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Admissions tests / ESAT / Physics / Mechanics

Demanding. 15 questions, 15 marks, about 28 minutes.

ESAT Physics: Mechanics, set 3

Scalars and vectors, distance, displacement, speed and velocity, acceleration, the equations of motion, graphs of motion, forces, Newton's laws, momentum, moments, work, energy and power.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    A cyclist accelerates uniformly from rest to a speed of 12 m/s over 4 s, then travels at this constant 12 m/s for a further 5 s, then decelerates uniformly to rest over a further 6 s. What total distance does the cyclist travel?

    1. A 90 m
    2. B 60 m
    3. C 120 m
    4. D 180 m
  2. 21 mark

    A car decelerates uniformly from an initial speed u to rest, coming to a stop after travelling 48 m. The magnitude of its deceleration is 6 m/s^2 throughout. What was its initial speed u?

    1. A 24 m/s
    2. B 576 m/s
    3. C 288 m/s
    4. D 48 m/s
  3. 31 mark

    A crate of mass 10 kg is pushed to the right with a force of 100 N. At the same time it is pulled to the left by a rope with a force of 30 N, and friction acts on it with a force of 20 N, also opposing its motion to the right. What is the magnitude of the crate's acceleration?

    1. A 7 m/s^2
    2. B 8 m/s^2
    3. C 15 m/s^2
    4. D 5 m/s^2
  4. 41 mark

    A spring with spring constant 100 N/m is stretched, within its limit of proportionality, from an extension of 0.02 m to an extension of 0.06 m. How much additional elastic potential energy is stored in the spring during this further stretching, from 0.02 m to 0.06 m?

    1. A 0.08 J
    2. B 0.16 J
    3. C 0.18 J
    4. D 0.32 J
  5. 51 mark

    A rocket of mass 5000 kg experiences an upward thrust force of 70000 N as it lifts off. Using g = 10 N/kg, what is its acceleration at this instant?

    1. A 14 m/s^2
    2. B 13 m/s^2
    3. C 0.4 m/s^2
    4. D 4 m/s^2
  6. 61 mark

    Two astronauts float at rest next to each other in deep space, well away from any other object: astronaut A has mass 60 kg and astronaut B has mass 90 kg. They push off from one another. By Newton's third law, the force astronaut A exerts on astronaut B is equal in magnitude to the force astronaut B exerts on astronaut A. If astronaut A has an acceleration of magnitude 9 m/s^2 during the push, what is the magnitude of astronaut B's acceleration at that instant?

    1. A 6 m/s^2
    2. B 9 m/s^2
    3. C 13.5 m/s^2
    4. D 3.6 m/s^2
  7. 71 mark

    A skydiver is in free fall, before reaching terminal velocity, and has a downward speed of 15 m/s at a certain instant. Using g = 10 N/kg as her acceleration during this interval (ignore air resistance here), how many more seconds does it take for her speed to increase to 55 m/s?

    1. A 5.5 s
    2. B 7 s
    3. C 4 s
    4. D 400 s
  8. 81 mark

    A ball of mass 0.5 kg travelling at 8 m/s horizontally strikes a wall and rebounds, moving back the way it came at 6 m/s. The collision with the wall lasts 0.02 s. What is the magnitude of the average force the wall exerts on the ball?

    1. A 50 N
    2. B 350 N
    3. C 200 N
    4. D 7 N
  9. 91 mark

    A firework shell of total mass 5 kg is at rest in mid-air when it explodes into two fragments. One fragment, of mass 1 kg, flies off at a speed of 20 m/s. What is the speed of the other fragment, of mass 4 kg, immediately after the explosion?

    1. A 5 m/s
    2. B 20 m/s
    3. C 80 m/s
    4. D 4 m/s
  10. 101 mark

    A car of mass 1000 kg, travelling at 20 m/s, brakes and comes to rest after travelling 50 m. Assuming a constant braking (friction) force is the only horizontal force acting on it, what is the magnitude of that force?

    1. A 8000 N
    2. B 200 N
    3. C 200000 N
    4. D 4000 N
  11. 111 mark

    A crane lifts a load of mass 300 kg through a vertical height of 8 m at constant speed, using g = 10 N/kg. The crane's motor supplies 30000 J of energy to do this. What is the percentage efficiency of the crane?

    1. A 125%
    2. B 8%
    3. C 80%
    4. D 0.8%
  12. 121 mark

    A weather balloon rises through the atmosphere at a constant velocity. Ignoring any horizontal wind, which statement correctly describes the vertical forces acting on it at this constant velocity?

    1. A Upthrust alone must be greater than the balloon's weight, since the balloon is still moving upward; this means there is still a nonzero resultant force acting on it even though its velocity is constant.
    2. B The upward upthrust on the balloon is exactly balanced by the combined downward forces of its weight and the drag resisting its upward motion, giving a zero resultant force, consistent with its constant velocity.
    3. C The balloon's weight and its upthrust are equal and opposite, but drag plays no part in this balance, since drag can only act on an object while it is accelerating.
    4. D Since the balloon's velocity is constant, no forces at all can be acting on it any more; the upthrust, weight and drag that acted on it earlier, while it was still speeding up, no longer exist.
  13. 131 mark

    Two supermarket trolleys, one empty and one loaded with heavy tins, are both initially at rest. A shopper gives each trolley an identical horizontal push, applying the same force for the same short time to each. Which of the following correctly compares their motion immediately afterwards, using Newton's first law and the idea of inertia?

    1. A The loaded trolley has greater inertia, meaning it resists the push more strongly, so it will end up moving faster than the empty trolley, to compensate for this greater resistance.
    2. B Since both trolleys receive exactly the same force for the same time, Newton's first law guarantees that they undergo exactly the same change in velocity, regardless of their different masses.
    3. C Inertia only resists a force acting on an object that is already moving; since both trolleys start at rest, neither trolley's inertia has any effect on its response to this push.
    4. D The loaded trolley has the greater mass and therefore the greater inertia, so for the same applied force, it undergoes a smaller change in velocity than the empty trolley does.
  14. 141 mark

    A drone flies in a straight line from its base to a checkpoint 300 m away, taking 60 s, then immediately flies back to base along the same straight-line path, taking a further 40 s. What is the drone's average speed for the whole trip, and what is the magnitude of its average velocity for the whole trip?

    1. A Average speed = 6 m/s, average velocity = 0 m/s.
    2. B Average speed = 6 m/s, average velocity = 6 m/s.
    3. C Average speed = 6 m/s, average velocity = 3 m/s.
    4. D Average speed = 6.25 m/s, average velocity = 0 m/s.
  15. 151 mark

    A spring obeys Hooke's law over the range shown on its force-extension graph, which is a straight line through the origin. A student reads off that a force of 12 N corresponds to an extension of 0.04 m on this straight-line section. What does the gradient of this straight-line section represent, and what is its value here?

    1. A The gradient represents the elastic potential energy stored in the spring at this point, found by multiplying the force by the extension; here that gives 12 x 0.04 = 0.48 J.
    2. B The gradient represents the spring constant, but found by dividing the extension by the force rather than the force by the extension; here that gives 0.04/12 = 1/300 N/m.
    3. C The gradient represents the spring constant, k, given by force divided by extension; here k = 12/0.04 = 300 N/m.
    4. D The gradient represents the spring constant, found by adding the force and the extension together rather than dividing one by the other; here that gives 12 + 0.04 = 12.04 N/m.

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. On a velocity-time graph, distance travelled is the area between the graph and the time axis, so each of the three phases must be treated separately and added together.
    2. The accelerating phase is a triangle rising from 0 to 12 m/s over 4 s: distance = (1/2) x 4 x 12 = 24 m.
    3. The constant-speed phase is a rectangle at 12 m/s for 5 s: distance = 12 x 5 = 60 m.
    4. The decelerating phase is a triangle falling from 12 m/s to 0 over 6 s: distance = (1/2) x 6 x 12 = 36 m.
    5. Total distance = 24 + 60 + 36 = 120 m, so the answer is C.
    • Why not A: Treats the whole 15 s journey as a single triangle rising from 0 to 12 m/s and back down again, using (1/2) x 15 x 12 = 90; this ignores that there is a separate 5 s section in the middle where the cyclist travels at a genuinely constant speed rather than continuing to accelerate.
    • Why not B: Counts only the middle, constant-speed section of the journey (12 x 5 = 60 m) and forgets that the accelerating and decelerating phases at the start and end also cover real distance.
    • Why not D: Treats the cyclist as travelling at the full 12 m/s for the entire 15 s (12 x 15 = 180), ignoring that her speed is only 12 m/s during the middle phase and is lower than this throughout the accelerating and decelerating phases.

  2. Question 2Answer: A

    1. The car's final velocity is v = 0, since it comes to rest. The relevant equation of motion is v^2 = u^2 + 2as, where a is taken as the deceleration acting against the direction of travel.
    2. Rearranging with v = 0: 0 = u^2 - 2 x 6 x 48, so u^2 = 2 x 6 x 48 = 576.
    3. Taking the square root of both sides gives u = sqrt(576) = 24 m/s.
    4. The car's initial speed was 24 m/s, so the answer is A.
    • Why not B: Correctly computes u^2 = 2 x 6 x 48 = 576 using the equation of motion, but then reports this squared value directly as the speed itself, forgetting that a final square root step is needed to get from u^2 to u.
    • Why not C: Leaves out the factor of 2 that the equation v^2 = u^2 + 2as requires, computing u^2 = 6 x 48 = 288 instead of u^2 = 2 x 6 x 48 = 576, and then also forgets to take the square root of this (already wrong) value.
    • Why not D: Confuses the given stopping distance of 48 m with the initial speed being asked for, and simply restates that number in m/s instead of working through the equation of motion at all.

  3. Question 3Answer: D

    1. The rope's pull (30 N) and friction (20 N) both act to the left, opposing the 100 N push to the right, so both must be subtracted from the push to find the resultant force.
    2. Resultant force = 100 - 30 - 20 = 50 N, acting to the right.
    3. By Newton's second law, F = ma, so a = F/m = 50 N / 10 kg = 5 m/s^2.
    4. The crate's acceleration is 5 m/s^2, so the answer is D.
    • Why not A: Subtracts only the rope's 30 N from the push (100 - 30 = 70 N) and forgets that friction, at 20 N, is also acting to oppose the crate's motion to the right and so must also be subtracted.
    • Why not B: Subtracts only friction's 20 N from the push (100 - 20 = 80 N) and forgets that the rope's 30 N pull is also acting in the opposing direction and so must also be subtracted.
    • Why not C: Adds all three forces together (100 + 30 + 20 = 150 N) instead of subtracting the two forces that oppose the push; forces acting in opposite directions must be subtracted from each other to find the resultant, not simply summed.

  4. Question 4Answer: B

    1. The elastic potential energy stored at any extension x, within the limit of proportionality, is E = (1/2)kx^2.
    2. At x = 0.02 m: E1 = (1/2) x 100 x 0.02^2 = 50 x 0.0004 = 0.02 J.
    3. At x = 0.06 m: E2 = (1/2) x 100 x 0.06^2 = 50 x 0.0036 = 0.18 J.
    4. The additional energy stored while stretching from 0.02 m to 0.06 m is E2 - E1 = 0.18 - 0.02 = 0.16 J, so the answer is B.
    • Why not A: Applies E = (1/2)kx^2 to just the extra extension, x = 0.06 - 0.02 = 0.04 m, as if the spring had been completely unstretched at the start of this further stretching; but the spring already held energy at 0.02 m, so the extra energy must be found as the DIFFERENCE between the energy at 0.06 m and the energy already present at 0.02 m, not from the extra extension alone.
    • Why not C: Reports the total energy stored at the final extension of 0.06 m, 0.18 J, forgetting to subtract the 0.02 J already stored in the spring at the initial extension of 0.02 m; the question asks for the ADDITIONAL energy stored, not the total.
    • Why not D: Omits the factor of 1/2 from E = (1/2)kx^2 when finding the energy at each extension, doubling both individual energy values and so also doubling the true difference between them.

  5. Question 5Answer: D

    1. The rocket's weight is found using w = mg = 5000 kg x 10 N/kg = 50000 N, acting downward.
    2. The resultant force is the thrust minus the weight, since they act in opposite directions: 70000 - 50000 = 20000 N, acting upward.
    3. By Newton's second law, a = F/m = 20000 N / 5000 kg = 4 m/s^2.
    4. The rocket's acceleration at this instant is 4 m/s^2, so the answer is D.
    • Why not A: Divides the thrust alone by the mass (70000 / 5000 = 14) and ignores the rocket's weight entirely; the resultant force driving the rocket's acceleration must account for gravity pulling down on it as well as the thrust pushing it up.
    • Why not B: Confuses mass with weight, subtracting the mass value itself, 5000, directly from the thrust (70000 - 5000 = 65000, then /5000 = 13) instead of first multiplying the mass by g to find the weight (5000 x 10 = 50000 N) that actually opposes the thrust.
    • Why not C: Divides the correctly found resultant force (20000 N) by the weight, 50000 N, instead of by the mass, 5000 kg; Newton's second law requires dividing force by MASS to find acceleration, not by weight.

  6. Question 6Answer: A

    1. By Newton's third law, the force astronaut A exerts on astronaut B has the same magnitude as the force astronaut B exerts on astronaut A: call this magnitude F.
    2. For astronaut A, Newton's second law gives F = m_A x a_A = 60 x 9 = 540 N.
    3. The same magnitude of force, 540 N, acts on astronaut B, so astronaut B's acceleration is a_B = F/m_B = 540/90 = 6 m/s^2.
    4. Astronaut B's acceleration has magnitude 6 m/s^2, so the answer is A.
    • Why not B: Assumes that because the forces on the two astronauts are equal in magnitude (Newton's third law), their accelerations must also be equal; this ignores that Newton's second law, a = F/m, means the same force produces a smaller acceleration on the more massive astronaut.
    • Why not C: Inverts the mass ratio, scaling astronaut A's acceleration by m_B/m_A (90/60) instead of m_A/m_B (60/90), giving an acceleration for astronaut B that is larger rather than smaller than astronaut A's, when the heavier astronaut must in fact experience the smaller acceleration.
    • Why not D: Uses the combined mass of both astronauts, 150 kg, as the denominator (60 x 9 / 150 = 3.6) instead of astronaut B's own mass, 90 kg; the equal-and-opposite force must be divided by astronaut B's OWN mass to find astronaut B's acceleration, not by the total mass of both astronauts together.

  7. Question 7Answer: C

    1. Acceleration is defined as the change in velocity divided by time: a = (v - u) / t, so t = (v - u) / a.
    2. The change in the skydiver's speed is v - u = 55 - 15 = 40 m/s.
    3. Using g = 10 N/kg as her acceleration during this interval: t = 40 / 10 = 4 s.
    4. It takes 4 more seconds for her speed to increase from 15 m/s to 55 m/s, so the answer is C.
    • Why not A: Divides the final speed of 55 m/s alone by g (55/10 = 5.5), forgetting that the skydiver already had a downward speed of 15 m/s at the start of this interval; only the CHANGE in speed should be divided by the acceleration, not the final speed on its own.
    • Why not B: Adds the two given speeds together (55 + 15 = 70) before dividing by g, instead of first finding the change in speed BETWEEN them (55 - 15 = 40); adding the speeds has no physical meaning here.
    • Why not D: Multiplies the change in speed by g instead of dividing by it (40 x 10 = 400), inverting the rearrangement of acceleration = change in velocity / time, which gives time = change in velocity / acceleration, not change in velocity x acceleration.

  8. Question 8Answer: B

    1. Taking the ball's initial direction of travel (towards the wall) as positive, its momentum before the collision is p1 = 0.5 x 8 = 4 kg m/s.
    2. After rebounding, the ball moves in the opposite direction, so its momentum is p2 = -(0.5 x 6) = -3 kg m/s.
    3. The change in momentum is p2 - p1 = -3 - 4 = -7 kg m/s, a magnitude of 7 kg m/s.
    4. Force = change in momentum / time = 7 / 0.02 = 350 N, so the answer is B.
    • Why not A: Treats the rebound as if the ball kept moving in its original direction, simply subtracting the two momenta (0.5 x 6 - 0.5 x 8 = -1, magnitude 1 kg m/s, giving 1/0.02 = 50 N), instead of recognising that the ball's momentum REVERSES direction on rebound, so the two momenta must be combined as opposing quantities, not simply subtracted as if they pointed the same way.
    • Why not C: Uses only the ball's momentum before the collision, 0.5 x 8 = 4 kg m/s, and ignores its momentum after rebounding entirely, giving 4/0.02 = 200 N; the force depends on the CHANGE in momentum, which requires both the before and after values.
    • Why not D: Correctly finds the change in momentum, 7 kg m/s, but reports this value directly as the force, forgetting that force = change in momentum / time, so this impulse-sized value must still be divided by the 0.02 s collision time to get the force in newtons.

  9. Question 9Answer: A

    1. Before the explosion, the shell is at rest, so its total momentum is zero.
    2. By conservation of momentum, the total momentum after the explosion must also be zero, so the two fragments' momenta must be equal in magnitude and opposite in direction.
    3. The first fragment's momentum has magnitude 1 x 20 = 20 kg m/s, so the second fragment's momentum must also have magnitude 20 kg m/s (in the opposite direction).
    4. The second fragment's speed is therefore 20 / 4 = 5 m/s, so the answer is A.
    • Why not B: Assumes both fragments must move off at the same speed, ignoring that conservation of momentum requires the lighter fragment to move faster than the heavier one so that their momenta can balance out to the shell's original zero momentum.
    • Why not C: Scales the first fragment's speed by the mass ratio the wrong way round, multiplying by 4 instead of dividing by 4 (20 x 4 = 80); the heavier fragment must move SLOWER, not faster, for the momenta to balance, so the ratio must reduce the speed, not increase it.
    • Why not D: Divides the first fragment's momentum by the total mass of the original shell, 5 kg, rather than by the second fragment's own mass, 4 kg (1 x 20 / 5 = 4); the second fragment's speed must come from dividing by ITS OWN mass alone, since after the explosion the two fragments are separate objects.

  10. Question 10Answer: D

    1. The car's kinetic energy before braking is KE = (1/2)mv^2 = (1/2) x 1000 x 20^2 = (1/2) x 1000 x 400 = 200000 J.
    2. As the car brakes to rest, all of this kinetic energy is transferred as work done against the braking force, over the 50 m stopping distance: work done = force x distance.
    3. So 200000 = F x 50, giving F = 200000 / 50 = 4000 N.
    4. The magnitude of the braking force is 4000 N, so the answer is D.
    • Why not A: Omits the factor of 1/2 from the kinetic energy formula, computing 1000 x 20^2 = 400000 J instead of (1/2) x 1000 x 20^2 = 200000 J, which doubles the true kinetic energy and so also doubles the calculated braking force.
    • Why not B: Uses the car's speed, not its square, in the kinetic energy formula ((1/2) x 1000 x 20 = 10000 J instead of (1/2) x 1000 x 20^2 = 200000 J), giving a far smaller and incorrect kinetic energy to work from.
    • Why not C: Reports the car's kinetic energy, 200000 J, directly as the force in newtons, forgetting that work = force x distance means this energy value must still be divided by the 50 m braking distance to find the force.

  11. Question 11Answer: C

    1. The useful energy output is the gravitational potential energy gained by the load: E = mgh = 300 x 10 x 8 = 24000 J.
    2. Percentage efficiency = (useful energy output / total energy input) x 100.
    3. Substituting the values: efficiency = (24000 / 30000) x 100 = 0.8 x 100 = 80%.
    4. The crane's percentage efficiency is 80%, so the answer is C.
    • Why not A: Inverts the efficiency fraction, dividing the total energy supplied by the useful energy output instead of the other way round (30000/24000 x 100 = 125); percentage efficiency = (useful output / total input) x 100, with the useful output as the numerator, and it can never exceed 100%.
    • Why not B: Omits g when finding the useful energy output, using mass x height alone (300 x 8 = 2400 J) instead of the gravitational potential energy gained, mgh = 300 x 10 x 8 = 24000 J; height risen alone is not energy, since g must convert the mass into a weight force first.
    • Why not D: Correctly finds the fraction of useful energy to total energy, 24000/30000 = 0.8, but forgets to multiply by 100 to express this as a percentage, leaving the decimal fraction itself as if it were already the percentage figure.

  12. Question 12Answer: B

    1. A constant velocity means zero acceleration, and by Newton's second law, zero acceleration means the resultant force on the balloon is zero.
    2. The forces acting on the rising balloon are upthrust (upward), weight (downward), and drag (downward, since it opposes the balloon's upward motion through the air).
    3. For the resultant to be zero, the upward upthrust must exactly balance the sum of the two downward forces: upthrust = weight + drag.
    4. This balance of all three forces, not the absence or partial presence of any of them, is what is consistent with the balloon's constant velocity, so the answer is B.
    • Why not A: Confuses constant velocity with the idea that motion needs an ongoing nonzero resultant force to sustain it; Newton's first law says the opposite, that a constant velocity is exactly what happens when the resultant force IS zero, with no further net push needed to keep the balloon rising steadily.
    • Why not C: Assumes drag can only exist while an object is accelerating; in fact drag acts on any object moving relative to the air around it, whether or not it is accelerating, and here it is precisely the drag opposing the balloon's steady upward motion that helps balance the upthrust.
    • Why not D: Assumes a constant velocity means the forces themselves have vanished; a zero RESULTANT force means the individual forces (upthrust, weight, drag) are still all present and still acting, but now balance each other exactly, rather than having disappeared.

  13. Question 13Answer: D

    1. Inertia is the property of an object, arising from its mass, that resists a change in its state of motion.
    2. The loaded trolley has a greater mass than the empty trolley, and so has greater inertia.
    3. For the same applied force over the same time, the trolley with greater inertia (the loaded one) undergoes a smaller change in velocity than the trolley with less inertia (the empty one).
    4. So the loaded trolley ends up moving more slowly than the empty trolley after the identical push, so the answer is D.
    • Why not A: Gets the direction of the effect backwards: a trolley with greater inertia resists a CHANGE in its motion more strongly, meaning it ends up moving SLOWER for the same push, not faster; there is no mechanism by which greater resistance to a force somehow produces a greater resulting speed.
    • Why not B: Confuses Newton's first law, which qualitatively states that a resultant force is needed to change an object's velocity, with the separate fact (from Newton's second law) that the SIZE of that change in velocity, for a given force and time, depends on the object's mass; equal forces on different masses do not produce equal changes in velocity.
    • Why not C: Wrongly restricts inertia to objects already in motion; inertia is a property of an object's mass that resists ANY change in its state of motion, including being set in motion from rest, so both stationary trolleys' inertia is fully relevant to how they respond to the push.

  14. Question 14Answer: A

    1. Average speed = total distance travelled / total time taken. Total distance = 300 + 300 = 600 m, and total time = 60 + 40 = 100 s, so average speed = 600/100 = 6 m/s.
    2. Average velocity = displacement / total time. Displacement is the straight-line distance from the drone's overall starting point to its overall finishing point.
    3. Since the drone flies out and then back along the same path to its base, its final position is the same as its starting position, so its overall displacement for the whole trip is 0 m.
    4. Average velocity = 0 m / 100 s = 0 m/s, so average speed is 6 m/s and average velocity is 0 m/s, matching option A.
    • Why not B: Treats average velocity as if it were always the same as average speed, ignoring that the drone returns to its exact starting point, so its overall DISPLACEMENT for the whole trip is zero, however far it actually travelled.
    • Why not C: Uses only the outward leg's displacement, 300 m, divided by the TOTAL time of 100 s, rather than recognising that the drone's overall displacement for the round trip is zero, since it ends back where it started.
    • Why not D: Finds average speed by simply averaging the two leg speeds (300/60 = 5 m/s and 300/40 = 7.5 m/s, averaging to 6.25 m/s) instead of using total distance divided by total time, which is the definition average speed actually requires and which gives a different value here.

  15. Question 15Answer: C

    1. For a straight-line force-extension graph passing through the origin, the gradient is (change in force) / (change in extension).
    2. Hooke's law states F = kx, so rearranging gives k = F/x, which is exactly this gradient.
    3. Substituting the values read from the graph: k = 12/0.04 = 300 N/m.
    4. The gradient of the graph represents the spring constant, and here it equals 300 N/m, so the answer is C.
    • Why not A: Confuses the GRADIENT of a force-extension graph with the AREA under it; the area under a force-extension graph (up to a given extension) represents the elastic potential energy stored, but the gradient itself represents something different, the spring constant.
    • Why not B: Correctly identifies that the gradient represents the spring constant, but inverts the ratio used to calculate it, dividing extension by force instead of force by extension; the gradient of a graph is always (change in the quantity on the vertical axis) / (change in the quantity on the horizontal axis), which here is force / extension, not the other way round.
    • Why not D: Replaces the division that defines a gradient with simple addition of the two axis values; a gradient is always found by dividing one change by another, never by adding the raw values read off the two axes together.

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