A-Level Physics: Mechanics and Waves Practice

A-Level Physics — Mechanics and Waves Practice

18 MCQ practice problems covering core A-Level Mechanics and Waves content.

What These Questions Test

These problems require you to apply the core mechanics and wave equations, interpret graphs, and reason about physical situations. Topics range from basic kinematics to quantum phenomena.

Typical question types:

• Kinematics (SUVAT): Find displacement, velocity, or time under constant acceleration. Interpret velocity-time and displacement-time graphs. Distinguish between distance and displacement, speed and velocity.
• Dynamics and Newton’s Laws: Identify forces acting on objects. Apply $F = ma$ to find acceleration or net force. Use free-body diagrams to resolve forces at angles.
• Momentum: Calculate momentum changes in collisions. Apply conservation of momentum. Distinguish elastic and inelastic collisions. Use impulse-momentum theorem.
• SHM: Relate displacement, velocity, and acceleration in SHM. Interpret graphs of these quantities. Use $a = -\omega^{2}x$ and energy transformations.
• Waves: Use $v = f\lambda$. Interpret wavefront diagrams and diffraction patterns. Understand the conditions for constructive and destructive interference.
• Quantum and Thermal: Apply $E = hf$ and the photoelectric equation. Use $Q = mc\Delta\theta$ and $Q = mL$ for thermal calculations.

Approach Strategy

1. List knowns and unknowns. For mechanics problems, write down $u$, $v$, $a$, $s$, $t$ and identify which three you know. This tells you which SUVAT equation to use.
2. Choose a sign convention. Decide which direction is positive and stick to it. Upward as positive is conventional; gravity then gives $a = -9.81 \, \text{m/s}^{2}$.
3. Draw diagrams. For forces, a free-body diagram is almost always necessary. For waves, sketch the wavefront or ray diagram.
4. Watch for hidden assumptions. “Smooth” means no friction. “Light” means massless. “Inextensible” means the string/rope does not stretch.

Intuition

Mechanics is fundamentally about predicting motion. If you know the forces, you know the acceleration ($F = ma$), and from acceleration you can work out velocity and position using kinematics. Think of it as a chain: force $\rightarrow$ acceleration $\rightarrow$ velocity $\rightarrow$ position.

For waves, the key insight is that the medium does not travel — only the disturbance does. Each particle of the medium oscillates locally while the wave pattern propagates. This is why you can send a signal across the ocean without the water moving from one side to the other.

Worked Examples

Example 1: SUVAT with a Twist

Problem: A ball is thrown vertically upward at 20 m/s from the edge of a 45 m cliff. Find the time when it hits the ground. ($g = 9.81 \, \text{m/s}^2$)

Solution: Step 1: Choose sign convention: upward = positive, so $a = -g = -9.81$

Step 2: When the ball hits the ground, displacement $s = -45$ m (below starting point)

Step 3: Use $s = ut + \frac{1}{2}at^2$: $-45 = 20t - 4.905t^2$

Step 4: Rearrange: $4.905t^2 - 20t - 45 = 0$

Step 5: Quadratic formula: $t = \frac{20 \pm \sqrt{400 + 4(4.905)(45)}}{9.81} = \frac{20 \pm \sqrt{400 + 882.9}}{9.81} = \frac{20 \pm 35.83}{9.81}$

Step 6: $t = \frac{55.83}{9.81} = 5.69$ s (reject negative root)

Key insight: The displacement is negative because the ground is below the starting point. Always define your sign convention before writing equations.

Example 2: Photoelectric Effect

Problem: Light of wavelength 200 nm falls on a metal with work function 3.0 eV. Find the maximum kinetic energy of emitted electrons.

Solution: Step 1: Calculate photon energy: $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9}} = 9.94 \times 10^{-19}$ J

Step 2: Convert to eV: $E = \frac{9.94 \times 10^{-19}}{1.60 \times 10^{-19}} = 6.21$ eV

Step 3: Photoelectric equation: $KE_{max} = hf - \phi = E - \phi$

Step 4: $KE_{max} = 6.21 - 3.0 = 3.21$ eV

Key insight: The photoelectric effect depends on frequency (energy), not intensity (brightness). Increasing intensity increases the number of electrons, not their maximum energy.

Example 3: SHM Energy

Problem: A mass on a spring has amplitude 0.10 m and maximum speed 0.50 m/s. Find the spring constant if the mass is 0.20 kg.

Solution: Step 1: Maximum kinetic energy equals total energy: $E = \frac{1}{2}mv_{max}^2 = \frac{1}{2}(0.20)(0.50)^2 = 0.025$ J

Step 2: Total energy in SHM: $E = \frac{1}{2}kA^2$

Step 3: Solve for $k$: $k = \frac{2E}{A^2} = \frac{2(0.025)}{(0.10)^2} = \frac{0.05}{0.01} = 5.0$ N/m

Step 4: Verify: $\omega = \sqrt{k/m} = \sqrt{5/0.2} = 5$ rad/s, $v_{max} = \omega A = 5 \times 0.10 = 0.50$ m/s ✓

Key insight: In SHM, energy oscillates between kinetic and potential. At maximum displacement, all energy is potential; at equilibrium, all energy is kinetic.

Common Mistakes

1. Confusing mass and weight. Mass ($m$) is in kilograms and is constant. Weight ($W = mg$) is a force in newtons and varies with gravitational field strength. A common error is using $W$ in $F = ma$ when $m$ is already the mass.
2. Using the wrong wave equation. $v = f\lambda$ applies to all waves, but the speed $v$ depends on the medium. Changing frequency changes wavelength, not speed (in the same medium). This is a frequent conceptual trap.
3. Forgetting that SHM acceleration is not constant. Unlike uniformly accelerated motion, in SHM the acceleration depends on position ($a = -\omega^{2}x$). You cannot use SUVAT for SHM.
4. Misinterpreting “threshold frequency.” It is the minimum frequency of incident light, not intensity, that determines whether electrons are emitted in the photoelectric effect.

Cross-References

• Electricity: Mechanics connects to electricity
• Nuclear: Nuclear physics uses mechanics
• Waves: Oscillations link mechanics and waves