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Admissions tests / ESAT / Physics / Mechanics

Test standard. 15 questions, 15 marks, about 25 minutes.

ESAT Physics: Mechanics, set 2

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

    An object's velocity-time graph is a straight line rising uniformly from 0 m/s at t = 0 to 24 m/s at t = 6 s, after which its velocity stays constant at 24 m/s from t = 6 s to t = 10 s. What is the total distance travelled by the object over the whole 10 seconds?

    1. A 240 m
    2. B 120 m
    3. C 168 m
    4. D 72 m
  2. 21 mark

    A runner jogs once around a rectangular field 80 m long and 60 m wide, starting and finishing at the same corner. The lap takes 70 seconds. What is her average speed, and what is the magnitude of her average velocity, for the whole lap?

    1. A Average speed = 4 m/s, average velocity = 0 m/s.
    2. B Average speed = average velocity = 4 m/s.
    3. C Average speed = 0 m/s, average velocity = 4 m/s.
    4. D Average speed = 4 m/s, average velocity = 2 m/s.
  3. 31 mark

    A car brakes to a stop from a velocity of 20 m/s, decelerating uniformly, and travels 40 m before coming to rest. What is the magnitude of the car's deceleration?

    1. A 10 m/s^2
    2. B 0.25 m/s^2
    3. C 0.5 m/s^2
    4. D 5 m/s^2
  4. 41 mark

    A lift cable exerts an upward tension of 9000 N on a lift of weight 8000 N. What is the resultant force acting on the lift, and in which direction does the lift accelerate?

    1. A 17000 N, upward
    2. B 1000 N, upward
    3. C 1000 N, downward
    4. D 9000 N, upward
  5. 51 mark

    A boat floats stationary on a still lake, with its engine switched off. Which pair of forces must be equal in magnitude for the boat to remain in vertical equilibrium?

    1. A The boat's weight and the horizontal thrust from its engine.
    2. B The boat's weight and the drag force from the water.
    3. C The upthrust from the water and the horizontal thrust from its engine.
    4. D The boat's weight and the upthrust from the water.
  6. 61 mark

    A spring, obeying Hooke's law, extends by 4 cm when a force of 20 N is applied to it. Assuming the spring remains within its limit of proportionality, what force would be needed to produce an extension of 10 cm?

    1. A 50 N
    2. B 8 N
    3. C 26 N
    4. D 200 N
  7. 71 mark

    A force of 12 N stretches a spring, within its limit of proportionality, by 5 cm. How much elastic potential energy is stored in the spring?

    1. A 0.6 J
    2. B 30 J
    3. C 0.3 J
    4. D 60 J
  8. 81 mark

    A resultant force acts on a trolley of mass 2 kg, causing it to accelerate uniformly from rest to a velocity of 12 m/s in 3 seconds. What is the magnitude of the resultant force?

    1. A 24 N
    2. B 0.5 N
    3. C 2 N
    4. D 8 N
  9. 91 mark

    A passenger stands on a bus moving at a constant velocity in a straight line. The bus suddenly brakes hard. Which of the following best explains, using Newton's first law, why the passenger lurches forward?

    1. A The braking action of the bus creates a new forward force that pushes the passenger forward.
    2. B The passenger's body tends to continue moving at its original velocity due to inertia, because no sufficiently large force acts on the passenger to decelerate them at the same rate as the bus.
    3. C The passenger lurches forward because gravity pulls them forward as the bus slows down.
    4. D The frictional force between the passenger's feet and the floor of the bus suddenly increases, pushing the passenger forward.
  10. 101 mark

    An object has a mass of 20 kg on Earth, where g = 10 N/kg. The object is transported to another planet, where the gravitational field strength is 4 N/kg. What is the object's weight on this new planet, and how does its mass there compare with its mass on Earth?

    1. A Weight = 80 N; the object's mass on the new planet is less than on Earth, because the gravitational field strength there is weaker.
    2. B Weight = 5 N; the object's mass is the same as on Earth.
    3. C Weight = 80 N; the object's mass on the new planet is the same as its mass on Earth (20 kg), because mass does not depend on gravitational field strength.
    4. D Weight = 20 N; the object's mass on the new planet is greater than on Earth, because a weaker gravitational field means more mass is needed to anchor the object.
  11. 111 mark

    Two steel balls of the same mass are dropped simultaneously from the same height in air. One is a compact sphere; the other has been flattened into a disc, giving it a much larger surface area exposed to the air. Which one reaches the ground first, and why?

    1. A The compact sphere reaches the ground first, because its smaller surface area means it experiences less air resistance at a given speed than the larger, flattened disc.
    2. B The flattened disc reaches the ground first, because its larger surface area increases the downward force of gravity acting on it.
    3. C Both reach the ground at exactly the same time, because they have the same mass, and objects of the same mass always fall at the same rate regardless of shape.
    4. D The flattened disc reaches the ground first, because its greater surface area lets it catch more air, pushing it downward faster, similar to how a sail catches wind.
  12. 121 mark

    A hockey ball of mass 0.15 kg is struck by a stick, and its velocity changes from 0 m/s to 20 m/s while it is in contact with the stick for 0.05 s. What is the average force exerted on the ball by the stick?

    1. A 3 N
    2. B 20 N
    3. C 0.15 N
    4. D 60 N
  13. 131 mark

    A stationary ice skater of mass 60 kg pushes away from a stationary skater of mass 40 kg, so both move off in opposite directions. If the 40 kg skater moves off at 3 m/s, what is the speed of the 60 kg skater?

    1. A 1.2 m/s
    2. B 3 m/s
    3. C 2 m/s
    4. D 4.5 m/s
  14. 141 mark

    A worker pushes a crate 6 m across a warehouse floor using a horizontal force of 80 N. A frictional force of 30 N acts on the crate as it moves, opposing the push. How much of the work done by the worker is usefully transferred to the crate's kinetic energy, and how much is wasted overcoming friction?

    1. A Useful = 480 J, wasted = 180 J
    2. B Useful = 300 J, wasted = 180 J
    3. C Useful = 480 J, wasted = 30 J
    4. D Useful = 180 J, wasted = 300 J
  15. 151 mark

    An electric heater transfers 9000 J of energy to a room over a time of 3 minutes. What is the power output of the heater?

    1. A 50 W
    2. B 3000 W
    3. C 0.02 W
    4. D 30 W

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. The area under a velocity-time graph gives the distance travelled.
    2. From t = 0 to t = 6 s, the graph is a straight line from 0 to 24 m/s, so this section forms a triangle: distance = 1/2 x base x height = 1/2 x 6 x 24 = 72 m.
    3. From t = 6 s to t = 10 s, the velocity is constant at 24 m/s for 4 s, forming a rectangle: distance = 24 x 4 = 96 m.
    4. Total distance = 72 + 96 = 168 m, so the answer is C.
    • Why not A: This treats the first 6 seconds as a rectangle of height 24 m/s (6 x 24 = 144 m) rather than a triangle, then adds the 96 m from the constant-velocity phase (4 x 24 = 96), giving 240 m; but during uniform acceleration from rest the area under the graph is a TRIANGLE, so it must be halved (1/2 x 6 x 24 = 72 m), not treated as a full rectangle.
    • Why not B: This treats the whole 10 seconds as if the object accelerated uniformly throughout, using the average of the initial and final velocities, (0+24)/2 = 12 m/s, multiplied by the full 10 s (12 x 10 = 120 m); but the object only accelerates for the first 6 s, after which it travels at a CONSTANT 24 m/s for the remaining 4 s, so this single average cannot be applied across the whole motion.
    • Why not D: This calculates only the distance covered during the accelerating phase (1/2 x 6 x 24 = 72 m) and forgets to add the distance covered during the second phase, when the object continues at a constant 24 m/s for a further 4 seconds (4 x 24 = 96 m).

  2. Question 2Answer: A

    1. Average speed = total distance travelled / time. The runner covers the full perimeter of the field: 2 x (80 + 60) = 280 m.
    2. Average speed = 280 m / 70 s = 4 m/s.
    3. Average velocity = displacement / time. Because the runner starts and finishes at the same corner, her displacement for the complete lap is 0 m, so average velocity = 0/70 = 0 m/s.
    4. Her average speed is 4 m/s but her average velocity is 0 m/s, so the answer is A.
    • Why not B: This assumes average velocity equals average speed, but average velocity depends on displacement, not distance travelled. Because the runner finishes at exactly her starting point, her displacement for the whole lap is zero, so her average velocity must be zero, not 4 m/s.
    • Why not C: This swaps the two definitions around: it assigns the (zero) displacement to speed and the distance-based value to velocity, the wrong way round from their correct definitions.
    • Why not D: This assumes the displacement for the lap is half the total distance travelled (140 m), giving 140/70 = 2 m/s; but because the runner starts and finishes at the same point, her displacement for the complete lap is zero, not half the perimeter.

  3. Question 3Answer: D

    1. The equation of motion linking velocity, acceleration and distance is v^2 = u^2 + 2as, where u is the initial velocity, v is the final velocity, a is the acceleration and s is the distance.
    2. Here u = 20 m/s, v = 0 (the car stops) and s = 40 m, so 0 = 20^2 + 2 x a x 40, giving 0 = 400 + 80a.
    3. Rearranging: a = -400/80 = -5 m/s^2. The negative sign shows the acceleration acts opposite to the direction of motion, that is, it is a deceleration.
    4. The magnitude of the deceleration is 5 m/s^2, so the answer is D.
    • Why not A: This uses a = (u^2 - v^2)/s, forgetting to double the distance in the denominator; the equation v^2 = u^2 + 2as requires 2s, not s, so this gives a value exactly twice the correct deceleration.
    • Why not B: This uses a = (u - v)/(2s), substituting u and v directly instead of their squares; the equation requires u^2 and v^2 (400 and 0), not u and v (20 and 0), in the numerator.
    • Why not C: This computes a = u/s directly, both forgetting to square u and forgetting to double s in the denominator, combining two separate errors.

  4. Question 4Answer: B

    1. The tension in the cable (9000 N) acts upward, and the lift's weight (8000 N) acts downward; these are the only two forces acting on the lift, in opposite directions.
    2. The resultant force is found by subtracting the smaller force from the larger: 9000 N - 8000 N = 1000 N.
    3. Because the tension, the larger force, acts upward, the resultant force of 1000 N also acts upward.
    4. So the lift accelerates upward under a resultant force of 1000 N, and the answer is B.
    • Why not A: This adds the tension and weight together (9000 + 8000 = 17000 N), treating them as acting in the same direction; but tension acts upward while weight acts downward, so these two forces act in OPPOSITE directions and must be subtracted, not added.
    • Why not C: This correctly finds the size of the resultant force (9000 - 8000 = 1000 N) but assigns it the wrong direction; because the tension (9000 N) is larger than the weight (8000 N), the resultant force must act in the tension's direction, upward, not downward.
    • Why not D: This reports the tension force alone (9000 N) as if it were the resultant, ignoring the lift's weight entirely; the resultant must account for both forces acting on the lift, so the weight (8000 N) must be subtracted from the tension, not ignored.

  5. Question 5Answer: D

    1. For the boat to be in vertical equilibrium (floating, stationary, neither sinking nor rising), the resultant vertical force on it must be zero.
    2. Weight acts vertically downward on the boat, due to gravity.
    3. Upthrust is the upward force exerted on the boat by the water it displaces.
    4. These are the only two forces acting vertically on the boat, so for equilibrium they must be equal in magnitude and opposite in direction, giving answer D.
    • Why not A: Thrust from an engine is a horizontal force that would drive the boat forward through the water; it plays no part in supporting the boat's weight vertically, which is instead balanced by the water's upthrust. (The engine is switched off here in any case.)
    • Why not B: Drag is a resistive horizontal force that opposes a boat's forward motion through the water; it does not act vertically, and so cannot balance the boat's weight, which acts vertically downward.
    • Why not C: Upthrust acts vertically while thrust acts horizontally; these two forces act in different, perpendicular directions and so cannot form a balancing pair with each other.

  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 = 20 N / 4 cm = 5 N/cm.
    3. For an extension of 10 cm, the required force is F = kx = 5 N/cm x 10 cm = 50 N.
    4. Alternatively, since 10 cm is 2.5 times the original 4 cm extension, the force is also 2.5 times as large: 20 N x 2.5 = 50 N, confirming the answer is A.
    • Why not B: This treats extension as inversely proportional to force, using force = 20 x (4/10) = 8 N; Hooke's law states force and extension are directly proportional (F = kx), so a larger extension requires a larger force, not a smaller one.
    • Why not C: This adds the difference in extension (10 - 4 = 6 cm) directly onto the original force (20 + 6 = 26), as if centimetres could simply be added to newtons; extension and force are different quantities, related through the spring constant, not additive to one another.
    • Why not D: This multiplies the original force by the new extension (20 x 10 = 200) without first finding the spring constant from the original data; this ignores that the original 4 cm extension is needed to establish the correct proportionality between force and extension.

  7. Question 7Answer: C

    1. The energy stored in a stretched spring, within its limit of proportionality, is given by E = (1/2)Fx, where F is the applied force and x is the extension.
    2. Converting the extension to metres: 5 cm = 0.05 m.
    3. Substituting the values: E = (1/2) x 12 x 0.05 = 6 x 0.05 = 0.3 J.
    4. The elastic potential energy stored in the spring is 0.3 J, so the answer is C.
    • Why not A: This uses E = Fx, omitting the factor of 1/2 from the correct formula E = (1/2)Fx, giving exactly double the true energy stored.
    • Why not B: This uses the extension in centimetres (5) directly in the formula, rather than converting to metres (0.05 m) first; using SI units in E = (1/2)Fx, the extension must be in metres, not centimetres.
    • Why not D: This compounds two errors: it uses E = Fx instead of E = (1/2)Fx, dropping the factor of 1/2, and it also uses the extension in centimetres rather than converting to metres, giving an answer far too large on both counts.

  8. Question 8Answer: D

    1. First find the acceleration: a = (v - u)/t = (12 - 0)/3 = 4 m/s^2.
    2. Newton's second law states F = ma.
    3. Substituting the values: F = 2 kg x 4 m/s^2 = 8 N.
    4. The magnitude of the resultant force is 8 N, so the answer is D.
    • Why not A: This uses the final velocity (12 m/s) directly as the acceleration in F = ma, instead of first calculating the acceleration from a = (v - u)/t = 12/3 = 4 m/s^2; velocity and acceleration are different quantities and must not be substituted for one another.
    • Why not B: This divides mass by acceleration (2/4 = 0.5) instead of multiplying them; Newton's second law is F = ma, a multiplicative relationship, so mass must be multiplied by acceleration, not divided by it.
    • Why not C: This divides acceleration by mass (4/2 = 2) instead of multiplying mass by acceleration; rearranging F = ma requires multiplying m and a together, not dividing one by the other.

  9. Question 9Answer: B

    1. Newton's first law states that a body continues in its state of rest or uniform motion in a straight line unless acted on by a resultant external force.
    2. Before braking, both the bus and the passenger move at the same constant velocity.
    3. When the bus brakes, a decelerating force acts on the bus itself (through the brakes and the road), slowing it down, but this force does not act directly on the passenger.
    4. Because no sufficiently large force decelerates the passenger at the same rate, their body's inertia carries them forward at close to their original velocity, so they lurch forward relative to the now-slower bus, giving answer B.
    • Why not A: Braking applies a backward, decelerating force to the bus itself, through the brakes acting on the wheels and the road; this does not create any new forward force acting on the passenger, who is a separate object from the bus.
    • Why not C: Gravity acts vertically downward on the passenger at all times, including before the bus braked; it does not act horizontally, so it cannot explain a sudden forward horizontal lurch.
    • Why not D: If anything, friction between the passenger's feet and the floor acts to try to decelerate the passenger along with the bus, opposing the lurch, not causing it; the lurch happens precisely because this frictional force is not large enough to slow the passenger down at the same rate as the bus.

  10. Question 10Answer: C

    1. Weight is calculated using w = mg, where m is mass and g is gravitational field strength.
    2. On the new planet: w = 20 kg x 4 N/kg = 80 N.
    3. Mass is a measure of the amount of matter in an object and does not depend on the gravitational field strength acting on it; it stays the same everywhere, including on the new planet.
    4. So the object's weight on the new planet is 80 N, but its mass remains 20 kg, unchanged from its mass on Earth, so the answer is C.
    • Why not A: Mass is the amount of matter in an object and does not change with location; it is the object's weight that changes because the gravitational field strength is different, not its mass. This distractor confuses the correctly changing weight with an incorrectly assumed changing mass.
    • Why not B: This divides the mass by the gravitational field strength (20/4 = 5) instead of multiplying them; weight is found using w = mg, a multiplication, not a division.
    • Why not D: This uses the mass value (20) directly as if it were already the weight in newtons, without multiplying by g at all, and also wrongly assumes mass increases under weaker gravity; mass is an intrinsic property of an object and stays constant regardless of the gravitational field strength acting on it.

  11. Question 11Answer: A

    1. Air resistance depends on the surface area of an object exposed to the air, among other factors: a larger surface area experiences a greater drag force at a given speed.
    2. The flattened disc, having a much larger surface area than the compact sphere, experiences greater air resistance as it falls.
    3. Because both objects have the same mass, and so the same weight, but the disc experiences more opposing drag, the disc is slowed more than the sphere.
    4. The compact sphere, experiencing less air resistance, falls faster and reaches the ground first, so the answer is A.
    • Why not B: Surface area does not affect the weight (gravitational force) acting on an object; weight depends only on mass and gravitational field strength (w = mg). A larger surface area increases air resistance, which opposes falling, rather than increasing the accelerating force.
    • Why not C: It is true that, with no air resistance, all objects fall with the same acceleration regardless of mass; but this scenario is in air, where air resistance depends on an object's surface area and shape, not its mass, so the two objects experience different amounts of drag and do not fall at the same rate.
    • Why not D: Air resistance (drag) always acts to oppose an object's motion through the air, in the direction opposite to its velocity; it cannot act in the same direction as the motion to speed an object up, unlike wind pushing a sail from behind, which is a different physical situation.

  12. Question 12Answer: D

    1. Force is the rate of change of momentum: force = change in momentum / time.
    2. The change in momentum is m(v - u) = 0.15 kg x (20 - 0) m/s = 3 kg m/s.
    3. Dividing by the contact time: force = 3 kg m/s / 0.05 s = 60 N.
    4. The average force exerted on the ball by the stick is 60 N, so the answer is D.
    • Why not A: This reports the change in momentum itself (0.15 x 20 = 3 kg m/s) as if it were the force; but force is the RATE of change of momentum, so this value must still be divided by the time over which it occurs (0.05 s), not left as it is.
    • Why not B: This divides the change in momentum by the mass again (3/0.15 = 20), rather than by the time taken; the rate of change of momentum requires dividing by TIME, and the mass has already been used once, in calculating the momentum change itself.
    • Why not C: This multiplies the change in momentum by the time (3 x 0.05 = 0.15) instead of dividing by it; force is momentum change divided by time, not multiplied by it, since force is a rate.

  13. Question 13Answer: C

    1. Before the push, both skaters are stationary, so the total momentum of the system is zero.
    2. Momentum is conserved, so after the push the two skaters' momenta must still sum to zero: they are equal in magnitude and opposite in direction.
    3. The 40 kg skater's momentum is 40 kg x 3 m/s = 120 kg m/s, so the 60 kg skater's momentum must also be 120 kg m/s, in the opposite direction.
    4. Her speed is 120 kg m/s / 60 kg = 2 m/s, so the answer is C.
    • Why not A: This divides the total momentum (40 x 3 = 120 kg m/s) by the combined mass of both skaters (60 + 40 = 100 kg), as if they stuck together and moved off as one object; but here the skaters push apart and each keeps their own mass, so the 60 kg skater's momentum must be divided by 60 kg alone, not the combined mass.
    • Why not B: This assumes both skaters must move off at the same speed because momentum is 'shared'; but conservation of momentum requires equal and opposite MOMENTA (mass x velocity), not equal speeds. Since the 60 kg skater has a greater mass, she must move off more slowly to have the same magnitude of momentum, not the same speed.
    • Why not D: This uses the mass ratio the wrong way round (3 x 60/40 = 4.5), giving the more massive skater the higher speed; a larger mass needs a SMALLER speed to produce the same momentum as a smaller mass moving faster, so the ratio must be inverted.

  14. Question 14Answer: B

    1. Total work done by the worker = force x distance = 80 N x 6 m = 480 J.
    2. Work done against friction (energy wasted) = frictional force x distance = 30 N x 6 m = 180 J.
    3. The useful energy transferred to the crate's kinetic energy is the total work done minus the energy wasted: 480 J - 180 J = 300 J.
    4. So the useful energy transfer is 300 J and the wasted energy is 180 J, so the answer is B.
    • Why not A: This reports the full work done by the worker (80 x 6 = 480 J) as the useful energy, without subtracting the energy wasted overcoming friction (180 J); the useful kinetic energy transferred is the total work done MINUS the energy wasted, not the total work done alone.
    • Why not C: This uses the frictional FORCE (30 N) directly as the wasted ENERGY, without multiplying by the distance moved (6 m); work, and so the energy wasted against a force, is force multiplied by distance, so the wasted energy is 30 N x 6 m = 180 J, not just 30 J.
    • Why not D: This swaps the useful and wasted energy figures around; the larger force (80 N, the worker's push) contributes the larger share of the energy transfer, most of which does useful work moving the crate, while the smaller opposing force (friction, 30 N) accounts for the smaller, wasted share.

  15. Question 15Answer: A

    1. Power = energy transferred / time taken.
    2. Convert 3 minutes to seconds: 3 x 60 = 180 s.
    3. Power = 9000 J / 180 s = 50 W.
    4. The power output of the heater is 50 W, so the answer is A.
    • Why not B: This divides the energy by 3 (the time in minutes) rather than converting the time to seconds first; power must be calculated using time in seconds, and 3 minutes is 180 seconds, not 3.
    • Why not C: This calculates time divided by energy (180/9000) instead of energy divided by time; power is defined as P = E/t, so the energy transferred must be divided by the time taken, not the other way round.
    • Why not D: This converts 3 minutes into seconds incorrectly, using 100 seconds per minute (3 x 100 = 300) instead of the correct 60 seconds per minute (3 x 60 = 180); this conversion error gives a time that is too large, and so a power output that is too small.

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