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Admissions tests / ESAT / Physics / Mechanics

Foundation. 15 questions, 15 marks, about 22 minutes.

ESAT Physics: Mechanics, set 1

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.

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

    A car travels around a circular test track at a constant speed. Which of the following correctly explains why the car's velocity is not constant, even though its speed is constant?

    1. A Velocity is a vector, so it depends on direction as well as magnitude; the car's direction of travel is continuously changing, so its velocity is continuously changing even though its speed is constant.
    2. B Velocity is a scalar, so it depends only on magnitude; because the magnitude (speed) is constant, the velocity must also be constant.
    3. C The car's acceleration is zero, so its velocity must be constant.
    4. D The car's velocity is not constant only because its speed is gradually decreasing due to friction, not because of its direction.
  2. 21 mark

    A drone flies 30 m due east from its launch point, then 40 m due north, landing at a point 50 m in a straight line from the launch point. The whole flight takes 10 seconds. What is the drone's average speed, and what is the magnitude of its average velocity?

    1. A Average speed = 5 m/s, average velocity = 7 m/s.
    2. B Average speed = 7 m/s, average velocity = 5 m/s.
    3. C Average speed = average velocity = 7 m/s.
    4. D Average speed = average velocity = 5 m/s.
  3. 31 mark

    A skateboarder starts down a ramp with an initial velocity of 8 m/s and accelerates uniformly at 3 m/s^2 over a distance of 6 m. What is her velocity at the bottom of the ramp?

    1. A 100 m/s
    2. B 6 m/s
    3. C 10 m/s
    4. D 26 m/s
  4. 41 mark

    A box of weight 40 N rests on a rough horizontal floor. A person pushes it with a horizontal force of 25 N, and friction acts on the box with a force of 15 N, opposing the push. The normal contact force from the floor is 40 N upward. What is the magnitude of the resultant (net) force acting on the box?

    1. A 0 N
    2. B 25 N
    3. C 80 N
    4. D 10 N
  5. 51 mark

    A cyclist increases her speed while cycling on a flat road. Which statement correctly describes what happens to the size of the air resistance (drag) force acting on her, and why?

    1. A Drag stays the same regardless of speed, because it depends only on the cyclist's frontal area and shape, not her speed.
    2. B Drag increases as speed increases, because at a higher speed the cyclist collides with more air particles per second and with greater force each time.
    3. C Drag decreases as speed increases, because the cyclist spends less time exposed to the air at each point along the road.
    4. D Drag decreases as speed increases, because moving faster reduces the density of air the cyclist experiences.
  6. 61 mark

    A spring obeys Hooke's law. A force of 15 N produces an extension of 3 cm. Assuming the spring remains within its limit of proportionality, what extension would a force of 60 N produce?

    1. A 12 cm
    2. B 6 cm
    3. C 0.75 cm
    4. D 3 cm
  7. 71 mark

    A spring, obeying Hooke's law, has a spring constant of 200 N/m. It is stretched by an extension of 0.05 m, within its limit of proportionality. How much elastic potential energy is stored in the spring?

    1. A 0.5 J
    2. B 10 J
    3. C 0.25 J
    4. D 5 J
  8. 81 mark

    A resultant force of 24 N acts on an object of mass 6 kg, initially at rest. What is the object's acceleration?

    1. A 144 m/s^2
    2. B 0.25 m/s^2
    3. C 18 m/s^2
    4. D 4 m/s^2
  9. 91 mark

    A book rests on a table. According to Newton's third law, which of the following correctly identifies the reaction force paired with the table's normal contact force pushing up on the book?

    1. A The book pushing down on the table, with an equal and opposite normal contact force, because Newton's third law pairs act on different objects and are of the same type of force.
    2. B The weight of the book, acting downward on the book, because the weight and the normal force are equal and opposite and act on the book itself.
    3. C The table's weight acting downward on the table, because both forces are examples of weight and must be equal by Newton's third law.
    4. D There is no reaction force in this case, because the book is stationary and Newton's third law applies only to accelerating objects.
  10. 101 mark

    An astronaut has a mass of 80 kg. Using g = 10 N/kg on Earth, what is her weight on Earth's surface?

    1. A 8 N
    2. B 800 N
    3. C 90 N
    4. D 80 kg
  11. 111 mark

    A skydiver jumps from a plane and falls, eventually reaching terminal velocity before opening her parachute. Which statement correctly describes the forces acting on her at the moment she reaches terminal velocity?

    1. A Her weight is greater than air resistance, so she continues to accelerate at a constant rate even at terminal velocity.
    2. B Air resistance is greater than her weight, so she begins to decelerate and eventually stops falling completely.
    3. C There are no forces acting on her at terminal velocity, which is why her velocity no longer changes.
    4. D Her weight and air resistance are equal in size and act in opposite directions, so the resultant force on her is zero and she falls at a constant velocity.
  12. 121 mark

    A car of mass 1200 kg travels at a velocity of 15 m/s. What is its momentum?

    1. A 1215 kg m/s
    2. B 80 kg m/s
    3. C 18000 kg m/s
    4. D 18 kg m/s
  13. 131 mark

    A trolley of mass 4 kg moving at 6 m/s collides with a stationary trolley of mass 2 kg. The two trolleys stick together after the collision. What is their common velocity immediately after the collision?

    1. A 12 m/s
    2. B 4 m/s
    3. C 8 m/s
    4. D 6 m/s
  14. 141 mark

    A crate of mass 5 kg is lifted vertically through a height of 4 m. Using g = 10 N/kg, how much gravitational potential energy does the crate gain?

    1. A 200 J
    2. B 20 J
    3. C 9 J
    4. D 50 J
  15. 151 mark

    An electric motor transfers 250 J of energy in total, of which 200 J is usefully transferred to kinetic energy and the rest is wasted as heat and sound. What is the percentage efficiency of the motor?

    1. A 20%
    2. B 200%
    3. C 80%
    4. D 125%

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: A

    1. Speed is a scalar quantity, describing only how fast an object is moving (its magnitude), with no reference to direction.
    2. Velocity is a vector quantity, describing both how fast an object is moving and the direction in which it moves.
    3. As the car travels around the circular track, its direction of travel constantly changes, even though the magnitude of its speed stays the same.
    4. Because velocity depends on direction as well as magnitude, a changing direction means the velocity is continuously changing, so the answer is A.
    • Why not B: This treats velocity as a scalar quantity, confusing it with speed. Velocity is a vector, defined by both magnitude and direction, so a constant magnitude alone does not guarantee a constant velocity.
    • Why not C: This assumes that constant speed means zero acceleration. But acceleration is the rate of change of velocity, not speed; since the direction of velocity is continuously changing, the car has a nonzero acceleration even though its speed is constant.
    • Why not D: This is wrong on two counts: the question states the speed is constant, so friction is not slowing the car, and even if it were, the reason for the changing velocity here is the changing direction, not a changing speed.

  2. Question 2Answer: B

    1. The total distance travelled is the sum of each straight leg of the journey: 30 m + 40 m = 70 m.
    2. Average speed = total distance / time = 70 m / 10 s = 7 m/s.
    3. The displacement is the straight-line distance from the start point to the end point, given directly here as 50 m (the hypotenuse of the 30 m, 40 m right-angled path).
    4. Average velocity = displacement / time = 50 m / 10 s = 5 m/s, matching average speed 7 m/s and average velocity 5 m/s, so the answer is B.
    • Why not A: This swaps the two definitions around: it uses the straight-line displacement (50 m) to find the speed and the total path length (70 m) to find the velocity, when average speed must use the total distance travelled and average velocity must use the displacement.
    • Why not C: This assumes distance and displacement are the same and both equal 70 m divided by time, but displacement is the straight-line distance from start to finish (50 m), not the total path length flown (70 m); ignoring this difference conflates speed with velocity.
    • Why not D: This uses the displacement (50 m) for both calculations, correctly getting the average velocity of 5 m/s, but forgets that average speed must be calculated using the total distance actually flown (70 m), not the straight-line displacement.

  3. Question 3Answer: C

    1. The equation of motion linking velocity, acceleration and distance is v^2 = u^2 + 2as, where u is the initial velocity, a is the acceleration and s is the distance travelled.
    2. Substituting the given values: v^2 = 8^2 + (2 x 3 x 6) = 64 + 36 = 100.
    3. Taking the square root of both sides gives v = sqrt(100) = 10 m/s.
    4. The skateboarder's velocity at the bottom of the ramp is 10 m/s, so the answer is C.
    • Why not A: This correctly calculates v^2 = u^2 + 2as = 8^2 + (2 x 3 x 6) = 64 + 36 = 100, but then reports 100 as the final velocity in m/s instead of taking the square root to find v itself; 100 is v^2, not v.
    • Why not B: This uses v = sqrt(2as) = sqrt(2 x 3 x 6) = sqrt(36) = 6 m/s, which finds the speed as if the skateboarder started from rest; it drops the u^2 term from the equation entirely, ignoring the given initial velocity of 8 m/s.
    • Why not D: This uses v = u + a x s = 8 + (3 x 6) = 26 m/s, mistaking the equation v = u + at and substituting the distance s where the time t belongs; s is a distance in metres, not a time in seconds, so this mixes two different equations of motion.

  4. Question 4Answer: D

    1. Forces acting in the same direction add together, and forces acting in opposite directions are subtracted to find the resultant in that direction.
    2. Vertically, the weight (40 N down) and the normal contact force (40 N up) are equal and opposite, so they cancel, giving zero resultant force in the vertical direction.
    3. Horizontally, the push (25 N forward) and friction (15 N backward, opposing the motion) act in opposite directions, so the resultant horizontal force is 25 N - 15 N = 10 N.
    4. Since the vertical resultant is zero, the overall resultant force on the box is 10 N, horizontally in the direction of the push, so the answer is D.
    • Why not A: This assumes that because the box is not accelerating vertically (weight and normal force cancel), the resultant force overall must be zero; but the push and friction act horizontally and do not balance each other (25 N vs 15 N), so there is a nonzero resultant force in the horizontal direction.
    • Why not B: This uses only the push force and ignores friction entirely; friction acts to oppose the push, so it must be subtracted from the push force to find the true resultant force in the horizontal direction, not ignored.
    • Why not C: This adds together the magnitudes of the weight, push and friction forces (40 + 25 + 15 = 80 N) without considering that these forces act in different directions; forces acting in opposite directions must be subtracted, not simply added, to find a resultant, and the box's weight is balanced separately by the normal contact force so does not contribute to the horizontal resultant at all.

  5. Question 5Answer: B

    1. Air resistance (drag) is a contact force caused by an object colliding with air particles as it moves through the air.
    2. Increasing speed increases both the number of air particles the cyclist meets each second and the force with which she collides with each one.
    3. Both effects act together to increase the total drag force acting on the cyclist as her speed increases.
    4. So the air resistance force increases as speed increases, so the answer is B.
    • Why not A: Frontal area and shape do affect the size of the drag force, but so does speed: moving faster through the air increases the rate and force of collisions with air particles, so drag is not independent of speed.
    • Why not C: This gets the relationship backwards: air resistance increases with speed, because at higher speeds the cyclist meets more air particles per second and pushes each one aside with greater force, not less.
    • Why not D: The density of the air does not change because the cyclist is moving faster; it is a property of the air itself (affected by altitude, temperature and so on), not by the cyclist's speed, so this reasoning is incorrect.

  6. Question 6Answer: A

    1. Hooke's law states that extension is directly proportional to the applied force, F = kx, provided the limit of proportionality is not exceeded.
    2. The spring constant is found from the first measurement: k = F/x = 15 N / 3 cm = 5 N/cm.
    3. Since 60 N is four times the original 15 N force, and extension is directly proportional to force, the new extension is four times the original: 3 cm x 4 = 12 cm.
    4. Alternatively, x2 = F2/k = 60 N / 5 N/cm = 12 cm, confirming the answer is A.
    • Why not B: This treats extension as proportional to the SQUARE ROOT of the applied force rather than directly proportional to it; Hooke's law states F = kx, a direct (linear) proportion between force and extension, not a square-root relationship.
    • Why not C: This treats extension as INVERSELY proportional to force, so that a larger force gives a smaller extension; Hooke's law states that extension increases as force increases (F = kx), so a force four times as large produces an extension four times as large, not four times smaller.
    • Why not D: This confuses the limit of proportionality, the point beyond which Hooke's law (F = kx) no longer holds and extension stops increasing in direct proportion to force, with a fixed maximum extension the spring can never exceed; the question states the spring stays within its limit of proportionality, so Hooke's law still applies and the extension does increase with the larger force.

  7. Question 7Answer: C

    1. The elastic potential energy stored in a stretched spring, within its limit of proportionality, is given by E = (1/2) k x^2, where k is the spring constant and x is the extension.
    2. Substituting the values given: E = (1/2) x 200 x (0.05)^2.
    3. 0.05^2 = 0.0025, so E = (1/2) x 200 x 0.0025 = 100 x 0.0025 = 0.25 J.
    4. The elastic potential energy stored in the spring is 0.25 J, so the answer is C.
    • Why not A: This uses E = k x^2 = 200 x 0.05^2 = 0.5 J, omitting the factor of 1/2 from the correct formula E = (1/2) k x^2, giving a value exactly double the true energy stored.
    • Why not B: This correctly calculates the FORCE needed to produce this extension, using F = kx = 200 x 0.05 = 10 N, but then reports this force, in newtons, as if it were the energy stored, in joules; force and energy are different quantities with different formulas and different units.
    • Why not D: This uses E = (1/2) k x = 0.5 x 200 x 0.05 = 5 J, using the extension itself instead of its square; the energy formula needs x^2 (extension squared), not x, because the force needed to stretch the spring increases as the extension increases, so the average force must be combined with the extension, not the extension alone.

  8. Question 8Answer: D

    1. Newton's second law states that the resultant force on an object equals its mass multiplied by its acceleration: F = ma.
    2. Rearranging this equation to find acceleration gives a = F/m.
    3. Substituting the given values: a = 24 N / 6 kg = 4 m/s^2.
    4. The object's acceleration is 4 m/s^2, so the answer is D.
    • Why not A: This multiplies the force and the mass together (24 x 6 = 144) instead of dividing; Newton's second law is F = ma, so acceleration is found by dividing force by mass, a = F/m, not by multiplying them.
    • Why not B: This inverts the formula, calculating mass divided by force (6/24 = 0.25) instead of force divided by mass; rearranging F = ma for acceleration correctly gives a = F/m, not m/F.
    • Why not C: This subtracts the mass from the force (24 - 6 = 18) as if acceleration were found by simple subtraction; force, mass and acceleration are related by the equation F = ma, a multiplicative relationship, not an additive or subtractive one.

  9. Question 9Answer: A

    1. Newton's third law states that if body A exerts a force on body B, then body B exerts an equal and opposite force of the same type on body A.
    2. The table exerts a normal contact force upward on the book; by Newton's third law, the paired reaction force must be the book exerting an equal and opposite normal contact force downward on the table.
    3. This is different from the weight of the book (the pull of gravity on the book) and the normal force from the table, which are merely two different forces that happen to balance on the SAME object (the book), keeping it in equilibrium; this is Newton's first law in action, not a third law pair.
    4. The correct Newton's third law pair to the table's normal force on the book is the book's normal force on the table, so the answer is A.
    • Why not B: This pairs the normal contact force with the book's weight, because they happen to be equal in magnitude and opposite in direction while the book is stationary; but a genuine Newton's third law pair must act on two DIFFERENT objects, and both weight and the normal force act on the same object, the book, so they cannot be a third law pair. They are simply balanced (this is why the book does not accelerate), not action-reaction partners.
    • Why not C: This assumes the reaction force must also be a weight force, because both are 'the same type' in a loose sense; but Newton's third law pairs must be the exact same TYPE of force acting between the same two objects in opposite directions (here, a normal contact force), not merely forces belonging to the same broad category on unrelated objects.
    • Why not D: Newton's third law applies to every pair of interacting objects regardless of whether they are moving or stationary; a stationary book still exerts a normal contact force down on the table, and the table exerts an equal and opposite normal contact force up on the book, which is why the book stays at rest.

  10. Question 10Answer: B

    1. Weight is the force of gravity acting on an object's mass, calculated using w = mg, where m is the mass in kg and g is the gravitational field strength in N/kg.
    2. On Earth's surface, g is approximately 10 N/kg.
    3. Substituting the given mass: w = 80 kg x 10 N/kg = 800 N.
    4. The astronaut's weight on Earth's surface is 800 N, so the answer is B.
    • Why not A: This divides the mass by g (80/10=8) instead of multiplying; the formula is w = mg, so mass must be MULTIPLIED by gravitational field strength, not divided by it.
    • Why not C: This adds the mass and g together (80+10=90) instead of multiplying them; weight is found by multiplying mass by gravitational field strength, w=mg, not by adding the two numbers.
    • Why not D: This confuses mass with weight, giving the answer in kg rather than newtons; mass (in kg) is the amount of matter in an object and stays constant everywhere, while weight (in N) is the force of gravity acting on that mass and depends on the gravitational field strength, so the two are different quantities with different units.

  11. Question 11Answer: D

    1. As the skydiver falls, her weight (a constant force) pulls her downward, while air resistance opposes her motion and increases as her speed increases.
    2. Initially her weight is greater than air resistance, so there is a resultant downward force and she accelerates, gaining speed.
    3. As her speed increases, air resistance increases too, until eventually it becomes equal in size to her weight.
    4. At this point the resultant force is zero, so she stops accelerating and falls at a constant velocity, called terminal velocity, so the answer is D.
    • Why not A: If her weight were still greater than air resistance, there would be a nonzero resultant force acting downward, and she would continue to accelerate; but 'terminal velocity' specifically means her velocity has stopped increasing, which only happens once the resultant force becomes zero, when the two forces become equal.
    • Why not B: If air resistance permanently exceeded her weight, the resultant force would act upward, decelerating her; in fact, as she speeds up, air resistance increases until it exactly equals her weight, at which point the forces balance and her velocity becomes constant, rather than her stopping or continuing to slow down further.
    • Why not C: Weight and air resistance are still both acting on the skydiver at terminal velocity; a constant velocity (zero acceleration) does not mean there are no forces acting, it means the forces are perfectly balanced, giving a zero RESULTANT force, which is different from there being no forces at all.

  12. Question 12Answer: C

    1. Momentum is defined as the product of an object's mass and its velocity: p = mv.
    2. The car has a mass of 1200 kg and a velocity of 15 m/s.
    3. Substituting these values: p = 1200 kg x 15 m/s = 18000 kg m/s.
    4. The car's momentum is 18000 kg m/s, so the answer is C.
    • Why not A: This adds the mass and velocity together (1200 + 15 = 1215) instead of multiplying; momentum is defined as p = mv, the PRODUCT of mass and velocity, not their sum.
    • Why not B: This divides the mass by the velocity (1200/15=80) instead of multiplying them; the definition of momentum is p=mv, so mass must be multiplied by velocity, not divided by it.
    • Why not D: This effectively uses a mass of 1.2 instead of 1200 (1.2 x 15 = 18), as if the mass had been given in tonnes rather than kilograms without converting back; momentum must be calculated using the mass in kilograms, giving an answer 1000 times larger than this.

  13. Question 13Answer: B

    1. In a collision where objects stick together, momentum is conserved: total momentum before the collision equals total momentum after.
    2. Momentum before the collision: the moving trolley has p = mv = 4 kg x 6 m/s = 24 kg m/s; the stationary trolley has p = 2 kg x 0 m/s = 0 kg m/s. Total momentum before = 24 kg m/s.
    3. After the collision the trolleys move together as one object of combined mass 4 kg + 2 kg = 6 kg.
    4. Using conservation of momentum, 24 kg m/s = 6 kg x v, so v = 24/6 = 4 m/s, giving answer B.
    • Why not A: This divides the total momentum (24 kg m/s) by the mass of the stationary trolley alone (2 kg), giving 12 m/s; but after the collision the trolleys move together as one combined object, so momentum must be divided by their TOTAL combined mass (4 kg + 2 kg = 6 kg), not by one trolley's mass alone.
    • Why not C: This averages the two masses, (4 + 2)/2 = 3 kg, and divides the total momentum by this average (24/3=8), instead of using the actual combined mass of 6 kg; when two objects stick together, their masses ADD, they are not averaged.
    • Why not D: This assumes the final velocity is simply the same as the moving trolley's original velocity (6 m/s), as if the stationary trolley had no effect at all; but momentum must be conserved and shared between the combined, larger mass, which slows the final velocity below the original 6 m/s.

  14. Question 14Answer: A

    1. Gravitational potential energy is calculated using E = mgh, where m is mass, g is the gravitational field strength, and h is the height risen.
    2. The crate has a mass of 5 kg and is lifted through a height of 4 m, with g = 10 N/kg.
    3. Substituting these values: E = 5 kg x 10 N/kg x 4 m = 200 J.
    4. The crate gains 200 J of gravitational potential energy, so the answer is A.
    • Why not B: This calculates m x h = 5 x 4 = 20 J, leaving out the gravitational field strength g entirely; the formula for gravitational potential energy is E = mgh, and all three quantities, mass, g and height, must be multiplied together.
    • Why not C: This adds the mass and the height together (5 + 4 = 9) instead of multiplying all three quantities in the formula E = mgh; gravitational potential energy is found by MULTIPLYING mass, gravitational field strength and height, not by adding any of them.
    • Why not D: This calculates m x g = 5 x 10 = 50 J, leaving out the height entirely; the formula for gravitational potential energy needs the height risen, h, as well as mass and g, since more energy is gained the higher the object is lifted.

  15. Question 15Answer: C

    1. Percentage efficiency is calculated using efficiency = (useful energy output / total energy input) x 100.
    2. Here, the useful energy output is 200 J (the kinetic energy transferred) and the total energy input is 250 J.
    3. Substituting these values: efficiency = (200/250) x 100 = 0.8 x 100 = 80%.
    4. The motor's percentage efficiency is 80%, so the answer is C.
    • Why not A: This calculates the WASTED energy's share of the total (50/250 x 100 = 20%), rather than the USEFUL energy's share; percentage efficiency is defined using the useful energy output, not the wasted energy.
    • Why not B: This reports the useful energy value itself (200) directly as a percentage, without dividing by the total energy input at all; percentage efficiency must always be calculated as a FRACTION of the total input, converted to a percentage, not simply read off as the raw joule value.
    • Why not D: This divides the total energy input by the useful energy output (250/200 x 100 = 125%) instead of the other way round; efficiency is defined as (useful output / total input) x 100, so the useful energy must be the numerator, not the denominator, and efficiency also cannot exceed 100%, since a device can never output more useful energy than it takes in.

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