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QCAA Physics practice questions

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Question 1External assessment Paper 1

A ball is thrown horizontally from a cliff and lands on level ground below. Ignoring air resistance, which statement describes the ball's horizontal velocity during the flight?

  1. It increases steadily because gravity accelerates the ball forward.
  2. It decreases steadily because the ball loses speed as it falls.
  3. It remains constant because no horizontal force acts on the ball. Correct answer.
  4. It becomes zero at the highest point of the trajectory.

Explanation

Once released, the only force acting on the ball is gravity, which acts vertically downward. With no horizontal force, Newton's first law means the horizontal component of velocity does not change. Gravity changes only the vertical component, so the horizontal motion stays uniform throughout the flight.

Question 2External assessment Paper 1

A car travelling at 20 m/s brakes uniformly and stops in 5.0 s. What distance does the car travel while braking?

  1. 50 m Correct answer.
  2. 20 m
  3. 100 m
  4. 4.0 m

Explanation

For uniform acceleration the average velocity equals the mean of initial and final velocities, here (20 + 0)/2 = 10 m/s. Multiplying average velocity by time gives 10 m/s × 5.0 s = 50 m. Alternatively, acceleration is -4.0 m/s² and s = ut + ½at² gives the same result.

Question 3External assessment Paper 1

A 2.0 kg object is pushed across a frictionless surface by a constant net force of 6.0 N. What is the object's acceleration?

  1. 3.0 m/s² Correct answer.
  2. 12 m/s²
  3. 8.0 m/s²
  4. 0.33 m/s²

Explanation

Newton's second law states that net force equals mass times acceleration, so a = F/m. Substituting gives 6.0 N ÷ 2.0 kg = 3.0 m/s². The direction of acceleration is the same as the net force.

Question 4External assessment Paper 1

Two isolated point charges are moved so that the distance between them doubles while their charges stay the same. How does the magnitude of the electrostatic force between them change?

  1. It doubles.
  2. It stays the same.
  3. It falls to one quarter of its original value. Correct answer.
  4. It halves.

Explanation

Coulomb's law gives force proportional to the product of the charges divided by the square of the separation. Doubling the separation squares the factor of two in the denominator, giving 1/4 of the original force. The inverse-square relationship makes the force fall by a factor of four.

Question 5External assessment Paper 1

Which quantity is a scalar?

  1. Magnetic field strength
  2. Kinetic energy Correct answer.
  3. Electric field strength
  4. Displacement

Explanation

A scalar has magnitude only, with no direction, and kinetic energy fits this description. Field strengths and displacement are vectors because each requires a direction to be fully specified. Recognising which quantities need direction is central to choosing the correct representation.

Question 6External assessment Paper 1

A wave has a frequency of 250 Hz and a wavelength of 1.2 m. What is its speed?

  1. 251 m/s
  2. 208 m/s
  3. 300 m/s Correct answer.
  4. 0.0048 m/s

Explanation

The wave equation relates speed, frequency and wavelength: v = fλ. Multiplying 250 Hz by 1.2 m gives 300 m/s. Frequency and wavelength are independent properties of the medium and source respectively, and their product gives the propagation speed.

Question 7External assessment Paper 1

In a transverse wave, what is the direction of particle oscillation relative to the direction of energy transfer?

  1. At 45 degrees to the direction of energy transfer
  2. Parallel to the direction of energy transfer
  3. Perpendicular to the direction of energy transfer Correct answer.
  4. Random, with no fixed relationship

Explanation

By definition, in a transverse wave the particles oscillate perpendicular to the direction in which the wave travels. A wave on a string is a common example, with particles moving up and down while energy moves along the string. Longitudinal waves instead have particles oscillating parallel to energy transfer.

Question 8External assessment Paper 1

Which statement correctly describes the strong nuclear force?

  1. It acts only between protons and is repulsive at all separations.
  2. It acts between quarks only and never between nucleons.
  3. It acts between nucleons, is attractive at typical nuclear separations and is very short range. Correct answer.
  4. It acts between all charged particles and has infinite range.

Explanation

The strong nuclear force binds protons and neutrons together inside the nucleus. It is attractive at separations of roughly a femtometre but becomes strongly repulsive at very short range, and it drops to negligible strength beyond a few femtometres. This short range explains why it holds small nuclei together despite electrostatic repulsion.

Question 9External assessment Paper 1

A 0.50 kg mass on a spring undergoes simple harmonic motion with a period of 0.80 s. What is the angular frequency of the oscillation?

  1. 0.80 rad/s
  2. 7.9 rad/s Correct answer.
  3. 5.0 rad/s
  4. 1.25 rad/s

Explanation

Angular frequency is related to period by ω = 2π/T. Substituting T = 0.80 s gives ω = 2π/0.80 ≈ 7.85 rad/s, which rounds to 7.9 rad/s. Angular frequency is measured in radians per second, not hertz.

Question 10External assessment Paper 1

A block slides down a rough incline at constant velocity. Which statement about the forces on the block is correct?

  1. The normal force equals the full weight of the block.
  2. Friction is the only force acting along the slope.
  3. The component of weight down the slope is greater than the friction force up the slope.
  4. The net force on the block is zero because its acceleration is zero. Correct answer.

Explanation

Constant velocity means zero acceleration, so Newton's second law requires the net force to be zero. Along the slope, the downhill component of weight is exactly balanced by kinetic friction acting uphill. The normal force balances only the perpendicular component of weight, which is less than the full weight.

Question 11External assessment Paper 1

A satellite orbits Earth in a circular orbit. If the satellite's orbital radius is increased, what happens to its orbital speed?

  1. It becomes zero because gravity no longer acts.
  2. It stays the same because gravitational force is unchanged.
  3. It decreases because gravitational force provides a smaller centripetal acceleration. Correct answer.
  4. It increases because the satellite has further to travel.

Explanation

For a circular orbit, gravity supplies the centripetal force: GMm/r² = mv²/r, giving v = √(GM/r). As r increases, the required speed decreases, so the satellite moves more slowly at greater altitude. The gravitational force also weakens with distance, consistent with the reduced speed.

Question 12External assessment Paper 1

A gas is compressed rapidly in an insulated cylinder. Which description best matches the process and the resulting temperature change?

  1. Isothermal compression, so the temperature stays constant.
  2. Adiabatic expansion, so the temperature falls.
  3. Adiabatic compression, so the temperature rises. Correct answer.
  4. Isobaric compression, so the temperature is unchanged.

Explanation

An insulated cylinder prevents heat exchange with the surroundings, so the process is adiabatic (Q = 0). Work is done on the gas during compression, and since Q = 0 this energy raises the internal energy and therefore the temperature. Rapid compression is treated as adiabatic because there is little time for heat transfer.

Question 13External assessment Paper 1

A resistor of 12 Ω carries a current of 0.50 A. What power is dissipated in the resistor?

  1. 24 W
  2. 0.042 W
  3. 3.0 W Correct answer.
  4. 6.0 W

Explanation

Electrical power in a resistor can be found using P = I²R. Substituting gives (0.50)² × 12 = 0.25 × 12 = 3.0 W. Alternatively, the voltage is 6.0 V and P = VI gives the same result.

Question 14External assessment Paper 1

Two resistors of 6.0 Ω and 3.0 Ω are connected in parallel across a 12 V supply. What is the total current drawn from the supply?

  1. 1.3 A
  2. 2.0 A
  3. 4.0 A
  4. 6.0 A Correct answer.

Explanation

For parallel resistors, 1/R = 1/6 + 1/3 = 1/2, so the equivalent resistance is 2.0 Ω. The total current is V/R = 12/2.0 = 6.0 A. Equivalently, the branch currents are 2.0 A and 4.0 A, which sum to 6.0 A.

Question 15External assessment Paper 1

A charged particle enters a uniform magnetic field moving parallel to the field lines. What is the magnetic force on the particle?

  1. Directed opposite to the particle's velocity.
  2. Zero, because the velocity is parallel to the field. Correct answer.
  3. Maximum, because the particle is moving fastest along the field.
  4. Directed along the magnetic field lines.

Explanation

The magnetic force is given by F = qvB sin θ, where θ is the angle between velocity and field. When the particle moves parallel to the field, θ = 0 and sin θ = 0, so the force is zero. The force is greatest when velocity and field are perpendicular.

Question 16External assessment Paper 1

Light of wavelength 500 nm is incident on a double slit with slit separation 0.25 mm. What is the angular separation between adjacent bright fringes?

  1. 1.25 × 10⁻⁷ rad
  2. 5.0 × 10⁻⁴ rad
  3. 2.0 × 10⁻³ rad Correct answer.
  4. 2.0 × 10⁻⁶ rad

Explanation

Adjacent bright fringes satisfy d sin θ = λ, so for small angles θ ≈ λ/d. Converting units gives 500 × 10⁻⁹ m ÷ 0.25 × 10⁻³ m = 2.0 × 10⁻³ rad. The small-angle approximation is valid because the angle is far below one radian.

Question 17External assessment Paper 1

An electron is accelerated from rest through a potential difference of 200 V. What is its kinetic energy in electronvolts?

  1. 3.2 × 10⁻¹⁷ eV
  2. 200 eV Correct answer.
  3. 100 eV
  4. 1.6 × 10⁻¹⁹ eV

Explanation

By definition, an electron accelerated through a potential difference of 1 V gains 1 eV of kinetic energy. Therefore a 200 V potential difference gives 200 eV of kinetic energy. This is equivalent to 3.2 × 10⁻¹⁷ J when converted using the joule value of the electronvolt.

Question 18External assessment Paper 1

A photon is emitted when a hydrogen atom transitions from the n = 3 level to the n = 2 level. How does the photon energy compare with the energy difference between these levels?

  1. The photon energy is greater than the energy difference, with the excess becoming kinetic energy.
  2. The photon energy depends only on the final level, not the difference.
  3. The photon energy equals the energy difference between the levels. Correct answer.
  4. The photon energy is half the energy difference.

Explanation

In a radiative transition, the energy lost by the atom is carried away by a single photon. Conservation of energy therefore makes the photon energy exactly equal to the difference between the two energy levels. The photon wavelength follows from E = hc/λ.

Question 19External assessment Paper 1

A moving observer measures the length of a rod that is at rest in a laboratory frame. Compared with the rod's proper length, the measured length is:

  1. Shorter only if the rod is oriented perpendicular to the motion.
  2. Shorter, because of length contraction along the direction of motion. Correct answer.
  3. Longer, because the observer is moving.
  4. The same, because length does not depend on the observer.

Explanation

Special relativity predicts that a moving observer measures a length contracted along the direction of relative motion: L = L₀√(1 − v²/c²). The proper length measured in the rod's rest frame is the maximum value. Contraction occurs only along the direction of motion, not perpendicular to it.

Question 20External assessment Paper 1

A 5.0 kg block moving at 4.0 m/s collides with and sticks to a stationary 3.0 kg block on a frictionless surface. What is the velocity of the combined blocks after the collision?

  1. 1.5 m/s
  2. 2.5 m/s Correct answer.
  3. 0.67 m/s
  4. 4.0 m/s

Explanation

Momentum is conserved in the perfectly inelastic collision: m₁u₁ = (m₁ + m₂)v. Substituting gives 5.0 × 4.0 = 8.0 × v, so v = 20/8.0 = 2.5 m/s. Kinetic energy is not conserved because the blocks deform and stick together.

Question 21External assessment Paper 1

A projectile is launched at 30 m/s at an angle of 40° above the horizontal from level ground. Taking g = 9.8 m/s² and ignoring air resistance, what is the maximum height reached?

  1. 46 m
  2. 19 m Correct answer.
  3. 7.4 m
  4. 30 m

Explanation

The vertical component of velocity is 30 sin 40° ≈ 19.3 m/s. At maximum height the vertical velocity is zero, so h = v_y²/(2g) = (19.3)²/(2 × 9.8) ≈ 19 m. The horizontal component does not affect the height reached.

Question 22External assessment Paper 1

A capacitor of 4.0 μF is charged to 50 V. What is the energy stored in the capacitor?

  1. 1.0 × 10⁻² J
  2. 2.0 × 10⁻⁴ J
  3. 5.0 × 10⁻³ J Correct answer.
  4. 1.0 × 10⁻⁴ J

Explanation

Energy stored in a capacitor is U = ½CV². Substituting C = 4.0 × 10⁻⁶ F and V = 50 V gives ½ × 4.0 × 10⁻⁶ × 2500 = 5.0 × 10⁻³ J. The charge on the capacitor is 2.0 × 10⁻⁴ C, and U = ½QV gives the same energy.

Question 23External assessment Paper 1

Monochromatic light is incident on a metal surface. The intensity of the light is doubled while the frequency stays the same. What is the effect on the maximum kinetic energy of the emitted photoelectrons?

  1. It is unchanged because the photon energy depends on frequency, not intensity. Correct answer.
  2. It falls to half because the photons are spread more thinly.
  3. It doubles because twice as many photons arrive.
  4. It increases by a factor of four because intensity is proportional to energy squared.

Explanation

The photoelectric equation is K_max = hf − W, which depends on the frequency of the light and the work function of the metal. Intensity determines the number of photons arriving per second, so doubling it doubles the photocurrent, not the maximum kinetic energy. Only increasing the frequency raises the maximum kinetic energy.

Question 24External assessment Paper 1

A current-carrying wire of length 0.40 m lies perpendicular to a uniform magnetic field of strength 0.30 T and carries a current of 2.0 A. What is the magnitude of the magnetic force on the wire?

  1. 0.60 N
  2. 0.024 N
  3. 0.24 N Correct answer.
  4. 0.15 N

Explanation

The force on a straight current-carrying wire is F = BIL sin θ. With the wire perpendicular to the field, sin θ = 1, so F = 0.30 × 2.0 × 0.40 = 0.24 N. The direction of the force is given by the right-hand rule.

Question 25External assessment Paper 1

An ideal transformer has 200 turns on the primary coil and 50 turns on the secondary coil. If the primary is connected to a 240 V alternating supply, what is the secondary voltage?

  1. 960 V
  2. 240 V
  3. 120 V
  4. 60 V Correct answer.

Explanation

For an ideal transformer the voltage ratio equals the turns ratio: V_s/V_p = N_s/N_p. Substituting gives V_s = 240 × 50/200 = 60 V. This is a step-down transformer because the secondary has fewer turns than the primary.

Question 26External assessment Paper 1

A ball is thrown vertically upward and returns to the thrower's hand. Ignoring air resistance, which quantity remains constant throughout the ball's flight?

  1. The displacement of the ball from the thrower
  2. The velocity of the ball
  3. The magnitude of the acceleration Correct answer.
  4. The speed of the ball

Explanation

Once the ball leaves the hand, the only force acting is gravity, so the acceleration has a constant magnitude of about 9.8 m/s² directed downward for the whole flight. Speed and velocity both change as the ball rises and falls, and displacement varies continuously.

Question 27External assessment Paper 1

A car of mass 1200 kg travelling at 20 m/s brakes to rest in 5.0 s. What is the magnitude of the average net force on the car?

  1. 240 N
  2. 24000 N
  3. 12000 N
  4. 4800 N Correct answer.

Explanation

The average acceleration is the change in velocity divided by time, giving 20 ÷ 5.0 = 4.0 m/s². Newton's second law then gives F = ma = 1200 × 4.0 = 4800 N.

Question 28External assessment Paper 1

Which quantity is a vector?

  1. Work
  2. Magnetic field strength Correct answer.
  3. Electric potential
  4. Temperature

Explanation

Magnetic field strength has both magnitude and direction, so it is a vector quantity. Work, temperature and electric potential are all scalars, each fully described by a magnitude alone.

Question 29External assessment Paper 1

Two identical resistors are connected in series across a battery. They are then reconnected in parallel across the same battery. How does the total power dissipated compare with the series case?

  1. It is twice as large
  2. It is half as large
  3. It is four times as large Correct answer.
  4. It is one quarter as large

Explanation

In series the total resistance is 2R, so power is V²/(2R). In parallel the total resistance is R/2, so power is V²/(R/2) = 2V²/R, which is four times the series value.

Question 30External assessment Paper 1

A sound wave travels from air into water, where its speed is greater. What happens to its frequency and wavelength?

  1. Frequency stays the same and wavelength increases Correct answer.
  2. Frequency increases and wavelength stays the same
  3. Frequency stays the same and wavelength decreases
  4. Frequency decreases and wavelength increases

Explanation

Frequency is set by the source and does not change when the wave crosses a boundary. Since v = fλ and the speed increases while f is fixed, the wavelength must increase.

Question 31External assessment Paper 1

A 2.0 kg object moving at 3.0 m/s collides head-on with a 4.0 kg object moving at 1.0 m/s in the opposite direction. They stick together. What is the velocity of the combined mass?

  1. 1.7 m/s in the direction of the 2.0 kg object
  2. 0.33 m/s in the direction of the 4.0 kg object Correct answer.
  3. 0.33 m/s in the direction of the 2.0 kg object
  4. 0 m/s

Explanation

Taking the 2.0 kg object's direction as positive, initial momentum is (2.0 × 3.0) + (4.0 × −1.0) = 2.0 kg·m/s. Dividing by the total mass of 6.0 kg gives 0.33 m/s, positive meaning it moves with the 2.0 kg object's original direction.

Question 32External assessment Paper 1

A power supply of negligible internal resistance delivers 2.0 A through a wire of resistance 0.50 Ω. What is the potential difference across the wire?

  1. 0.25 V
  2. 4.0 V
  3. 1.0 V Correct answer.
  4. 2.5 V

Explanation

Ohm's law gives V = IR = 2.0 × 0.50 = 1.0 V across the resistance. The internal resistance is negligible, so no additional drop occurs inside the supply.

Question 33External assessment Paper 1

A charged particle moves in a circle in a uniform magnetic field. If the particle's speed is doubled while the field strength is unchanged, what happens to the radius of its path?

  1. It quadruples
  2. It stays the same
  3. It halves
  4. It doubles Correct answer.

Explanation

The magnetic force provides the centripetal force, so qvB = mv²/r, giving r = mv/(qB). The radius is directly proportional to speed, so doubling the speed doubles the radius.

Question 34External assessment Paper 1

A student measures the period of a pendulum several times and obtains values that are all very close to each other but consistently 5% larger than the accepted value. Which statement best describes the measurements?

  1. They are both accurate and precise
  2. They are accurate but not precise
  3. They are neither accurate nor precise
  4. They are precise but not accurate Correct answer.

Explanation

Precision refers to how closely repeated measurements agree with each other, while accuracy refers to how close they are to the true value. Clustered values that are all offset from the accepted value are precise but inaccurate, indicating a systematic error.

Question 35External assessment Paper 1

An ideal gas undergoes an isothermal expansion. Which statement about the gas is correct?

  1. Its internal energy increases as it expands
  2. It absorbs heat and does work on its surroundings Correct answer.
  3. It releases heat and has work done on it
  4. No heat is exchanged with the surroundings

Explanation

For an isothermal process the internal energy of an ideal gas is constant, so the first law gives Q = W. During expansion the gas does positive work, so it must absorb an equal amount of heat from its surroundings.

Question 36External assessment Paper 1

Light of frequency 6.0 × 10¹⁴ Hz is incident on a metal with work function 1.5 eV. Using h = 6.63 × 10⁻³⁴ J·s and 1 eV = 1.6 × 10⁻¹⁹ J, what is the maximum kinetic energy of the emitted photoelectrons?

  1. 2.5 eV
  2. 0.49 eV
  3. 1.5 eV
  4. 0.99 eV Correct answer.

Explanation

The photon energy is hf = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ ≈ 3.98 × 10⁻¹⁹ J, which is about 2.49 eV. Subtracting the work function of 1.5 eV gives a maximum kinetic energy of about 0.99 eV.

Question 37External assessment Paper 1

In a nuclear fission reaction, the total mass of the products is slightly less than the total mass of the reactants. What does this mass difference represent?

  1. Mass converted into electric charge
  2. Mass lost as stray neutrons
  3. Mass converted to energy released in the reaction Correct answer.
  4. Mass that becomes a magnetic field

Explanation

The mass defect is equivalent to the energy released, according to E = mc². This binding energy appears as kinetic energy of the fission fragments and neutrons.

Question 38External assessment Paper 1

A satellite in a low circular orbit experiences a small atmospheric drag that slowly reduces its orbital radius. What happens to its speed and total mechanical energy?

  1. Speed increases and total energy increases
  2. Speed increases and total energy decreases Correct answer.
  3. Speed decreases and total energy decreases
  4. Speed decreases and total energy stays the same

Explanation

For a circular orbit v = √(GM/r), so a smaller radius means a larger speed. The total energy E = −GMm/(2r) becomes more negative as r decreases, meaning the total energy decreases, with the lost energy dissipated as heat by drag.

Question 39External assessment Paper 1

Unpolarised light of intensity I₀ passes through two ideal polarising filters whose transmission axes are at 60° to each other. What is the intensity of the light emerging from the second filter?

  1. I₀/8 Correct answer.
  2. 3I₀/8
  3. I₀/4
  4. I₀/2

Explanation

The first filter transmits half the unpolarised intensity, giving I₀/2. Malus's law then gives (I₀/2)cos²60° = (I₀/2)(1/4) = I₀/8.

Question 40External assessment Paper 1

A capacitor is charged through a resistor by a constant-voltage supply. Which graph shape best describes how the charging current varies with time after the switch is closed?

  1. A linear rise from zero
  2. A straight line of constant negative gradient
  3. A straight horizontal line above zero
  4. An exponential decay towards zero Correct answer.

Explanation

The charging current is I = (V/R)e^(−t/RC), so it starts at V/R and decays exponentially towards zero as the capacitor voltage approaches the supply voltage. The rate of decay is set by the time constant RC.

Question 41External assessment Paper 2

A car travels 120 m due east in 8.0 s and then 80 m due west in 4.0 s. What is the magnitude of the car's average velocity for the whole journey?

  1. 5.0 m/s
  2. 16.7 m/s
  3. 2.5 m/s
  4. 3.3 m/s Correct answer.

Explanation

Average velocity is total displacement divided by total time. Displacement is 120 m east minus 80 m west, giving 40 m east, and total time is 12 s. Dividing 40 m by 12 s gives about 3.3 m/s.

Question 42External assessment Paper 2

An object is projected horizontally from a cliff and air resistance is negligible. Which statement about its motion is correct?

  1. Horizontal acceleration equals the acceleration due to gravity.
  2. Horizontal velocity stays constant while vertical acceleration is constant. Correct answer.
  3. Vertical velocity remains constant while horizontal acceleration is zero.
  4. Both horizontal and vertical velocities increase at the same rate.

Explanation

With negligible air resistance, no horizontal force acts, so horizontal velocity is constant. The only force is weight, giving a constant downward acceleration equal to g. The vertical velocity therefore changes uniformly while the horizontal component does not.

Question 43External assessment Paper 2

A 5.0 kg block rests on a horizontal surface with coefficient of static friction 0.40. What is the minimum horizontal force needed to just start the block moving? Take g = 9.8 m/s^2.

  1. 12.3 N
  2. 19.6 N Correct answer.
  3. 49 N
  4. 2.0 N

Explanation

The normal force on a horizontal surface equals the weight, mg = 5.0 x 9.8 = 49 N. The maximum static friction is the product of the coefficient and the normal force, 0.40 x 49 = 19.6 N. A slightly larger applied force is needed to overcome this limiting friction.

Question 44External assessment Paper 2

A moving trolley collides with a stationary trolley on a frictionless track and they stick together. Which quantity is conserved during the collision?

  1. Total kinetic energy of the two-trolley system
  2. Total momentum of the two-trolley system Correct answer.
  3. Total speed of the two trolleys
  4. Total acceleration of the two trolleys

Explanation

In any collision where external horizontal forces are negligible, total momentum of the system is conserved. A perfectly inelastic collision converts some kinetic energy into other forms, so kinetic energy is not conserved. Speed and acceleration are not conserved quantities in this sense.

Question 45External assessment Paper 2

A wave has a frequency of 250 Hz and a wavelength of 1.4 m. What is its speed?

  1. 0.0056 m/s
  2. 251 m/s
  3. 179 m/s
  4. 350 m/s Correct answer.

Explanation

The wave equation relates speed to frequency and wavelength by v = f times wavelength. Multiplying 250 Hz by 1.4 m gives 350 m/s. The units are consistent because hertz is per second and metres times per second gives metres per second.

Question 46External assessment Paper 2

Which statement correctly describes the electric field between two oppositely charged parallel plates?

  1. It is uniform in magnitude and points from the positive plate to the negative plate. Correct answer.
  2. It is strongest near the plate edges and weakest in the centre.
  3. It points from the negative plate to the positive plate everywhere.
  4. It decreases in magnitude as the distance from the positive plate increases.

Explanation

Between large parallel plates the field is approximately uniform, meaning constant magnitude and direction. Field lines run from positive charge to negative charge. Edge effects occur only near the boundaries, not in the central region.

Question 47External assessment Paper 2

A resistor of 12 ohms carries a current of 0.50 A. What is the power dissipated in the resistor?

  1. 24 W
  2. 6.0 W
  3. 0.042 W
  4. 3.0 W Correct answer.

Explanation

Power in a resistor can be found from P = I squared times R. Squaring 0.50 A gives 0.25, and multiplying by 12 ohms gives 3.0 W. The same result follows from first finding voltage and then using P = VI.

Question 48External assessment Paper 2

A 1500 kg car accelerates uniformly from rest to 20 m/s in 5.0 s. What is the net force acting on the car?

  1. 300 N
  2. 30 000 N
  3. 1500 N
  4. 6000 N Correct answer.

Explanation

Acceleration is change in velocity divided by time, giving 20 divided by 5.0, or 4.0 m/s^2. Applying Newton's second law, net force equals mass times acceleration, so 1500 kg times 4.0 m/s^2 equals 6000 N.

Question 49External assessment Paper 2

A ball is thrown vertically upward with an initial speed of 15 m/s. Taking g = 9.8 m/s^2 and ignoring air resistance, what is the maximum height reached?

  1. 22.9 m
  2. 1.53 m
  3. 11.5 m Correct answer.
  4. 7.5 m

Explanation

At maximum height the velocity is zero, so using v squared = u squared minus 2gh gives h = u squared divided by 2g. Substituting 15 squared over twice 9.8 gives 225 over 19.6, approximately 11.5 m.

Question 50External assessment Paper 2

A 2.0 kg object moving at 6.0 m/s collides head-on with a stationary 4.0 kg object and they stick together. What is their common velocity after the collision?

  1. 6.0 m/s
  2. 3.0 m/s
  3. 1.0 m/s
  4. 2.0 m/s Correct answer.

Explanation

Total momentum before the collision is 2.0 times 6.0, giving 12 kg m/s. After sticking together the combined mass is 6.0 kg, so the common velocity is 12 divided by 6.0, which is 2.0 m/s.