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NCEA Level 2 Physics study materials
NCEA Level 2 Physics study resources

NCEA Level 2 Physics practice questions

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Question 1Mechanics

A runner completes one full lap of a 400 m circular track in 50 s and finishes at the starting point. Which pair gives the runner's average speed and the magnitude of the average velocity for the lap?

  1. 8.0 m s⁻¹ and 400 m s⁻¹
  2. 8.0 m s⁻¹ and 8.0 m s⁻¹
  3. 8.0 m s⁻¹ and 0 m s⁻¹ Correct answer.
  4. 0 m s⁻¹ and 8.0 m s⁻¹

Explanation

Average speed is total distance divided by time, so 400 m ÷ 50 s = 8.0 m s⁻¹. Velocity is a vector, so average velocity uses displacement divided by time; the runner returns to the start, meaning displacement is zero and the average velocity is zero. The two quantities only agree when motion is in a straight line without reversing.

Question 2Mechanics

A book rests on a table. Which pair of forces is an action–reaction pair under Newton's third law?

  1. the weight of the book and the upward normal force of the table on the book
  2. Earth's pull on the book and the book's pull on Earth Correct answer.
  3. the push of the book on the table and the weight of the book
  4. the normal force of the table on the book and the weight of the table

Explanation

Newton's third law pairs act on two different bodies, are the same type of force, and are equal in size and opposite in direction. Earth pulls the book gravitationally and the book pulls Earth gravitationally with the same size force. The weight and the normal force are equal and opposite but both act on the book, so they are a balanced pair rather than a third-law pair.

Question 3Mechanics

A student pushes a 20 kg box 3.0 m across a rough floor at constant speed with a constant horizontal force of 40 N. Which statement correctly describes the work done by the student and the kinetic energy gained by the box?

  1. 120 J of work is done and 120 J of kinetic energy is gained
  2. 0 J of work is done and no kinetic energy is gained
  3. 120 J of work is done and no kinetic energy is gained Correct answer.
  4. 600 J of work is done and no kinetic energy is gained

Explanation

Work done by a force equals force × distance moved in the direction of the force, so 40 N × 3.0 m = 120 J. Because the box moves at constant speed, its kinetic energy does not change and the net work on it is zero; the 120 J supplied is dissipated by friction as heat. A force acting at 90° to the motion would do no work, but this force is horizontal and parallel to the displacement.

Question 4Mechanics

Two horizontal forces of 6.0 N and 8.0 N act at right angles to each other on a trolley. Which option gives the resultant force on the trolley?

  1. 2.0 N at 53° to the 6.0 N force
  2. 14 N, along the direction of the 6.0 N force
  3. 10 N at 37° to the 6.0 N force
  4. 10 N at 53° to the 6.0 N force Correct answer.

Explanation

Perpendicular vectors are added tip-to-tail, so the resultant magnitude is √(6.0² + 8.0²) = 10 N. The angle to the 6.0 N force is tan⁻¹(8.0/6.0) = 53°, because the 8.0 N force is the opposite side and the 6.0 N force is the adjacent side. Simply adding the magnitudes to get 14 N is only valid when the forces act along the same line.

Question 5Mechanics

A marble rolls horizontally at 3.0 m s⁻¹ off the edge of a bench that is 1.8 m above the floor. Take g = 10 N kg⁻¹ and ignore air resistance. How far horizontally from the edge of the bench does the marble land?

  1. 1.8 m Correct answer.
  2. 0.60 m
  3. 5.4 m
  4. 3.0 m

Explanation

Vertically the marble starts from rest, so 1.8 m = ½ × 10 × t² gives t = 0.60 s of fall time. With no air resistance the horizontal velocity stays at 3.0 m s⁻¹, so the horizontal distance is 3.0 × 0.60 = 1.8 m. The vertical and horizontal motions are independent of each other.

Question 6Mechanics

A 2.0 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 4.0 kg trolley. The trolleys lock together and move off as one. What is their common velocity immediately after the collision?

  1. 0.50 m s⁻¹ in the opposite direction
  2. 1.0 m s⁻¹ in the original direction Correct answer.
  3. 2.0 m s⁻¹ in the original direction
  4. 1.5 m s⁻¹ in the original direction

Explanation

Total momentum before the collision is (2.0 × 3.0) + (4.0 × 0) = 6.0 kg m s⁻¹, and momentum is conserved when no external horizontal forces act. The combined mass afterwards is 6.0 kg, so the common velocity is 6.0 ÷ 6.0 = 1.0 m s⁻¹ in the direction of the original motion. Kinetic energy is not conserved here because the trolleys lock together in an inelastic collision.

Question 7Mechanics

A 1200 kg car travels at a constant speed of 15 m s⁻¹ around a flat circular corner of radius 50 m. What is the magnitude of the centripetal force acting on the car?

  1. 5.4 × 10³ N Correct answer.
  2. 1.1 × 10⁴ N
  3. 360 N
  4. 2.7 × 10⁵ N

Explanation

The centripetal force points towards the centre of the circle and has magnitude F = mv²/r. Substituting gives F = 1200 × 15² / 50 = 5.4 × 10³ N. It is supplied by friction between the tyres and the road, and no outward force is needed on the car.

Question 8Mechanics

A worker pushes a 25 kg crate across a horizontal floor at constant speed using a constant horizontal force of 80 N. Take g = 10 N kg⁻¹. What is the coefficient of kinetic friction between the crate and the floor?

  1. 0.32 Correct answer.
  2. 0.080
  3. 3.2
  4. 0.64

Explanation

At constant speed the horizontal forces balance, so the kinetic friction force equals the 80 N push. The normal force equals the crate's weight, 25 × 10 = 250 N, so the coefficient is μ = Ffriction / N = 80 / 250 = 0.32. A coefficient of friction is a ratio of two forces and therefore has no units.

Question 9Mechanics

A 50 kg crate starts from rest and slides 5.0 m down a ramp inclined at 30° to the horizontal. A constant friction force of 80 N acts on the crate as it slides. Take g = 10 N kg⁻¹. What is the speed of the crate at the bottom of the ramp?

  1. 4.0 m s⁻¹
  2. 5.8 m s⁻¹ Correct answer.
  3. 9.0 m s⁻¹
  4. 7.1 m s⁻¹

Explanation

The vertical height dropped is 5.0 sin 30° = 2.5 m, so the gravitational potential energy lost is 50 × 10 × 2.5 = 1250 J. Friction removes 80 × 5.0 = 400 J as heat, leaving 850 J of kinetic energy at the bottom. Solving ½ × 50 × v² = 850 gives v ≈ 5.8 m s⁻¹.

Question 10Mechanics

A uniform plank 3.0 m long has a mass of 12 kg and pivots freely at one end. A 10 kg mass hangs from the far end, and a vertical rope attached 2.0 m from the pivot keeps the plank horizontal. Take g = 10 N kg⁻¹. What upward force does the rope exert on the plank?

  1. 240 N Correct answer.
  2. 200 N
  3. 120 N
  4. 470 N

Explanation

Take torques about the pivot, where the pivot force acts with zero lever arm. The plank's weight of 120 N acts at its centre, 1.5 m from the pivot, giving 180 N m, and the hanging mass gives 100 N at 3.0 m, or 300 N m. The rope at 2.0 m must balance the total 480 N m, so F = 480 / 2.0 = 240 N.

Question 11Mechanics

A student walks 30 m north, then 40 m east. Which option correctly gives the distance travelled and the magnitude of the displacement?

  1. Distance 70 m, displacement 50 m Correct answer.
  2. Distance 50 m, displacement 50 m
  3. Distance 70 m, displacement 70 m
  4. Distance 50 m, displacement 70 m

Explanation

Distance is the total length of the path, so 30 m + 40 m = 70 m. Displacement is the straight-line vector from start to finish; the two legs are perpendicular, so its magnitude is √(30² + 40²) = 50 m.

Question 12Mechanics

A cyclist rides around a circular track at a constant speed of 8.0 m s⁻¹. Which quantity remains constant during the motion?

  1. velocity
  2. speed Correct answer.
  3. acceleration
  4. net force

Explanation

At constant speed around a circle, the speed (magnitude of velocity) is fixed, but the direction of motion changes continuously. Velocity, acceleration and net force all change direction as the cyclist moves around the track.

Question 13Mechanics

A passenger standing in a bus is thrown forward when the bus brakes suddenly. Which statement best explains this?

  1. A forward force from the braking bus acts on the passenger.
  2. Inertia keeps the passenger moving forward as the bus slows. Correct answer.
  3. The passenger's weight acts forward during braking.
  4. Air resistance inside the bus pushes the passenger forward.

Explanation

By Newton's first law, an object continues in its state of motion unless acted on by a net force. When the bus brakes, the passenger keeps moving forward while the bus slows, which feels like being thrown forward.

Question 14Mechanics

A rugby ball is kicked with an initial velocity of 20 m s⁻¹ at 30° above the horizontal. What is the vertical component of its initial velocity? Take sin 30° = 0.50 and cos 30° = 0.87.

  1. 10 m s⁻¹ Correct answer.
  2. 23 m s⁻¹
  3. 17 m s⁻¹
  4. 20 m s⁻¹

Explanation

The vertical component is found using the sine of the angle: v_y = v sin θ = 20 × sin 30° = 20 × 0.50 = 10 m s⁻¹. The horizontal component would be 20 cos 30° ≈ 17 m s⁻¹.

Question 15Mechanics

A cricket ball of mass 0.16 kg approaches a bat horizontally at 20 m s⁻¹ and leaves the bat horizontally at 15 m s⁻¹ in the opposite direction. The contact time is 0.020 s. What is the magnitude of the average force exerted by the bat on the ball?

  1. 40 N
  2. 160 N
  3. 280 N Correct answer.
  4. 560 N

Explanation

Take one direction as positive. The change in velocity is 15 − (−20) = 35 m s⁻¹, so the impulse magnitude is 0.16 × 35 = 5.6 N s. Average force is impulse divided by contact time: 5.6 / 0.020 = 280 N.

Question 16Mechanics

A 1200 kg car accelerates from 10 m s⁻¹ to 20 m s⁻¹ on a level road. What is the minimum work done on the car by the net force?

  1. 3.6 × 10⁵ J
  2. 6.0 × 10⁴ J
  3. 2.4 × 10⁵ J
  4. 1.8 × 10⁵ J Correct answer.

Explanation

The work done by the net force equals the change in kinetic energy. ΔKE = ½ × 1200 × (20² − 10²) = ½ × 1200 × 300 = 1.8 × 10⁵ J.

Question 17Mechanics

A 0.20 kg ball on a string of length 0.80 m is swung in a vertical circle. At the highest point its speed is 4.0 m s⁻¹. Take g = 10 m s⁻². What is the tension in the string at the highest point?

  1. 2.0 N Correct answer.
  2. 4.0 N
  3. 6.0 N
  4. 0 N

Explanation

At the highest point, both tension and weight act downwards towards the centre, so T + mg = mv²/r. Rearranging gives T = m(v²/r − g) = 0.20(4.0²/0.80 − 10) = 2.0 N.

Question 18Mechanics

A uniform metre ruler is pivoted at its centre. A 2.0 N weight hangs from the 20 cm mark. At which mark must a 4.0 N weight be hung to balance the ruler horizontally?

  1. 15 cm mark
  2. 35 cm mark
  3. 65 cm mark Correct answer.
  4. 80 cm mark

Explanation

For balance, clockwise moments equal anticlockwise moments about the pivot. The 2.0 N weight is 0.30 m from the centre, giving a moment of 0.60 N m. A 4.0 N weight must be 0.60/4.0 = 0.15 m from the centre on the other side, which is the 65 cm mark.

Question 19Mechanics

Two masses, 3.0 kg and 5.0 kg, are connected by a light string over a frictionless pulley. The system is released from rest. Take g = 10 m s⁻². What is the acceleration of the masses?

  1. 5.0 m s⁻²
  2. 10 m s⁻²
  3. 7.5 m s⁻²
  4. 2.5 m s⁻² Correct answer.

Explanation

The net force on the system is the difference in weights: (5.0 − 3.0) × 10 = 20 N. The total mass accelerated is 8.0 kg, so a = F_net/m_total = 20/8.0 = 2.5 m s⁻².

Question 20Mechanics

A car travels around a circular curve of radius 80 m on a road banked at an angle θ to the horizontal. The curve is designed so that no friction is needed when the speed is 20 m s⁻¹. Take g = 10 m s⁻². What is the value of tan θ?

  1. 0.50 Correct answer.
  2. 1.0
  3. 0.25
  4. 2.0

Explanation

For a frictionless banked curve, the horizontal component of the normal force provides the centripetal force, giving tan θ = v²/(rg). Substituting v = 20 m s⁻¹, r = 80 m and g = 10 m s⁻² gives tan θ = 400/800 = 0.50.

Question 21Mechanics

In mechanics, which of the following quantities is classified as a vector?

  1. Time
  2. Mass
  3. Temperature
  4. Weight Correct answer.

Explanation

A vector quantity has both magnitude and direction. Weight is the gravitational force on an object and acts vertically downward, so it is a vector. Mass, time and temperature have magnitude only and are scalars.

Question 22Mechanics

A net force of 15 N acts on a 5.0 kg trolley on a frictionless horizontal surface. What is the magnitude of the trolley's acceleration?

  1. 10 m s⁻²
  2. 75 m s⁻²
  3. 0.33 m s⁻²
  4. 3.0 m s⁻² Correct answer.

Explanation

Newton's second law gives a = F_net / m. Substituting 15 N and 5.0 kg gives 3.0 m s⁻². The surface is frictionless, so the net horizontal force is the 15 N applied force.

Question 23Mechanics

A motor raises a 20 kg load vertically at a constant speed of 1.5 m s⁻¹. Taking g = 10 N kg⁻¹, what useful power does the motor develop?

  1. 3000 W
  2. 200 W
  3. 300 W Correct answer.
  4. 30 W

Explanation

At constant speed the upward force equals the load's weight, mg = 20 × 10 = 200 N. Power is P = Fv = 200 × 1.5 = 300 W. The 20 kg value is mass, so using it directly as a force would be incorrect.

Question 24Mechanics

Two ice skaters stand at rest and push apart. Skater A has mass 60 kg and moves east at 1.2 m s⁻¹. Skater B has mass 40 kg. What is the magnitude of B's velocity immediately after the push?

  1. 0.80 m s⁻¹
  2. 1.8 m s⁻¹ Correct answer.
  3. 1.2 m s⁻¹
  4. 2.4 m s⁻¹

Explanation

Total momentum before the push is zero, so the skaters' momenta after the push must be equal and opposite. Skater A's momentum is 60 × 1.2 = 72 kg m s⁻¹ east, so B's speed is 72 / 40 = 1.8 m s⁻¹.

Question 25Mechanics

A 0.40 kg toy car moves at a constant speed of 3.0 m s⁻¹ around a horizontal circular track of radius 0.50 m. What is the centripetal force on the car?

  1. 7.2 N Correct answer.
  2. 2.4 N
  3. 18 N
  4. 1.8 N

Explanation

Centripetal force is F = mv²/r. Substituting m = 0.40 kg, v = 3.0 m s⁻¹ and r = 0.50 m gives F = (0.40 × 9.0) / 0.50 = 7.2 N. This force acts toward the centre of the circular track.

Question 26Mechanics

A 0.50 kg ball is released from rest 5.0 m above the ground. Ignoring air resistance and taking g = 10 N kg⁻¹, what is its speed just before impact?

  1. 5.0 m s⁻¹
  2. 50 m s⁻¹
  3. 100 m s⁻¹
  4. 10 m s⁻¹ Correct answer.

Explanation

Conservation of energy gives mgh = ½mv², so v = √(2gh). Substituting g = 10 N kg⁻¹ and h = 5.0 m gives v = √100 = 10 m s⁻¹. Mass cancels, so the 0.50 kg value does not change the answer.

Question 27Mechanics

A uniform beam 4.0 m long has a weight of 300 N and is supported at each end. A 200 N load is placed 1.0 m from the left end. What upward force must the right support provide?

  1. 500 N
  2. 150 N
  3. 125 N
  4. 200 N Correct answer.

Explanation

Take moments about the left support, so its unknown force has zero moment. The beam's 300 N weight acts at 2.0 m and the 200 N load at 1.0 m, giving total clockwise moment 300×2.0 + 200×1.0 = 800 N m. The right support at 4.0 m must provide 800/4.0 = 200 N.

Question 28Mechanics

A ball is kicked from level ground with speed 25 m s⁻¹ at 53° above the horizontal. Taking g = 10 N kg⁻¹ and sin 53° = 0.80, how long is the ball in the air?

  1. 2.5 s
  2. 5.0 s
  3. 2.0 s
  4. 4.0 s Correct answer.

Explanation

For a projectile on level ground, time of flight is T = 2v sinθ / g. The vertical component is 25 × 0.80 = 20 m s⁻¹, so T = 2 × 20 / 10 = 4.0 s. Horizontal motion does not affect the time in the air.

Question 29Mechanics

A 4.0 kg block on a rough horizontal table is connected by a light string over a frictionless pulley to a hanging 2.0 kg mass. The coefficient of kinetic friction between the block and the table is 0.20. Taking g = 10 N kg⁻¹, what is the acceleration of the system?

  1. 3.3 m s⁻²
  2. 0.67 m s⁻²
  3. 1.3 m s⁻²
  4. 2.0 m s⁻² Correct answer.

Explanation

The hanging mass provides a driving force of 2.0 × 10 = 20 N. Friction on the 4.0 kg block is μmg = 0.20 × 4.0 × 10 = 8.0 N, so the net force is 12 N. The total mass is 6.0 kg, giving a = 12/6.0 = 2.0 m s⁻².

Question 30Mechanics

A 800 kg car travels at 15 m s⁻¹ over the top of a circular hump of radius 50 m. Taking g = 10 N kg⁻¹, what is the magnitude of the normal force exerted by the road on the car at the top?

  1. 4400 N Correct answer.
  2. 3600 N
  3. 11600 N
  4. 8000 N

Explanation

At the top of the hump the centripetal acceleration is downward, so mg − N = mv²/r. Rearranging gives N = mg − mv²/r = 8000 − (800 × 15²)/50 = 8000 − 3600 = 4400 N. The road pushes up on the car with this normal force.

Question 31Mechanics

A 2.0 kg box is lifted vertically at constant speed through a height of 3.0 m. Taking g = 10 N kg⁻¹, what is the gain in gravitational potential energy of the box?

  1. 600 J
  2. 20 J
  3. 60 J Correct answer.
  4. 6.0 J

Explanation

Gravitational potential energy gained is ΔE_p = m g Δh. Substituting gives 2.0 kg × 10 N kg⁻¹ × 3.0 m = 60 J. The constant-speed condition only means the lifting force equals the weight; it does not change the energy gain.

Question 32Mechanics

A book rests on a horizontal table. Which pair of forces is an action–reaction pair according to Newton’s third law?

  1. The normal force of the table on the book and the weight of the table
  2. The weight of the book and the normal force of the table on the book
  3. The weight of the book and the weight of the table
  4. The normal force of the table on the book and the normal force of the book on the table Correct answer.

Explanation

Newton’s third law pairs are equal in magnitude, opposite in direction, and act on different objects. The table pushes up on the book while the book pushes down on the table with an equal and opposite normal force. Weight and normal force on the book are not a third-law pair because both act on the same object.

Question 33Mechanics

For an object moving in a straight line, the area under its velocity–time graph between two times gives which quantity?

  1. Displacement Correct answer.
  2. Acceleration
  3. Net force
  4. Average speed

Explanation

The area under a velocity–time graph is found by multiplying velocity by time, which gives displacement. Acceleration is the gradient of the graph, not its area. Average speed would require total distance, and net force is not obtained from this area.

Question 34Mechanics

A crate is pulled 10 m along a horizontal floor by a rope that makes an angle of 30° above the horizontal. The tension in the rope is 50 N. Taking cos 30° = 0.87, what is the work done by the tension on the crate?

  1. 870 J
  2. 250 J
  3. 435 J Correct answer.
  4. 500 J

Explanation

Work done by a force is W = F d cos θ, where θ is the angle between the force and the displacement. Here W = 50 N × 10 m × cos 30° = 50 × 10 × 0.87 = 435 J. Only the horizontal component of the tension does work because the displacement is horizontal.

Question 35Mechanics

A 1.5 kg block is pulled across a rough horizontal floor by a horizontal force of 12 N. The block accelerates at 4.0 m s⁻². What is the magnitude of the friction force acting on the block?

  1. 6 N Correct answer.
  2. 18 N
  3. 12 N
  4. 2 N

Explanation

The net horizontal force is F_net = ma = 1.5 kg × 4.0 m s⁻² = 6.0 N. The applied force of 12 N must overcome friction, so 12 N − friction = 6.0 N, giving friction = 6 N. Thus the friction force is equal in magnitude to the net force in this case.

Question 36Mechanics

A pendulum bob of mass 0.50 kg is released from rest from a height of 0.80 m above its lowest point. Ignoring air resistance, what is the speed of the bob at the lowest point? Take g = 10 N kg⁻¹.

  1. 8.0 m s⁻¹
  2. 4.0 m s⁻¹ Correct answer.
  3. 16 m s⁻¹
  4. 2.0 m s⁻¹

Explanation

Mechanical energy is conserved, so the loss in gravitational potential energy equals the gain in kinetic energy: mgh = ½mv². Cancelling mass gives v = √(2gh) = √(2 × 10 × 0.80) = √16 = 4.0 m s⁻¹. The mass of the bob therefore does not affect the speed.

Question 37Mechanics

A stone is thrown horizontally from the top of a 45 m high cliff with a speed of 15 m s⁻¹. Taking g = 10 N kg⁻¹ and ignoring air resistance, how far from the base of the cliff does the stone land?

  1. 30 m
  2. 15 m
  3. 90 m
  4. 45 m Correct answer.

Explanation

The vertical motion determines the time of flight: h = ½gt², so t = √(2h/g) = √(2 × 45 / 10) = 3.0 s. The horizontal speed stays constant at 15 m s⁻¹, so horizontal range = 15 × 3.0 = 45 m. Air resistance is ignored, so no horizontal deceleration occurs.

Question 38Mechanics

A 2.0 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 4.0 kg trolley. The trolleys stick together. What is their common velocity immediately after the collision?

  1. 0.50 m s⁻¹
  2. 2.0 m s⁻¹
  3. 1.5 m s⁻¹
  4. 1.0 m s⁻¹ Correct answer.

Explanation

Total momentum is conserved in the collision. Initial momentum = 2.0 kg × 3.0 m s⁻¹ = 6.0 kg m s⁻¹, and the combined mass is 2.0 + 4.0 = 6.0 kg. Therefore the common velocity is 6.0 ÷ 6.0 = 1.0 m s⁻¹.

Question 39Mechanics

A 2.0 kg ball moving east at 3.0 m s⁻¹ collides with a stationary 1.0 kg ball on a frictionless surface. After the collision the 2.0 kg ball moves north at 1.0 m s⁻¹. What is the speed of the 1.0 kg ball immediately after the collision?

  1. 4.0 m s⁻¹
  2. 2.0 m s⁻¹
  3. 8.0 m s⁻¹
  4. 6.3 m s⁻¹ Correct answer.

Explanation

Momentum is a vector and is conserved separately in each direction. Initially the total momentum is 6.0 kg m s⁻¹ east; after the collision the 2.0 kg ball has 2.0 kg m s⁻¹ north, so the 1.0 kg ball must have 6.0 kg m s⁻¹ east and 2.0 kg m s⁻¹ south. Its speed is √(6.0² + 2.0²) = √40 ≈ 6.3 m s⁻¹.

Question 40Mechanics

A 0.20 kg mass on a string of length 0.50 m moves in a horizontal circle at constant speed, with the string making an angle of 30° with the vertical. Taking g = 10 N kg⁻¹ and tan 30° = 0.58, what is the speed of the mass?

  1. 4.8 m s⁻¹
  2. 1.2 m s⁻¹ Correct answer.
  3. 2.4 m s⁻¹
  4. 0.60 m s⁻¹

Explanation

For a conical pendulum, the vertical component of tension balances weight and the horizontal component provides centripetal force. Dividing these equations gives tan θ = v²/(rg), where r = L sin θ = 0.50 × 0.50 = 0.25 m. Thus v = √(r g tan θ) = √(0.25 × 10 × 0.58) ≈ 1.2 m s⁻¹.

Question 41Mechanics

A student walks 3.0 km east and then 4.0 km north. Which statement correctly compares the distance travelled with the magnitude of the displacement?

  1. Distance travelled is 7.0 km and displacement magnitude is 5.0 km. Correct answer.
  2. Distance travelled is 7.0 km and displacement magnitude is 7.0 km.
  3. Distance travelled is 5.0 km and displacement magnitude is 5.0 km.
  4. Distance travelled is 5.0 km and displacement magnitude is 7.0 km.

Explanation

Distance is the total length of the path followed, so it is 3.0 + 4.0 = 7.0 km. Displacement is the straight-line vector from start to finish, and because the two legs are perpendicular its magnitude is √(3.0² + 4.0²) = 5.0 km.

Question 42Mechanics

A ball is thrown vertically upward and moves freely under gravity. At the highest point of its motion, what is the ball's acceleration?

  1. Zero
  2. 10 m s⁻² upward
  3. 10 m s⁻² in the direction of motion
  4. 10 m s⁻² downward Correct answer.

Explanation

The only force acting on the ball is gravity, which acts downward throughout the motion. At the highest point the velocity is momentarily zero, but the acceleration due to gravity is still about 10 m s⁻² downward.

Question 43Mechanics

A runner completes one full lap of a circular track of circumference 400 m in 50 s. What is the runner's average velocity for the lap?

  1. 0 m s⁻¹ Correct answer.
  2. 4.0 m s⁻¹
  3. 400 m s⁻¹
  4. 8.0 m s⁻¹

Explanation

Average velocity is displacement divided by time. After one complete lap the runner returns to the starting point, so the displacement is zero and the average velocity is zero. The average speed is 400 ÷ 50 = 8.0 m s⁻¹, which is a different quantity.

Question 44Mechanics

A force of 50 N is applied to a crate at an angle of 30° above the horizontal. What is the horizontal component of this force? (sin 30° = 0.50, cos 30° = 0.87)

  1. 29 N
  2. 43 N Correct answer.
  3. 58 N
  4. 25 N

Explanation

The horizontal component is found using F cos θ, because the angle is measured from the horizontal. Substituting gives 50 × 0.87 = 43.5 N, which rounds to 43 N.

Question 45Mechanics

A 60 kg student stands on a bathroom scale in a lift. The lift accelerates upward at 2.0 m s⁻². Taking g = 10 N kg⁻¹, what does the scale read?

  1. 480 N
  2. 720 N Correct answer.
  3. 120 N
  4. 600 N

Explanation

The scale reads the normal force on the student. Taking upward as positive, N − mg = ma, so N = m(g + a) = 60 × (10 + 2.0) = 720 N.

Question 46Mechanics

A cricket ball of mass 0.15 kg travels horizontally at 20 m s⁻¹ and is struck by a bat. It rebounds horizontally at 15 m s⁻¹. If the contact time is 0.010 s, what is the magnitude of the average force exerted by the bat on the ball?

  1. 525 N Correct answer.
  2. 300 N
  3. 750 N
  4. 75 N

Explanation

Impulse equals change in momentum, and the ball reverses direction, so the change in velocity has magnitude 20 + 15 = 35 m s⁻¹. Thus Δp = 0.15 × 35 = 5.25 kg m s⁻¹, and the average force is 5.25 ÷ 0.010 = 525 N.

Question 47Mechanics

A 5.0 kg box slides from rest down a ramp and reaches the bottom with a speed of 4.0 m s⁻¹. The vertical height of the ramp is 2.0 m. Taking g = 10 N kg⁻¹, how much mechanical energy is dissipated by friction?

  1. 20 J
  2. 140 J
  3. 100 J
  4. 60 J Correct answer.

Explanation

The initial gravitational potential energy is mgh = 5.0 × 10 × 2.0 = 100 J. The final kinetic energy is ½mv² = ½ × 5.0 × 4.0² = 40 J. The difference, 100 − 40 = 60 J, is the energy dissipated by friction.

Question 48Mechanics

A car rounds a frictionless banked corner of radius 80 m at a speed of 20 m s⁻¹. Taking g = 10 N kg⁻¹, what banking angle is required?

  1. 27° Correct answer.
  2. 7.0°
  3. 14°
  4. 45°

Explanation

For a frictionless banked curve, tan θ = v²/(rg). Substituting gives tan θ = 20²/(80 × 10) = 400/800 = 0.50, so θ = tan⁻¹(0.50) ≈ 27°.

Question 49Mechanics

A uniform metre ruler is pivoted at its 50.0 cm mark and balances horizontally when a 20 g mass is hung at the 10.0 cm mark and a 50 g mass is hung on the other side. At which mark must the 50 g mass be hung? (Ignore the mass of the ruler.)

  1. 66.0 cm Correct answer.
  2. 34.0 cm
  3. 82.0 cm
  4. 16.0 cm

Explanation

For balance, the moments about the pivot must be equal. The 20 g mass is 40.0 cm from the pivot, giving a moment of 20 × 40 = 800 g cm. The 50 g mass must therefore be 800 ÷ 50 = 16.0 cm from the pivot on the opposite side, which is the 50.0 + 16.0 = 66.0 cm mark.

Question 50Mechanics

A car accelerates uniformly from 10 m s⁻¹ to 30 m s⁻¹ in 8.0 s. What distance does it travel during this time?

  1. 160 m Correct answer.
  2. 80 m
  3. 240 m
  4. 320 m

Explanation

For uniform acceleration, the average speed is (10 + 30) ÷ 2 = 20 m s⁻¹. The distance travelled is average speed × time = 20 × 8.0 = 160 m.