HSC Exam Practice

Sample response. The time of flight is the duration a charged particle spends inside the region between two parallel plates before exiting. It is calculated using \(t = L/v_x\), where \(L\) is the length of the plates and \(v_x\) is the horizontal component of the particle’s velocity. This formula is valid when \(v_x\) is constant (i.e., no horizontal force acts) and the particle exits the plates without striking them.

Marking notes. 1 mark for a clear definition of time of flight (time inside the field region); 1 mark for stating \(t = L/v_x\) and identifying at least one validity condition (uniform v_x OR particle doesn’t hit plate).

Sample response. The trajectory is parabolic because the particle experiences two independent, simultaneous motions: (1) constant horizontal velocity (\(v_x = \text{const}\), no horizontal force), so horizontal displacement \(x = v_x t\); and (2) constant vertical acceleration \(a = qE/m\) due to the uniform electric field, so vertical displacement \(y = \frac{1}{2}at^2\). Eliminating \(t\): \(y = \frac{a}{2v_x^2} x^2\), which is the equation of a parabola (\(y \propto x^2\)).

Marking notes. 1 mark for identifying horizontal motion as constant velocity with no horizontal force; 1 mark for identifying vertical motion as constant acceleration due to the electric force; 1 mark for deriving or stating the \(y \propto x^2\) parabolic relationship by eliminating \(t\).

Sample response. Both particles follow parabolic trajectories, but they curve in opposite directions. The proton (positive charge) is accelerated toward the negative (bottom) plate — downward in a downward field. The electron (negative charge) is accelerated toward the positive (top) plate — upward, opposing the field direction. The magnitude of the electron’s acceleration is far greater: \(a_e = eE/m_e \gg a_p = eE/m_p\) because \(m_e \ll m_p\) (electron mass is ~1836 times smaller than proton mass). Both have the same charge magnitude \(e\) and the same \(v_x\), so the electron is deflected far more for the same plate length.

Marking notes. 1 mark for correctly stating both follow parabolas but in opposite directions; 1 mark for correctly identifying the direction for each particle (electron toward + plate, proton toward − plate); 1 mark for comparing magnitude of accelerations using \(a = qE/m\) and stating the electron has far larger acceleration due to its much smaller mass.

(a) Time of flight. \(t = L/v_x = 0.090\,\text{m} / (3.0\times10^6\,\text{m/s}) = 3.0\times10^{-8}\,\text{s}\). [1 mark]

(b) Vertical deflection. \(a = qE/m_e = (1.60\times10^{-19}\times8000)/(9.11\times10^{-31}) = 1.404\times10^{15}\,\text{m/s}^2\). \(y = \frac{1}{2}at^2 = \frac{1}{2}\times1.404\times10^{15}\times(3.0\times10^{-8})^2 = 6.32\times10^{-1}\,\text{m}\). Wait — checking: \(\frac{1}{2}\times1.404\times10^{15}\times9\times10^{-16} = \frac{1}{2}\times1.264\times10^{-1} = 6.32\times10^{-2}\,\text{m} = 6.32\,\text{cm}\). [2 marks: 1 for correct a, 1 for correct y]

(c) Exit velocity. \(v_y = at = 1.404\times10^{15}\times3.0\times10^{-8} = 4.21\times10^7\,\text{m/s}\). \(v = \sqrt{(3.0\times10^6)^2+(4.21\times10^7)^2} = \sqrt{9.0\times10^{12}+1.773\times10^{15}} \approx 4.22\times10^7\,\text{m/s}\). [2 marks: 1 for correct v_y, 1 for correct resultant]

Marking notes. Accept answers within 2% rounding tolerance. Penalise only once for a carried arithmetic error. Must show formula substitution for full marks.

Sample response. The deflection increases by a factor of 4. Using \(y = \frac{1}{2}at^2\) and \(t = L/v_x\): \(y = \frac{aL^2}{2v_x^2}\). If \(L\) is doubled (\(L \to 2L\)), then \(y \to \frac{a(2L)^2}{2v_x^2} = \frac{4aL^2}{2v_x^2} = 4y\). Because \(t = L/v_x\) doubles, and \(y \propto t^2\), the deflection quadruples. The acceleration \(a = qE/m = qV/(md)\) is unchanged because the potential difference and separation are both constant.

Marking notes. 1 mark for correctly predicting the deflection increases by a factor of 4 (quadruples); 1 mark for correctly deriving the factor-of-4 relationship using \(y \propto L^2\); 1 mark for explicitly stating that the acceleration is unchanged (V and d are constant, so E and hence a are constant).

Completed table. \(t = L/v_x = 0.050/(4.0\times10^7) = 1.25\times10^{-9}\,\text{s}\) (constant for all rows). \(d = 0.010\,\text{m}\).

Row 1 (V=50): E = 50/0.010 = 5000 V/m; a = (1.60×10⁻¹⁹×5000)/(9.11×10⁻³¹) = 8.78×10¹⁴ m/s²; y = ½×8.78×10¹⁴×(1.25×10⁻⁹)² = 6.86×10⁻⁴ m ≈ 0.69 mm.

Row 2 (V=100): E = 10000 V/m; a = 1.756×10¹⁵ m/s²; y = 1.37×10⁻³ m ≈ 1.37 mm.

Row 3 (V=150): E = 15000 V/m; a = 2.634×10¹⁵ m/s²; y = 2.06×10⁻³ m ≈ 2.06 mm.

Row 4 (V=200): E = 20000 V/m; a = 3.512×10¹⁵ m/s²; y = 2.74×10⁻³ m ≈ 2.74 mm.

(b) \(y\) is directly proportional to \(V\) (linear relationship). From \(y = \frac{1}{2}at^2 = \frac{1}{2}\cdot\frac{qV}{md}\cdot\frac{L^2}{v_x^2}\), all quantities except V are constant, so \(y = kV\) where \(k = \frac{qL^2}{2mdv_x^2}\) is a constant. [2 marks: 1 for stating directly proportional/linear; 1 for deriving from the formula]

(c) The maximum voltage is limited by the plate separation \(d\): if the deflection exceeds half the plate separation (\(|y| > d/2\)), the electron strikes one of the plates and does not reach the screen. When this limit is exceeded, the beam is lost — it hits the plate rather than the detector, and no signal appears on screen. [2 marks: 1 for identifying the physical constraint (d/2 limit); 1 for explaining what happens when exceeded (electron hits plate, no signal)]

Marking notes for (a). 4 marks for table: 1 mark for correct method (E = V/d, a = qE/m, t calculation, y formula) demonstrated in the sample row; 1 mark for two or more correct E values; 1 mark for two or more correct a values; 1 mark for two or more correct y values. Accept rounding to 2 significant figures.

Sample Band 6 response (annotated).

Similarities. Both projectile motion under gravity and charged-particle motion in a uniform electric field involve a constant force acting perpendicular to the initial velocity, with no force along the horizontal direction. In both cases, horizontal motion is uniform (\(v_x = \text{const}\)) and vertical motion is uniformly accelerated (\(a = \text{const}\)). The SUVAT equations apply identically to both: \(y = u_y t + \frac{1}{2}at^2\), \(v_y = u_y + at\), and eliminating \(t\) gives \(y = (a/2v_x^2)x^2\) — a parabola in both cases. [2 marks]

Differences (at least two). (1) Gravity acts on all matter equally, with a direction fixed as “downward,” and its acceleration \(g = 9.8\,\text{m/s}^2\) is the same regardless of mass. Electric acceleration \(a = qE/m\) depends on the charge-to-mass ratio of the specific particle: an electron has \(q/m \approx 1.76\times10^{11}\,\text{C/kg}\) versus \(g = 9.8\,\text{m/s}^2\) — electric acceleration is roughly \(10^{10}\) times larger under typical laboratory fields. (2) The direction of electric acceleration depends on the sign of the charge: a positive particle accelerates in the direction of the field; a negative particle accelerates opposite to the field. Gravity has no such sign dependence — all massive objects are accelerated downward. (3) The magnitude of electric acceleration can be controlled by adjusting voltage; gravitational acceleration cannot be adjusted. [2 marks for two well-explained differences]

Numerical example. Consider an electron (\(m_e = 9.11\times10^{-31}\,\text{kg}\)) in field \(E = 1000\,\text{V/m}\). Electric acceleration: \(a_E = eE/m_e = (1.60\times10^{-19}\times1000)/(9.11\times10^{-31}) = 1.76\times10^{14}\,\text{m/s}^2\). Compare to gravitational acceleration \(g = 9.8\,\text{m/s}^2\). Ratio: \(a_E/g \approx 1.8\times10^{13}\). For the same entry speed and plate geometry, the electric deflection would be ~18 trillion times greater than the gravitational deflection — demonstrating why gravity is negligible in electron beam calculations. [2 marks: 1 for numerical calculation of both accelerations; 1 for comparison and conclusion that gravity is negligible]

Technological application. In a cathode ray oscilloscope (CRO), the Y-deflection plates exploit the parabolic trajectory of electrons. Electrons accelerated through a high voltage enter a pair of deflection plates with an applied signal voltage. The vertical deflection of the beam on the phosphor screen is proportional to the plate voltage (as derived above: \(y \propto V\)). This allows the CRO to display time-varying electrical signals as a visible trace. The precise mathematical control of deflection via \(y = \frac{aL^2}{2v_x^2}\) allows calibrated measurement of voltage amplitude and frequency from the deflection pattern. [2 marks: 1 for correctly naming and describing a real application; 1 for explaining how trajectory analysis (\(y \propto V\) or parabolic path) is used in that application]

Conclusion. The analogy between projectile motion and charged-particle motion in electric fields is mathematically exact: both follow parabolic paths described by identical SUVAT equations. However, electric acceleration vastly exceeds gravitational acceleration and depends on particle charge sign, making the analogy powerful but physically very different in scale and directionality. [1 mark for explicit evaluative conclusion integrating similarities and differences]

Marking criteria (9 marks): 1 = two or more similarities stated with reference to the mathematical form of equations; 1 = second similarity (e.g. independence of motion, SUVAT applies equally); 1 = first difference (direction depends on charge sign — gravity always down); 1 = second difference (magnitude: electric \(a \gg g\), or controllability of E); 1 = correct numerical calculation of electric acceleration for a specific particle; 1 = quantitative comparison with g and statement that gravity is negligible; 1 = names a real technological application and describes it accurately; 1 = explains how the parabolic/proportional trajectory analysis underlies that technology; 1 = explicit evaluative conclusion integrating similarities and differences with a judgement on the quality/limits of the analogy.