AP Physics 1 2017 Free Response Question 1

Question: This question tests students’ knowledge of electrical circuits.

(a) In the first part of the question, students have to rank the magnitudes of the potential differences across five light bulbs (A, B, C, D, and E) from highest to lowest.

In circuit 1, there is only a light bulb that is connected to a battery. In circuit 2, there are two lightbulbs that are connected in series in which the same electric current flows through the circuit. In circuit 3, there are two lightbulbs that are connected in parallel in which the same potential difference is applied across both light bulbs.

In a sense, the flow of charge-carriers through light bulbs in the three circuits is similar to the flow of water through sponge filters in tubes as shown below:

We may assume the same water level is maintained in the three water tanks as shown above. As an example, there could be a total of 4 water drops per second that flows through the two tubes (a sponge filter each) as shown in Fig 1A. If there are one tube and a sponge filter, we expect only 2 water drops per second as shown in Fig 1B. However, if there are two sponge filters in a tube, we expect an increase in the resistance to the flow of water and there could be only 1 water drop per second as shown in Fig 1C.

(b) In the second part of the question, students have to identify the circuit that runs out of electrical energy first and the circuit that run out of electrical energy last. Students need to assume the batteries have the same amount of electrical energy and they are all connected to the light bulbs at the same time.

Possible answers:

1 (a)

In the first circuit, the light bulb A has the same potential difference (V) as the battery.

In the second circuit, the two light bulbs divide the potential difference (V) of the battery equally. Thus, the potential differences of light bulbs B and C are both equal to V/2.

In the third circuit, the two light bulbs D and E have the same potential difference (V) because they are connected in parallel to each other.

Thus, V_A = V_D = V_E = V, whereas V_B = V_C = V/2. (We assume this is an ideal battery that has no internal resistance.)

(b)

For example, we may assume the potential difference of the battery is 1.0 V and the electrical resistance of the light bulb is 1 ohm.

In circuit 1, the electric current = V/R = (1.0)/1 = 1.0 A.

The power consumed in circuit 1 = IV = (1.0)(1.0) = 1.0 W.

In circuit 2, the electric current = V/R = (1.0)/(1 + 1) = 0.5 A.

The power consumed in circuit 2 = (0.5)(0.5) + (0.5)(0.5)  = 0.5 W.

(The potential difference is halved because it is “shared” by the two lightbulbs.)

In circuit 3, the electric current = V/R = 1.0 A.

The power consumed in circuit 3 = (1.0)(1.0) + (1.0)(1.0)  = 2.0 W.

In summary, P₁ = 1.0 W, P₂ = 0.5 W, P₃ = 2.0 W.

The battery will run out of usable energy first in circuit 3 because it consumes relatively more power.

The battery will run out of usable energy last in circuit 2 because it consumes relatively lesser power.

Below is a comparison of a series circuit and parallel circuit.

Feynman’s insights or goofs?:

Feynman explains a series circuit as follows: “Suppose we have a more complicated circuit which has two pieces, which by themselves have certain impedances, Z₁ and Z₂ and we put them in series and apply a voltage. What happens? It is now a little more complicated, but if I is the current through Z₁, the voltage difference across Z₁, is V₁ = IZ₁; similarly, the voltage across Z₂ is V₂ = IZ₂. The same current goes through both. Therefore the total voltage is the sum of the voltages across the two sections and is equal to V = V₁ + V₂ = (Z₁ + Z₂)I (Feynman et al., 1963, section 25–5 Series and parallel impedances).” Essentially, the electric current is the same through any components (resistors, capacitors, or inductors) that are in series. Thus, the total voltage through two components can be expressed as V = (Z₁ + Z₂) I.

In general, when we say that two resistors are in series, it means that they are connected end to end and the same electric current passes through them. In a sense, an electrical circuit that consists of only a battery and a resistor are also in series and it can be redrawn as shown below. Students should realize that the electric current that emerges from the battery (I_A) is equal to the electric current after passing through the resistor (I_B) and the electric current before passing through the resistor (I_C).

       In addition, Feynman also explains a parallel circuit as follows: “[w]e may also connect them in another way, called a parallel connection. Now we see that a given voltage across the terminals if the connecting wires are perfect conductors, is effectively applied to both of the impedances, and will cause currents in each independently. Therefore the current through Z₁ is equal to I₁= V/Z₁. The current in Z₂ is I₂ = V/Z₂. It is the same voltage. Now the total current which is supplied to the terminals is the sum of the currents in the two sections: I = V/Z₁ + V/Z₂ (Feynman et al., 1963, section 25–5 Series and parallel impedances).” In other words, it is the potential differences (or voltage differences) across the components in parallel that are the same instead of the electric current.

       On the other hand, when we say that two resistors are in parallel, it means that the terminals of the two resistors are connected to the same two points such that their potential differences are the same. Interestingly, the electrical circuit that consists of a battery and a resistor are also in parallel and it can be redrawn as shown below. Students should realize that the potential differences across the battery and the resistor are the same. Furthermore, the potential difference between any point in blue and any point in red are the same.

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.

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CIE 9702 Physics 2015 Jun Question 7

Question: 7 (a) Explain the meaning of an electric field.

(b) There is a uniform electric field between two vertical metal plates AB and CD that are placed in parallel. The potential difference between the two metal plates is 450 V and the separation of the plates is 16 mm. An alpha particle is accelerated from plate AB to plate CD.

(i) Draw electric field lines between the plate AB and plate CD.

(ii) Determine the electric field between the plate AB and plate CD.

(iii) Determine the work done by the electric field on the alpha particle as it moves a distance of 0.016 m from plate AB to plate CD.

(iv) A beta particle has moved from plate AB to plate CD. Determine the ratio of the work done by the electric field on the alpha particle to the work done by the same electric field on the beta particle.

Mark Scheme:

7 (a) It is a region in space where a (stationary) charge experiences an (electric) force [1]

(b) (i) Students should draw at least four parallel straight lines that are perpendicular to the two plates. The lines representing electric field should also be equally spaced. [1]

There should be an arrow on the lines from left to right that indicates the direction of the electric field. [1]

(ii) Use the equation of electric field, E = V/d [1]

E = (450 / 16 × 10^–3) = (28 125) V/m   [1]

(iii) The work done, W = Eqd or Vq   [1]

q = 3.2 × 10^–19 C   [1]

Thus, W = 450 × 3.2 × 10^–19 = 1.4(4) × 10^–16 J   [1]

(iv) Ratio of work done = (450 × 3.2 ×10⁻¹⁹) / (450 × -1.6 ×10⁻¹⁹)  = -2   [1]

Possible answers:

7 (a) An electric field is a region in space where an infinitely small test charge that is stationary experiences an electric (or electrostatic) force.

(b) (i) The electric field between the plate AB and plate CD is shown below:

(ii) The electric field between two plates is uniform.

Thus, E = V / d = (450 / 16 × 10^–3) = 28,125 V/m

(iii) The work done by the electric field on the alpha particle can be calculated by using the formula, W = Eqd = qV.

Thus, W = qV = (3.2 × 10^–19) × 450 = 1.44 × 10^–16 J

(iv) The ratio of the “work done by the electric field on the alpha particle” to “work done by the same electric field on the beta particle” = q_aV / q_bV = q_a / q_b

Thus, the ratio = (3.2 ×10⁻¹⁹) / (-1.6 ×10⁻¹⁹) = -2

Feynman’s insights or goofs?:

In the words of Feynman, “the existence of the positive charge, in some sense, distorts, or creates a ‘condition’ in space, so that when we put the negative charge in, it feels a force. This potentiality for producing a force is called an electric field. When we put an electron in an electric field, we say it is “pulled.” We then have two rules: (a) charges make a field, and (b) charges in fields have forces on them and move (Feynman et al., 1963, section 2.2 Physics before 1920).” That is, Feynman visualizes a positively charged particle creates a “condition” in space such that when we put another charged particle there, it feels an electric force.

Furthermore, Feynman elaborates that the nature of electric field can be illustrated by an analogy as follows: “If we are in a pool of water and there is a floating cork very close by, we can move it “directly” by pushing the water with another cork. If you looked only at the two corks, all you would see would be that one moved immediately in response to the motion of the other—there is some kind of “interaction” between them. Of course, what we really do is to disturb the water; the water then disturbs the other cork (Feynman et al., 1963, section 2.2 Physics before 1920).” In other words, there is a disturbance in the water (an electric field) caused by a cork (a charged particle), and the water then disturbs the other cork directly.

n general, an electric field can be defined as an electric force per unit charge acting on a test charge that is stationary. To be more accurate, we can define the electric field as the electric force per unit charge experienced by an infinitely small test charge that is stationary and it is placed in a vacuum. Mathematically, it can be expressed by the equation E = F/q, where F is the electric force experienced and q is the infinitely small test charge. This condition is necessary because a large test charge can influence the charge carriers in the surrounding and distort the electric fields.

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. 

CIE 9702 Physics 2015 Jun Question 6

Question: 6 (a) State the meaning of diffraction and interference.

(b) A diffraction grating has 650 lines per millimetre and light waves (7.06 × 10¹⁴ Hz) from a light source S1 are incident on a diffraction grating. Determine the maximum order (n) of fringes formed by the diffraction grating.

(c) State and explain whether more orders of fringes are formed if the light source S1 is replaced by another light source S2 that has a lower frequency (as compared to S1).

Mark scheme:

6 (a) Diffraction: spreading of a wave as it passes through a slit or an edge. [1] Interference: when two or more waves superpose, [1] resultant displacement is the sum of the displacement of the waves. [1]

(b) Using the equation d sin θ = nλ and v = fλ   [1]

Determine maximum order (n) by setting θ = 90° 

Deduce n = 7.06 × 10¹⁴ / (3 × 10⁸ × 650 × 10³)   [1]

n = 3.6 and hence maximum order of fringes, n = 3   [1]

(c) Longer wavelength implies fewer orders of fringes observed   [1]

Possible answers:

6 (a)

A definition of diffraction: “when waves encounter an edge, an obstacle, or an aperture the size of which is comparable to the wavelength of the waves, those waves spread out as they travel and, as a result, undergo interference (Halliday, Resnick, & Walker, 2005, p. 1012).”

A definition of interference: “in a region of overlap, two waves of the same frequency can combine constructively or destructively, depending on their relative phase, to produce a redistribution of energy in that area (Hecht, 2003, p. 922).”

(b) Using the diffraction grating equation, d sin θ = nλ and v = fλ

Putting them together, d sin θ = n(v/f) or n = (f/v) d sin θ

For maximum order n, we may let θ = 90° and thus, sin θ = 1

Substituting into the equation, n = (f/v) d = (7.06 × 10¹⁴/3 × 10⁸) (1/650 × 10³) = 3.6

Thus, the maximum order of fringes is 3.

(c) By using the diffraction grating equation d sin θ = nλ again,

We have n = d sin θ/λ and lower frequency (or longer wavelength) implies smaller n or fewer orders of fringes tend to be observed.

Feynman’s insights or goofs?:

In Feynman’s own words, “[n]o one has ever been able to define the difference between interference and diffraction satisfactorily. It is just a question of usage, and there is no specific, important physical difference between them. The best we can do, roughly speaking, is to say that when there are only a few sources, say two, interfering, then the result is usually called interference, but if there is a large number of them, it seems that the word diffraction is more often used (Feynman et al., 1963, p. 30-1).” In essence, there is no “pure diffraction” as a spreading of waves without the phenomenon interference, and there is no “pure interference” as a superposition of waves without the phenomenon, diffraction. In general, if there are only “two sources,” the phenomenon is usually known as interference. On the other hand, if there are a “large number of sources,” the phenomenon is called diffraction instead.

Historically speaking, Young (1802) states a general law of interference as follows: “wherever two portions of the same light arrive at the eye by different routes, either exactly or very nearly in the same direction, the light becomes more intense when the difference of the routes is any multiple of a certain length, and least intense in the intermediate state of the interfering portions; and this length is different for light of different colors (p. 387).” In other words, constructive interference and destructive interference can be observed if the light waves have the same frequency and the path differences between two sources of light waves are integral numbers of wavelengths and an integral number of wavelengths plus half a wavelength respectively.

Despite the difficulty of defining the difference between diffraction and interference as explained by Feynman, the definitions of diffraction and interference can be more comprehensive as follows: Diffraction of light is a spreading of light waves passing through a slit (or obstacle) whose size is comparable to the wavelength of the light waves and this results in bright and dark fringes instead of light rays moving in a straight line. Interference of light is a superposition of light waves from two or more sources that this results in a redistribution of energy such that bright fringes and dark fringes can be observed; bright fringes are observed when the path differences are multiples of wavelengths (∆x = nλ) and dark fringes are observed when the path differences are odd multiples of half-wavelength (∆x = [2n + 1][½λ]) respectively. The resultant displacement of light waves is the sum of individual displacement of each wave.

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. Halliday, D., Resnick, R., & Walker, J. (2005). Fundamentals of Physics (7th ed.). New York: Wiley.

3. Hecht, E. (2003). Physics: Algebra /Trigonometry (3rd ed.). Pacific Grove, California: Brooks/Cole Publishing.

4. Young, T. (1802). An Account of Some Cases of the Production of Colors, not Hitherto Described. Philosophical Transactions, 92, 387–92.

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.