E=(6.20\times10^3)(2.50)/(1.385\times10^{-7})=1.12\times10^{11}\,\text{Pa}=112\,\text{GPa}$ . Diameter can vary along the wire and the cross-section may not be perfectly circular; readings at several positions and orientations give a representative mean and reduce the large area uncertainty caused by squaring the diameter.	6
02.1	$1.70\,\text{mm}$ and $7.06\times10^4\,\text{N m}^{-1}$	For each wire, $\Delta L=FL/(AE)$ . Steel extends by $120(1.20)/[(0.800\times10^{-6})(2.00\times10^{11})]=9.00\times10^{-4}\,\text{m}$ . Brass extends by $120(0.800)/[(1.20\times10^{-6})(1.00\times10^{11})]=8.00\times10^{-4}\,\text{m}$ . Total extension is $1.70\times10^{-3}\,\text{m}=1.70\,\text{mm}$ . The effective stiffness is $k=F/\Delta L=120/(1.70\times10^{-3})=7.06\times10^4\,\text{N m}^{-1}$ to three significant figures.	5
03.1	$1.94\times10^{11}\,\text{Pa}$ and $13.0\%$	The corrected extension is $2.46-0.320=2.14\,\text{mm}=2.14\times10^{-3}\,\text{m}$ . The area is $A=\pi(0.350\times10^{-3})^2=3.848\times10^{-7}\,\text{m}^2$ . Hence $E=FL/(A\Delta L)=80.0(2.00)/[(3.848\times10^{-7})(2.14\times10^{-3})]=1.94\times10^{11}\,\text{Pa}$ . Ignoring support movement would give $E_{\rm uncorrected}=80.0(2.00)/[(3.848\times10^{-7})(2.46\times10^{-3})]=1.69\times10^{11}\,\text{Pa}$ . The underestimate is $[(1.94-1.69)/1.94]100=13.0\%$ , using unrounded values.	5
04.1	$0.903\,\text{mm}$ ; steel carries $34.0\,\text{N}$ and copper carries $46.0\,\text{N}$	For each wire, $k=EA/L$ . The areas are $A_s=\pi(0.600\times10^{-3})^2/4=2.82743\times10^{-7}\,\text{m}^2$ and $A_c=\pi(0.900\times10^{-3})^2/4=6.36173\times10^{-7}\,\text{m}^2$ . Thus $k_s=3.76991\times10^4\,\text{N m}^{-1}$ and $k_c=5.08938\times10^4\,\text{N m}^{-1}$ . Parallel wires have the same extension, so $x=80.0/(k_s+k_c)=9.03007\times10^{-4}\,\text{m}=0.903\,\text{mm}$ . The forces are $F_s=k_sx=34.0426\,\text{N}=34.0\,\text{N}$ and $F_c=k_cx=45.9574\,\text{N}=46.0\,\text{N}$ .	6
05.1	$(5.25\pm0.23)\times10^{10}\,\text{Pa}$ ; the diameter contributes most because area is proportional to diameter squared	The area is $A=\pi d^2/4=\pi(0.460\times10^{-3})^2/4=1.66190\times10^{-7}\,\text{m}^2$ . Since the graph gradient is $k=EA/L$ , $E=kL/A=(4.85\times10^3)(1.800)/(1.66190\times10^{-7})=5.25302\times10^{10}\,\text{Pa}$ . The maximum fractional uncertainty is $0.08/4.85+0.002/1.800+2(0.006/0.460)=0.0436929$ , or $4.37\%$ . Hence $\Delta E=0.0436929(5.25302\times10^{10})=2.2952\times10^9\,\text{Pa}$ , giving $(5.25\pm0.23)\times10^{10}\,\text{Pa}$ . The diameter term is $2(0.006/0.460)=2.61\%$ , the largest single contribution.	6