CIE 9702 Physics 2015 Jun Question 5

Question: 5. A graph of potential difference (V) with respect to the electric current (I) for a semiconductor diode is shown below.

(a) Describe the variation of the electrical resistance of the semiconductor diode from V = −0.5 V to V = 0.8 V.

(b) Sketch a graph of potential difference (V) with respect to the electric current (I) for a filament lamp.

(c) A battery is connected to a filament lamp and a switch (See figure below). The electromotive force of the battery is 12.0 V and its internal electrical resistance is 0.50 W. If the potential difference applied across the filament lamp is 12.0 V, the electrical power consumption is 36.0 W.

(i) Determine the electrical resistance of the filament lamp when the potential difference across it is 12.0 V.

(ii) Determine the electrical resistance of the filament lamp when the switch is closed and the electric current through the lamp is 2.8 A.

(d) Explain how the two values of electrical resistance determined in part (c) provide evidence for the shape of the graph that you have sketched in part (b).

Mark scheme:

5 (a) The electrical resistance of the diode is very high or infinite for negative voltages to about 0.4V [1]

The electrical resistance of the diode decreases from 0.4 V onwards [1]

(b) The graph is initially a straight line through the origin, and it changes into a curve at a higher voltage with decreasing gradient. [1]

A similar (inverted) graph is shown in the negative voltage region [1]

(c) (i) R = 12² / 36 = 4.0 Ω   [1]

(ii) E = 12 = 2.8 × (R + r)   [1]

(R + r) = 4.29 Ω   [1]

R = 3.8 Ω   [1]

(d) The electrical resistance of the filament lamp increases with an increase of potential difference and electrical current. [1] 

Possible answers:

(a) For negative voltages and up to about 0.45V, the electrical resistance is very high (or approaches infinity).

For positive voltages beyond 0.45 V, the electrical resistance decreases gradually.

(b) The graph of potential difference (V) with respect to the electric current (I) for a filament lamp is shown below.

(c) (i) P = VI = V²/R Þ R = V²/P = 12²/36 = 4.0 Ω

(ii) Assuming the electrical resistance of the wire is zero,

E = I (R + r) Þ 12 = 2.8 (R + 0.5)

Thus, R = 12/2.8 – 0.5 = 3.79 Ω (or 3.8 Ω)

(d) The electrical resistance of the lamp increases from about 3.8 Ω to 4.0 Ω when the potential difference or electric current increases. 

Feynman’s insights or goofs:

In his lectures on physics, Feynman explains that “[t]he voltage-current characteristic of Eq. (14.14) I = I₀(e^+qDV/kT – 1) is shown in Fig. 14–10. It shows the typical behavior of solid state diodes—such as those used in modern computers. We should remark that Eq. (14.14) is true only for small voltages. For voltages comparable to or larger than the natural internal voltage difference V, other effects come into play and the current no longer obeys the simple equation (Feynman et al., 1966, section 14–5 Rectification at a semiconductor junction).” That is, we should not expect the semiconductor diode to obey the simple equation in the real world especially for a higher potential difference applied. However, physics students should know that there are many different models of the diode.

Below are four simple models of the diode:

1. Model 1 (Ideal diode):

An ideal diode behaves like a perfect conductor (0 W and 0 V) when an applied voltage is forward-biased (“positive” voltage) and behaves like a perfect insulator (¥ W) when the applied voltage is reverse-biased (“negative” voltage). In other words, if the ideal diode is reverse-biased, the electric current through the diode is zero because its electrical resistance is very high (or infinite). On the other hand, this diode conducts at 0 V and allows any electric current passing through it. The current-voltage graph of the ideal diode is shown below.

2. Model 2 (An idealized diode that has a cut-in voltage):

This model includes the concept of cut-in voltage (or knee voltage). When a diode is formed by joining a p-type silicon to an n-type silicon, there are diffusions of electrons into the p-type silicon and holes into the n-type silicon such that it results in a potential hill or cut-in voltage. To be realistic, one may adopt a model of idealized diode that has a cut-in voltage of about 0.7 V. This voltage may vary and it is dependent on materials such as silicon or germanium used. The current-voltage graph of the ideal diode having the cut-in voltage is shown below.

3. Model 3 (An idealized diode that has a cut-in voltage and electrical resistance): This model includes the concept of cut-in voltage and electrical resistance. To be more realistic, one may prefer a model of idealized diode that also has a resistor. However, this diode is not a linear device and its electrical resistance is not constant because the electric current through it does not directly proportional to the applied voltage. The current-voltage graph of the ideal diode that has an electrical resistance is shown below.

4. Model 4 (Shockley diode):

The current-voltage characteristic of Shockley diode is based on the equation: I = I₀(e^+qDV/kT – 1) where II₀ is the saturation current of the diode, q is the electric charge of an electron, k is Boltzmann’s constant, and T is the absolute temperature (in Kelvin). In this model, the saturation current is proportional to the cross-sectional area of the diode. The current-voltage graph of a Shockley diode is shown below.

Importantly, in his lectures on computation, Feynman (1996) elaborates that “[i]n the real world, I(V) cannot just keep on growing exponentially with V; other phenomena will come into play, and the potential difference across the junction will differ from that applied. Note also that the current trickle that exists in the reverse-biased case, catastrophically increases (negatively) at a certain voltage, the so-called breakdown voltage (p. 221).” For instance, a breakdown-induced “meltdown” due to a higher potential difference applied such that the diode does not approximately obey Shockley’s equation. However, physics teachers should also have a deeper understanding how physicists and electrical engineers idealize the current-voltage characteristics of a semiconductor diode under different temperature conditions.

References:

1. Feynman, R. P. (1996). Feynman lectures on computation. Reading, Massachusetts: Addison-Wesley.

2. Feynman, R. P., Leighton, R. B., & Sands, M. (1966). The Feynman lectures on physics Vol III: Quantum Mechanics. Reading, MA: Addison-Wesley.

CIE 9702 Physics 2015 Jun Question 3 & 4

Question:
3. In this question, two spheres are supported by strings. When the two spheres are pulled back and released, they collide and rebound at a position as shown below.

The table below provides data for the two spheres X and Y during this collision. (Convention: the positive direction is horizontal and it is to the right.)

(a) By using the law of conservation of linear momentum to determine the mass of sphere Y. [3]
(b) State and explain whether the collision of the two spheres is elastic. [1] 
(c) Use Newton’s second law of motion and Newton’s third law of motion to explain why the magnitudes of the change in momentum of both spheres have the same value. [3]

4. A spring is attached to a movable slider at A and a fixed support at B as shown below. A cart (m = 1.7 kg) moves with a velocity v towards the slider. The cart collides with the slider and the spring is compressed such that they are momentarily at rest. 

A graph of compression x of the spring is varied with force F exerted on the spring is shown below.

(a) The cart and the spring are momentarily at rest when the restoring force (F) of the spring is 4.5 N.
(i) Show that the restoring force (compression) of the spring follows Hooke’s law. [2]
(ii) Calculate the elastic constant (or spring constant, k) of the spring. [2]
(iii) Determine the elastic potential energy (E_P) stored in the spring due to the cart when it is momentarily at rest. [3]

(b) Calculate the speed of the cart immediately after it collides with the slider. You may assume that the kinetic energy of the cart (after the collision) is completely converted to the elastic potential energy of the spring. [2]

Mark Scheme:
3(a) Left hand side: 50 × (+4.5) + M × (– 2.8)   [1]. 
Right hand side: 50 × (–1.8 ) + M × (+1.4)   [1]. 
M = 75 g   [1]. 
(b) The relative speed of approach does not equal to the relative speed of separation. Inelastic collision [1]. 
(c) Newton’s third law of motion: the magnitude of force on X is equal to the magnitude of force on Y [1]. 
Newton’s second law of motion: force is equal to the rate of change of momentum [1]. 
The time of collision is the same for both balls [1]. 

4 (a) (i) The spring constant is determined by two sets of coordinates [1] 
F / x is shown to be constant [1] 
Alternative: the gradient is calculated by using at least one point on the graph [1] 
Show that the y-intercept is zero [1]. 
(ii) The gradient is calculated, e.g., from 4.5 / 1.8 × 10^–2   [1]. 
The spring constant is 250 N m^–1   [1]. 
(iii) work done = ½Fx or ½kx²   [1] 
E_p = ½Fx = 0.5 × 4.5 × 1.8 × 10^–2 or E_p = ½kx² = 0.5 × 250 × (1.8 × 10^–2)²   [1] 
Energy stored = 0.0405 J   [1]. 

(b) ½mv² = 0.0405  [1]  v = 0.218 m s^–1  [1]

Possible answers: 
3(a) Using principle of conservation of linear momentum, mu_X + Mu_Y = mv_X + Mv_Y.
We have 50 × (+4.5) + M × (– 2.8) = 50 × (–1.8 ) + M × (+1.4)    
Thus, M = 75 g = 0. 075 kg. 

(b) For elastic collision, the relative speed of approach is equal to the relative speed of separation. 
However, (u_Y – u_X) = (–2.8 – [+4.5]) = –7.3 and (v_Y – v_X) = (+1.4 –[–1.8]) = +3.2.
That is, the magnitudes of (u_Y – u_X) and (v_Y – v_X) are not equal.
Thus, the collision is inelastic.

(c) By using Newton’s third law of motion, the magnitude of force on X is equal to the magnitude of force on Y, but they are in opposite direction (F_X= –F_Y). 
By using Newton’s second law of motion, force is equal to the rate of change of momentum. Thus, F_X = Δp_X/Δt_X = –F_Y = –Δp_Y/Δt_Y). 
Since the time of collision is the same for both balls, hence the change in momentum is the same. (i.e. Δt_X = Δt_Y implies Δp_X = –Δp_Y or |Δp_X| = |Δp_Y|)    

4 (a) (i) The straight line graph can be represented by y = mx + c.

The gradient can be calculated by using two coordinates:

m = (y₂ – y₁)/(x₂ – x₁) = (4.5 – 1.5)/(0.018 – 0.006) = 250

The y-intercept can be calculated as follows:

c = y – mx = 1.5 – (250)(0.006) = 0

The graph (F = kx) is a straight line having a constant gradient and it passes through the origin. Thus, it obeys Hooke’s law.

(ii) From the graph, when x₁ = 1.8 cm and F₁ = 4.5 N, k = 4.5/0.018 = 250 N/m. 

Alternatively, we may use x₂ = 0.6 cm and F₂ = 1.5 N, k = 1.5/0.006 = 250 N/m.

(iii) The elastic potential energy E_P stored in the spring = ½kx² = 0.5 × 250 × (0.018)² = 0.0405 J

(b) KE = ½mv² = 0.0405  
Thus, v = (2× KE/m)^1/2 = [2 × 0.0405 / 1.7]^1/2 =0.218 m s^–1 

Feynman’s insights and goofs: 
In his lectures, Feynman uses terms such as “elastic,” “effectively perfectly elastic,” “very nearly elastic,” and “various degrees of elasticity.” For example, he states that “[i]n these circumstances the velocities of rebound are practically equal to the initial velocities; such a collision is called elastic (Feynman et al., 1963, section 10–4 Momentum and energy).” In addition, Feynman mentions that “when equilibrium has set in, the net result is that the collisions are effectively perfectly elastic (Feynman et al., 1963, section 39–2 The pressure of a gas).” However, inelastic collision is sometimes defined as a collision in which the coefficient of restitution is between 0 and 1. (A definition of coefficient of restitution, e, is the ratio of the relative velocity after a collision to relative velocity before the collision.) Thus, it is potentially confusing that a collision having a coefficient of restitution of 0.9 may be described as inelastic, nearly elastic, or practically elastic.

On the other hand, Feynman explains that “[f]or a large number of materials, experiments show that for sufficiently small extensions the force is proportional to the extension F ∝ Δl. This relation is known as Hooke’s law (Feynman et al., 1964, section 38–1 Hooke’s law).” In general, Hooke’s law is also known as an empirical law because it is formulated by experiments. An important experimental condition for Hooke’s law is that it is applicable to an elastic spring or string when it is not stretched or compressed beyond the elastic limit. In other words, this empirical law is a good approximation of the elastic behavior of a substance when the deformations are relatively small. Mathematically, Hooke’s law is more precisely expressed as F = -kx because the direction of the restoring force exerted by the spring on an object is opposite to the displacement of the object. Strictly speaking, a real spring does not obey this law beyond a limited (linear) range.

References: 
1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.
2. Feynman, R. P., Leighton, R. B., & Sands, M. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley.

About the author

I am a fan of Feynman, with over ten years of experience teaching introductory physics. I don’t really like travelling, but I visited the following countries/places: Australia (Perth), Austria (Salzburg, Vienna), Bosnia, Brunei, China (Guangzhou, Hong Kong, Macau), Croatia (Split), Egypt (Cairo, Mount Sinai), France (Toulouse, Marseille, Nice, Paris), Germany (Frankfurt), Indonesia (Batam, Jakarta, Pulau Bintan), Israel (Mount Carmel, Golan Heights, Jerusalem), Italy (Assisi, Milan, Rome, Venice), Japan (Tokyo), Malaysia (Kenyir Lake, Kuala Lumpur, Malacca, Pulau Redang, Pulau Pemanggil), Portugal, South Korea (Gwangju, Mokpo, Seoul), Spain (Barcelona), Switzerland (Bern, Interlaken, Jungfrau), Taiwan (Taipei, Mount Alishan, Kaohsiung), Thailand, United Kingdom (London), United States (Hawaii, Pittsburgh), Vatican, and Vietnam (Ho Chi Minh, Quảng Trị).

My blogs:

Meditating Feynman’s Lectures
How would Feynman answer examination questions?

Problems of physics assessment

Selected Publications:
1. Wong, C. L., Chu, H. E., & Yap, K. C. (2016). Are Alternative Conceptions dependent on Researchers’ Methodology and Definition? : A Review of Empirical Studies related to Concepts of Heat. International Journal of Science and Mathematics Education. 14(3), 499-526.

2. Wong, C. L., Chu, H. E., & Yap, K. C. (2014). Developing a Framework for analyzing definitions: A Study of The Feynman Lectures. International Journal of Science Education. 36(15), 2481-2513.

3. Wong, C. L. (2013). Is there a best definition of weight? The Physics Teacher, 51(9), 516-7.

4. Lim, K. M., Lee, G. C. F., Sheppard, C. J. R., Phang, J. C. H., Wong, C. L., & Chen, X. (2011). Effect of polarization on a solid immersion lens of arbitrary thickness. Journal of the Optical Society of America A, 28(5), 903-911.

5. Yap, K. C., & Wong, C. L. (2007). Assessing conceptual learning from quantitative problem solving of a plane mirror problem. Physics Education, 42(1), 50-55.

CIE 9702 Physics 2015 Jun Question 1 & 2

Question:

1 (a) Students are expected to use the definition of power to show that its base SI units are kg m² s^–3. [2 marks]

(b) Students have to use a mathematical expression for electrical power to determine the base SI units of potential difference. [2 marks]

2 (a) Students are expected to provide definitions of speed and velocity and make use these definitions to explain why speed is a scalar and velocity is a vector. [2 marks]

(b) A ball is released from rest. The ball falls vertically, hits the ground and rebounds vertically. Students may assume there is no air resistance.

(i) Students have to describe the variation of the velocity of the ball with time from t = 0 to t = 2.1 s. [3 marks]

(ii) Students have to calculate the acceleration of the ball after it rebounds from the ground. [3 marks]  
(iii) Students have to calculate the distance and displacement from the initial position moved by the ball from t = 0 to t = 2.1 s. [5 marks]

(iv) Students have to plot a graph that shows the variation of the speed of the ball with time. [2 marks]

Mark Scheme: 

1 (a) power = work / time = (force × distance) / time   [1 mark] 
[power] = (kg m s^–2 × m) s^–1 = kg m² s^–3   [1 mark] 

(b) power = VI   [1 mark]
[V] = kg m² s^–3 A^–1   [1 mark] 

2 (a) speed = distance / time and velocity = displacement / time   [1 mark] 

speed is a scalar as distance has no direction and velocity is a vector as displacement has direction [1 mark] 

(b) (i) Indicate that the ball has a constant acceleration or linear/uniform increase in velocity until 1.1 s [1 mark] 

Indicate that the ball rebounds or bounces or changes direction [1 mark] 

Indicate that the ball decelerates to zero velocity having the same acceleration as the initial value [1 mark] 

(ii) Use a = (v – u) / t or gradient of the graph   [1 mark] 

Student’s working shows that (8.8 + 8.8) / 1.8 or appropriate values from line or (8.6 + 8.6) / 1.8   [1 mark] 

Student’s working shows that 9.8 (9.78) m s^–2 or = 9.6 m s^–2   [1 mark] 

(iii) 1. distance = first area above graph + second area below graph   [1 mark] 

First area = (1.1 × 10.8) / 2 + Second area = (0.9 × 8.8) / 2. 

Total area = 5.94 + 3.96   [1 mark] 

Distance = 9.9 m   [1 mark] 

2. Displacement = first area above graph – second area below graph   [1 mark] 

Displacement = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m   [1 mark] 

(iv) The graph has a correct shape with straight lines and all lines are above the time axis   [1 mark] 

The graph shows correct times for zero speeds (0.0, 1.15 s, 2.1 s) and peak speeds (10.8 m s^–1 at 1.1 s and 8.8 m s^–1 at 1.2 s and 3.0 s)   [1 mark] 

Possible answers: 

1 (a) power = work / time = (force × distance) / time. 

Base units of power = (kg m s^–2 × m) s^–1 = kg m² s^–3 

(b) power = VI   

Base SI units of VI = kg m² s^–3   (as shown above) 

Base SI units of V = kg m² s^–3 / A = kg m² s^–3 A^–1 

2. (a) Speed is the rate of change of distance travelled with respect to time. 

Velocity is the rate of change of displacement with respect to time. 

Speed is a scalar quantity because distance has no direction, whereas velocity is a vector quantity because displacement has a direction. 

Alternative answers: 

Speed is a scalar quantity defined by the word equation = distance traveled/time taken. Velocity is a vector quantity defined by the word equation = change of displacement/time taken. 

Speed is a scalar that is always positive because the distance is the numerator and velocity is a vector that can be positive or negative because the displacement is the numerator. 

(b) (i) The graph has a constant acceleration from 0.0 sec to 1.1 sec. Then, it changes direction and decelerates to zero velocity having the same magnitude of acceleration as the initial value. 

(ii) Using a = (v – u) / t = (8.8 + 8.8) / 1.8 = 9.78) m s^–2   

(iii) 1. Distance travelled = first area above graph + second area below graph 

= (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 = 5.94 + 3.96 = 9.9 m 

2. Displacement from the initial position = first area above graph – second area below graph = (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2 = 2.0 (1.98) m 

(iv) Graph (variation of the speed of the ball with time)

Feynman’s insights or goofs?: 

The mark scheme simply states that “speed = distance/time.” However, Feynman elaborates that “we can define the speed in this way: We ask, how far do we go in a very short time? We divide the distance by the time, and that gives the speed. But the time should be made as short as possible, the shorter the better… (Feynman et al., 1963, section 8–2 Speed).” To explain the subtleties in a definition of speed, Feynman provides the following joke: A lady driving a car is caught for high-speed driving, and a cop says, “Lady, you were going 60 miles an hour!” She argues that this is impossible because she was traveling for only seven minutes. Thus, the cop explains that “lady, if you kept on going the same speed in one hour you would go 60 miles.” Interestingly, her answer was, “Well, the car was slowing down and it would not go 60 miles.” Alternatively, she could argue that if the car moves at the same speed, then she would run into a wall! More important, we need to explain the concept of instantaneous speed and how speed can be measured by using an accurate speedometer.

The mark scheme states that velocity = displacement/time instead of velocity = (change of displacement)/(time taken). Nevertheless, Feynman provides a mathematical definition of velocity as follows: “[l]et us try to define velocity a little better. Suppose that in a short time, ϵ, the car or other body goes a short distance x; then the velocity, v, is defined as v = x/ϵ, an approximation that becomes better and better as the ϵ is taken smaller and smaller (Feynman et al., 1963, section 8–2 Speed). This definition of velocity is based on the ratio of an infinitesimal distance to the corresponding infinitesimal time. Theoretically speaking, we imagine what happens to the ratio as the time we use is shorter and shorter. In other words, we take a limit of the displacement traveled by an object divided by the time elapsed, as the time taken is assumed to be shorter and shorter, ad infinitum. This idea was independently invented by Newton and Leibnitz and it is now known as calculus.

Importantly, Feynman clarifies that “[o]rdinarily we think of speed and velocity as being the same, and in ordinary language they are the same. But in physics, we have taken advantage of the fact that there are two words and have chosen to use them to distinguish two ideas. We carefully distinguish velocity, which has both magnitude and direction, from speed, which we choose to mean the magnitude of the velocity, but which does not include the direction (Feynman et al., 1963, section 9–2 Speed and velocity).” In short, velocity is speed in a specified direction. Thus, we can also define velocity by describing how the x-, y-, and z-coordinates of an object change with time, as well as write v_x = Δx/Δt, v_y = Δy/Δt and v_z = Δz/Δt. 

Reference: 

Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.