Physics Practice (Interactive)

Scottish Highers — Physics Practice

10 auto-graded practice problems. Select an answer, submit, and review the explanation.

Mechanics: Equations of Motion, Projectiles, Forces, and Energy

Q1. A car accelerates uniformly from rest at 3 m/s squared for 8 seconds. Using the equations of motion, what distance does the car travel during this time?

A. The car travels 96 metres. B. The car travels 24 metres. C. The car travels 48 metres. D. The car travels 72 metres.’,

Show answer — A

Answer: A — Using the equations of motion (kinematic equations): s = ut + (1/2)at^2, where s is displacement, u is initial velocity, a is acceleration, and t is time. Given: u = 0 m/s (starting from rest), a = 3 m/s^2, t = 8 s. Substituting: s = (0)(8) + (1/2)(3)(8^2) = 0 + (1.5)(64) = 96 metres. Alternatively, first find the final velocity: v = u + at = 0 + 3(8) = 24 m/s, then use v^2 = u^2 + 2as to get s = (v^2 - u^2)/(2a) = (576 - 0)/6 = 96 m. Both methods give the same result. The car reaches a final speed of 24 m/s after covering 96 metres.

Q2. A projectile is launched horizontally from a cliff edge at 15 m/s and takes 4 seconds to reach the ground. How far horizontally from the cliff base does the projectile land, and what is its vertical velocity just before impact?

A. It lands 60 m horizontally from the base and has a vertical velocity of 39.2 m/s just before impact (ignoring air resistance). B. It lands 30 m horizontally from the base and has a vertical velocity of 15 m/s just before impact. C. It lands 60 m horizontally from the base and has a vertical velocity of 15 m/s just before impact. D. It lands 120 m horizontally from the base and has a vertical velocity of 78.4 m/s just before impact.’,

Show answer — A

Answer: A — For projectile motion, horizontal and vertical components are treated independently (assuming no air resistance). Horizontal: the horizontal velocity is constant at 15 m/s (no horizontal acceleration). Horizontal distance = horizontal velocity x time = 15 x 4 = 60 m. Vertical: the initial vertical velocity is 0 m/s (launched horizontally), and the vertical acceleration is g = 9.8 m/s^2 (downward). Vertical velocity at impact: v = u + at = 0 + (9.8)(4) = 39.2 m/s (downward). The horizontal and vertical motions are independent, so the constant horizontal velocity of 15 m/s does not affect the vertical acceleration. The resultant velocity at impact would be the vector sum of horizontal (15 m/s) and vertical (39.2 m/s) components.

Q3. A 5 kg object is pulled across a horizontal surface with a force of 40 N at an angle of 30 degrees above the horizontal. If the coefficient of friction is 0.3, what is the acceleration of the object?

A. “The acceleration is approximately 3.78 m/s squared, calculated by resolving the applied force into horizontal and vertical components, accounting for the normal reaction force, and applying Newton’s second law.”, ‘The acceleration is 8 m/s squared, calculated directly by dividing the applied force by the mass. B. The acceleration is 5 m/s squared, calculated by subtracting the friction force from the applied force and dividing by mass. C. The acceleration is 2 m/s squared, calculated using only the horizontal component of the applied force.’,

Show answer — A

Answer: A — Resolve the applied force into components: horizontal = 40cos(30) = 34.64 N, vertical = 40sin(30) = 20 N (upward). The vertical component reduces the normal reaction force: N = mg - 20 = (5)(9.8) - 20 = 49 - 20 = 29 N. Friction force = coefficient of friction x N = 0.3 x 29 = 8.7 N. Net horizontal force = 34.64 - 8.7 = 25.94 N. Using Newton’s second law: F = ma, so a = F/m = 25.94/5 = 5.19 m/s^2. The precise answer depends on the value used for g (9.8 or 10). Using g = 9.8: a is approximately 5.2 m/s^2. The key point is that the angled force affects both the horizontal driving force and the normal reaction, which in turn affects the friction force. This is a multi-step problem requiring careful resolution of forces.

Dynamics and Space: Gravitation, Circular Motion, and Impulse

Q4. A satellite orbits the Earth at an altitude where the gravitational field strength is 4.5 N/kg. If the satellite has a mass of 800 kg and orbits at a radius of 8.0 x 10 to the power of 6 metres from the Earth’s centre, what is the centripetal force acting on the satellite?

A. The centripetal force is 3600 N, provided by the gravitational attraction: F = mg = 800 x 4.5 = 3600 N. B. The centripetal force is 800 N, calculated using F = ma where a = g/5. C. The centripetal force is 1800 N, calculated using F = mg/2 at the given altitude. D. The centripetal force is 7200 N, calculated using F = 2mg to account for orbital motion.’,

Show answer — A

Answer: A — The centripetal force required to keep the satellite in circular orbit is provided entirely by the gravitational force. In orbital motion, the gravitational force equals the centripetal force: F = mg, where g is the gravitational field strength at that altitude. With m = 800 kg and g = 4.5 N/kg: F = 800 x 4.5 = 3600 N. This force acts towards the centre of the Earth and is perpendicular to the satellite’s velocity, causing it to follow a circular path without changing speed. We can verify this is consistent with circular motion: using F = mv^2/r, we can find the orbital speed v = sqrt(Fr/m) = sqrt(3600 x 8 x 10^6 / 800) = sqrt(36000) = approximately 190 m/s. The principle that gravitational force equals centripetal force is fundamental to understanding orbital mechanics.

Q5. A 0.15 kg cricket ball travelling at 30 m/s is hit by a bat and returns in the opposite direction at 20 m/s. If the contact time between bat and ball is 0.05 seconds, what is the average force exerted on the ball by the bat?

A. “The average force is 150 N in the direction of the ball’s return travel, calculated using impulse = change in momentum.”, ‘The average force is 50 N, calculated by dividing the average speed by the contact time. B. The average force is 75 N, calculated using the change in speed divided by the contact time and mass. C. The average force is 300 N, calculated by multiplying the total distance by the mass.’,

Show answer — A

Answer: A — Impulse (F x delta t) equals the change in momentum (delta p = m x delta v). Initial momentum: p1 = mu = 0.15 x 30 = 4.5 kg m/s (taking the initial direction as positive). Final momentum: p2 = 0.15 x (-20) = -3.0 kg m/s (negative because opposite direction). Change in momentum: delta p = p2 - p1 = -3.0 - 4.5 = -7.5 kg m/s. The negative sign indicates the impulse acts in the opposite direction to the initial motion (i.e., the direction the ball is sent back). Average force F = delta p / delta t = 7.5 / 0.05 = 150 N. The magnitude is 150 N and it acts in the direction of the ball’s return. This demonstrates Newton’s second law in its impulse-momentum form and highlights why taking direction into account is essential.

Electricity: Ohm’s Law, Kirchhoff’s Laws, and Circuits

Q6. A circuit contains a 12 V battery with internal resistance 0.5 ohms connected to an external resistor of 5.5 ohms. What is the current in the circuit and the terminal potential difference (p.d.) across the battery?

A. The current is 2.0 A and the terminal p.d. is 11 V. B. The current is 2.2 A and the terminal p.d. is 12 V. C. The current is 1.8 A and the terminal p.d. is 10.8 V. D. The current is 2.4 A and the terminal p.d. is 13.2 V.’,

Show answer — A

Answer: A — The total resistance in the circuit is the sum of internal resistance and external resistance: R_total = r + R = 0.5 + 5.5 = 6.0 ohms. Using Ohm’s law: current I = EMF / R_total = 12 / 6.0 = 2.0 A. The terminal p.d. (the voltage across the battery terminals, which is also the voltage across the external resistor) is: V = EMF - Ir = 12 - (2.0)(0.5) = 12 - 1 = 11 V. Alternatively, V = IR = 2.0 x 5.5 = 11 V. The difference between the EMF (12 V) and terminal p.d. (11 V) is the lost volts (1 V) across the internal resistance. This demonstrates how internal resistance reduces the useful voltage available to the external circuit.

Q7. In a series circuit with two resistors of 4 ohms and 6 ohms, a current of 2 A flows. What is the potential difference across each resistor and the total potential difference supplied by the battery?

A. The p.d. across the 4 ohm resistor is 8 V, across the 6 ohm resistor is 12 V, and the total p.d. is 20 V. B. The p.d. across each resistor is 10 V and the total p.d. is 10 V. C. The p.d. across the 4 ohm resistor is 12 V, across the 6 ohm resistor is 8 V, and the total p.d. is 20 V. D. The p.d. across each resistor is 2 V and the total p.d. is 4 V.’,

Show answer — A

Answer: A — In a series circuit, the current is the same through all components. Using Ohm’s law V = IR for each resistor: p.d. across 4 ohm resistor = 2 x 4 = 8 V; p.d. across 6 ohm resistor = 2 x 6 = 12 V. Kirchhoff’s second law (voltage law) states that the sum of p.d.s around a closed loop equals the sum of EMFs. Therefore, the total p.d. supplied by the battery = 8 + 12 = 20 V. This confirms Kirchhoff’s voltage law: the EMF equals the sum of potential differences across all components in the series circuit. The total resistance is 4 + 6 = 10 ohms, and V = IR = 2 x 10 = 20 V, consistent with the result. Kirchhoff’s laws are fundamental to circuit analysis.

Q8. In the UK mains supply, the root mean square (r.m.s.) voltage is 230 V and the frequency is 50 Hz. What is the peak voltage of the supply and what is the time for one complete cycle?

A. The peak voltage is approximately 325 V and one complete cycle takes 0.02 seconds (20 milliseconds). B. The peak voltage is 230 V and one complete cycle takes 0.05 seconds (50 milliseconds). C. The peak voltage is approximately 163 V and one complete cycle takes 0.01 seconds (10 milliseconds). D. The peak voltage is 460 V and one complete cycle takes 0.04 seconds (40 milliseconds).‘,

Show answer — A

Answer: A — For a sinusoidal AC supply, the peak voltage V_peak = V_rms x sqrt(2). With V_rms = 230 V: V_peak = 230 x 1.414 = approximately 325 V. The r.m.s. value is the DC equivalent that would deliver the same average power to a resistive load. The period (time for one complete cycle) T = 1/f, where f = 50 Hz. Therefore T = 1/50 = 0.02 seconds = 20 milliseconds. During one cycle, the voltage varies from +325 V through 0 to -325 V and back. The UK mains frequency of 50 Hz means the voltage completes 50 full cycles every second. Understanding the relationship between peak and r.m.s. values and between frequency and period is essential for AC circuit analysis at Higher level.

Particles and Waves: Photoelectric Effect and Interference

Q9. In the photoelectric effect, light of frequency 8.0 x 10 to the power of 14 Hz is incident on a metal surface with a work function of 4.0 x 10 to the power of minus 19 joules. The maximum kinetic energy of the emitted photoelectrons is approximately 1.30 x 10 to the power of minus 19 J. Which statement best explains this result?

A. The photon energy (5.30 x 10 to the power of minus 19 J) exceeds the work function, so the excess energy becomes the maximum kinetic energy of the photoelectron: E_k(max) = hf - phi. B. The photon energy is less than the work function, so no photoelectrons are emitted regardless of intensity. C. The maximum kinetic energy depends on the intensity of the light, not the frequency, so increasing the intensity would increase the kinetic energy. D. The photoelectric effect only occurs with ultraviolet light, so visible light cannot cause electron emission from any metal surface.’,

Show answer — A

Answer: A — Einstein’s photoelectric equation: hf = phi + E_k(max), where hf is the photon energy, phi is the work function (minimum energy needed to release an electron from the surface), and E_k(max) is the maximum kinetic energy of emitted photoelectrons. Photon energy: E = hf = (6.63 x 10^-34)(8.0 x 10^14) = 5.30 x 10^-19 J. Since this exceeds the work function (4.0 x 10^-19 J), photoelectrons are emitted. The excess energy becomes kinetic energy: E_k(max) = hf - phi = (5.30 - 4.0) x 10^-19 = 1.30 x 10^-19 J. Key points: the effect depends on frequency (not intensity), there is a threshold frequency f_0 = phi/h below which no emission occurs, and each photon can eject at most one electron. This supports the particle model of light.

Q10. In a double-slit interference experiment, light of wavelength 600 nm produces bright fringes 3.0 mm apart on a screen. If the slit-to-screen distance is 1.5 m, what is the slit separation?

A. The slit separation is 3.0 x 10 to the power of minus 4 metres (0.30 mm), calculated using the interference formula. B. The slit separation is 6.0 x 10 to the power of minus 4 metres (0.60 mm), calculated using the diffraction formula.’, “The slit separation is 1.5 x 10 to the power of minus 4 metres (0.15 mm), calculated using Young’s double slit formula.”, ‘The slit separation is 9.0 x 10 to the power of minus 4 metres (0.90 mm), calculated using the grating equation.’,

Show answer — A

Answer: A — For double-slit interference, the fringe spacing is given by: wavelength = (fringe spacing x slit separation) / slit-to-screen distance, rearranged as: slit separation d = (wavelength x slit-to-screen distance) / fringe spacing. Substituting: d = (600 x 10^-9 x 1.5) / (3.0 x 10^-3) = (9.0 x 10^-7) / (3.0 x 10^-3) = 3.0 x 10^-4 m = 0.30 mm. This is the standard Young’s double-slit formula: lambda = dx/W, where lambda is wavelength, d is slit separation, x is fringe spacing, and W is the distance from slits to screen. The interference pattern shows bright fringes where path difference equals whole numbers of wavelengths, and dark fringes where it equals half-wavelengths. This experiment provided key evidence for the wave nature of light.

Intuition

Physics connects mathematical models to physical reality: Every physics equation describes a real relationship between measurable quantities. The art of physics is knowing which model applies to which situation — Newtonian mechanics for everyday objects, quantum mechanics for atoms, relativity for near-light speeds.

Why it matters: Physics understanding enables technological innovation, from designing bridges to developing new medical imaging techniques.

The key insight: Units are your sanity check — if the units on both sides of an equation do not match, the equation is wrong or has been applied incorrectly.

Common Mistakes

Confusing scalar and vector quantities: Speed is scalar; velocity is vector. Force, acceleration, and momentum are vectors. Using scalar addition for vector quantities (e.g., adding speeds instead of velocities) gives wrong answers in multi-directional problems.

Forgetting unit conversions: Physics problems mix SI and non-SI units. Converting km/h to m/s requires dividing by 3.6, not multiplying. Getting the conversion direction wrong doubles the error. Always convert to SI base units at the start.

Using the wrong kinematic equation: Four kinematic equations exist, each missing one variable. Using the equation that contains the unknown you want to find avoids unnecessary steps. If you need final velocity without time, use $v^2 = u^2 + 2as$ .

Cross-References

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