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PHYC10001 Physics 1: Advanced Examination Answers – Semester 1, 2018

These are answers to the exam questions, not complete solutions, although there are hints given concerning possible solution methods. (There is often more than one way to solve a problem.) The answers provided here for questions requiring explanations are not necessarily complete, and might not receive the full marks allocated on the exam paper. Many marks are allocated for clearly showing your ‘working’ (i.e. for showing the examiners that you understand the relevant physics). Writing down a numerical answer or mathematical expression alone is often not sufficient to gain full marks. 1. (a)

An inertial frame of reference is one in which Newton’s 1st law of motion holds. It is a frame in which any object that is not subject to a net force is observed to move in a straight line at constant speed or, if it is stationary, to remain stationary. All of the laws of physics hold without modification in all inertial frames, without the need to introduce additional inertial (imaginary/fictitious) forces. To test if a laboratory you are in is an inertial frame, in principle you could carry out any experiment whose result is predictable using well-understood physics. If the experimental observations can be explained without fictitious forces, then the lab is inertial. From a Newtonian perspective, a lab on the Earth’s surface can be treated as inertial if the effects of the apparent accelerations of objects in the lab due to the Earth’s rotation (the centrifugal and Coriolis accelerations) are small compared to the effects of any other accelerations that affect the results of experimental measurements in the lab. From the perspective of General Relativity, a lab on the surface of the Earth cannot be treated as an inertial frame unless the effects of gravity on an experiment can be neglected, because in the GR picture, gravity in the lab is a fictitious force.

(b) (i)

The free-body diagram should show the weight force on the person directed downwards towards the centre of the Earth, and the normal force from the ground in the opposite direction.

(ii)

The person standing on the equator is in uniform circular motion. Any object moving in a circle is constantly accelerating (centripetal acceleration). A non-zero net force is needed to produce that acceleration. The net force is directed towards the Earth’s axis of rotation (it is a centripetal force).

(iii)

Proof required. Start with Newton’s 2nd law. The net force is the centripetal force. (Note: the vector 𝑟 on the exam paper should probably be 𝑟 instead since we need a unit vector here. Usually, we define 𝑟 = |𝑟|𝑟 = 𝑟𝑟 . )

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2. (a)

The diagram should show the weight force downwards, the normal force upwards and a friction force in the negative 𝑥 direction.

(b)

Both vectors are directed into the page. For the angular acceleration, note that the wheel’s angular velocity is increasing after the anti-lock system releases the brakes.

(c)

The angular acceleration is 𝛼= The angular velocity is then

𝜇𝑘 𝑚𝑐𝑎𝑟 𝑔𝑅 5𝜇𝑘 𝑚𝑐𝑎𝑟𝑔 . = 16𝑚𝑅 4𝐼

𝜔 = 𝜔0 + 𝛼𝑡 = 𝜔0 +

5𝜇𝑘 𝑚𝑐𝑎𝑟 𝑔 16𝑚𝑅

𝑡.

Note: 𝜔 could, alternatively, be the negative of the given expression, if you chose to use the convention that clockwise rotations correspond to negative angular velocities. (d)

The total friction force on the car is 4 times the friction force on one wheel. We find: 𝑣𝑐𝑜𝑚 = 𝑣0 − 𝑎𝑐𝑜𝑚 𝑡 = 𝑣0 − 𝜇𝑘 𝑔𝑡.

(e)

In general, the point on the wheel that is in contact with the ground has linear speed 𝜔𝑅 relative to the centre of the wheel, directed opposite to the direction of motion of the car (in the −𝑥 direction.) Relative to the road, the centre of mass of the wheel has speed 𝑣𝑐𝑜𝑚 in the positive 𝑥 direction. So, in general, the point on the wheel in contact with the road has velocity 𝑣𝑐𝑜𝑚 − 𝜔𝑅 relative to the road surface. The condition for rolling without slipping is that the point on the wheel that in contact with the road is instantaneously stationary (has zero velocity) with respect to the road, so we require 𝑣𝑐𝑜𝑚 − 𝜔𝑅 = 0.

(f)

Start with 𝑣𝑐𝑜𝑚 = 𝜔𝑅 and substitute the expressions obtained in parts (c) and (d). 𝑡=

𝑣0 − 𝜔0 𝑅 . 5𝑚𝑐𝑎𝑟 ) 𝜇𝑘 𝑔 (1 + 16𝑚

(g) (i)

(ii)

The angular momentum of the wheel is 𝐿 = 𝐼𝜔. Since |𝜔| is increasing during the skid, the angular momentum also increases in magnitude while the wheel is skidding. This is due to the frictional torque. Angular momentum is not conserved during the skid because there is a net torque on the wheel, but once the wheel stops skidding and resumes rolling without slipping, angular momentum is conserved (𝜔 is constant). Once the anti-lock system applies the brakes again, the angular momentum decreases, of course. While the wheel is skidding, some of the linear kinetic energy of the car is transformed partly into rotational kinetic energy of the wheels and partly into heat (also sound and perhaps some elastic energy in the tyres etc.) The total energy of the car-road-atmosphere system is conserved (as always). Considering just the car, during the skid the linear kinetic energy

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decreases while rotational kinetic energy (in the rotation of the wheels) increases. Overall, the kinetic energy of the car decreases because some energy is lost to heat etc. 3. (a) (i) (ii)

8.75 m s−2 . The ISS and objects inside it are all in free fall around the Earth. They share the same acceleration towards the centre of the Earth due to gravity. The perception of weight is usually due to contact forces (e.g. normal forces). Since no such forces act between, say, the walls of the ISS and an astronaut floating inside, astronauts feel weightless. The equivalence principle of general relativity holds that a reference frame in gravitational free-fall is locally equivalent to (experimentally indistinguishable from) an inertial frame far from any source of gravity, and the interior of the ISS is a frame in gravitational free-fall. ALTERNATIVELY, the equivalence principle says that inertial mass and gravitational mass are indistinguishable (there is really only one type of mass: inertial mass), and this is why the ISS and objects inside all have the same acceleration. (If it wasn’t true, different objects could have different accelerations due to gravity. The acceleration would be 𝑎 = 𝑚𝑔 𝑔/𝑚𝑖 , where 𝑚𝑔 and 𝑚𝑖 are the gravitational and inertial masses of the relevant object.)

(iii)

Weight is the force of gravity exerted on an object (in the Newtonian picture, anyway). This is not the same as (inertial) mass, which is a measure of an object’s resistance to being accelerated. Changes in the local value of 𝑔 don’t affect the inertial mass 𝑚 that appears in Newton’s 2nd law of motion (𝐹 = 𝑚𝑎), so objects exhibit the same inertia regardless of variations in 𝑔.

(b) 𝐺𝑀𝑒 𝑣=√ . 𝑟

(c)

Calculating using Kepler’s 3rd law or using the result from part (b), we get 15.7 orbits per day, which means that the ISS makes 15 complete orbits per day. (Note: perhaps the fastest way to solve this problem is to note that 𝑣 = 𝜔𝑟 = 2𝜋𝑟𝑓, where 𝑓 is the orbital frequency, then to equate the 𝑣 here with part (b).)

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(d) (i) 2𝐺𝑀𝑒 . 𝑣𝑒𝑠𝑐 = √ 𝑟

𝑣𝑚𝑖𝑛 = 𝑣𝑒𝑠𝑐 − 𝑣𝐼𝑆𝑆 , where 𝑣𝐼𝑆𝑆 is the orbital speed of the ISS given in part (b), above. We also

(ii)

have 𝑉𝑒𝑠𝑐 = 𝑣𝐼𝑆𝑆 √2, so 𝑣𝑚𝑖𝑛 = (√2 − 1)𝑣𝐼𝑆𝑆 = 3.18 × 103 m s−1 .

(iii)

The direction of launch affects the required speed. The probe should be launched in the same direction as the ISS is orbiting the Earth (i.e. in the ‘forwards’ direction tangential to the direction of the ISS orbit). The escape speed, 𝑣𝑒𝑠𝑐 , relative to the Earth, for a given value of 𝑟, is a constant regardless of the initial direction of the escaping object (because the kinetic energy is a scalar quantity that depends on the magnitude of the velocity but not its direction). Because the probe shares the ISS’s velocity just before launch, burning fuel could initially either increase or decrease the probe’s speed relative to the Earth, depending on the launch direction. If, for example, it was launched in the opposite direction to the direction of the ISS’s orbit, then initially the probe would slow relative to the Earth, then more fuel would be required to bring the speed back up to the required escape speed.

4. (a)

(b)

Proof required. Start with 𝐹 = 𝑚𝑎 = −𝑘𝑥. Use the definition of acceleration 𝑎 = 𝑑 2 𝑥/𝑑𝑡 2 and define 𝜔 = √𝑘/𝑚.

Proof required. Differentiate the given expression for x(t) twice with respect to time and substitute back into the equation in part (a).

(c) (i) (ii)

Determine the period of oscillation, 𝑇 from the graph. Then 𝜔 =

2𝜋 𝑇

= 8.4 rad s−1 .

𝜙 = 𝜋/2. For example, note that for curve B when 𝑡 = 0, 𝑥 = 𝐴, then use the expression in (b) to find the phase constant.

(d) (i)

B. The largest potential energy corresponds to the largest displacement of the mass from equilibrium.

(ii)

A. The total mechanical energy (potential energy plus kinetic energy) is constant throughout the simple harmonic motion. The oscillations with the largest amplitude will have the largest total energy.

(iii)

A. The speed of the mass, in general, is 𝑑𝑥/𝑑𝑡, the gradient of the 𝑥 vs. 𝑡 graph.

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5. (a)

The principle of superposition states that the net displacement of the medium due to two waves at a particular location is the sum of the displacements due to each wave considered individually. For example, if we have two waves 𝑦1 (𝑥, 𝑡) and 𝑦2 (𝑥, 𝑡), then the displacement caused by these two waves is 𝑦(𝑥, 𝑡) = 𝑦1 + 𝑦2 .

(b) (i)

Proof required. Put 𝐵 = 𝐴. The superposition of the two travelling waves can be expressed as a product of sin() and cos() functions using the following identity from the formula sheet: 1 1 sin 𝑎 + sin 𝑏 = 2 sin ( (𝑎 + 𝑏 )) cos ( (𝑎 − 𝑏 )), 2 2

With 𝑎 = 𝑘𝑥 − 𝜔𝑡, 𝑏 = 𝑘𝑥 + 𝜔𝑡.

(ii)

(iii)

No. (It is a standing wave.) The function 𝑦(𝑥, 𝑡) is not a function of 𝑘𝑥 ± 𝜔𝑡, which is characteristic of travelling waves. Similarly, it is not true that 𝑦(𝑥 ± 𝑣𝑡, 𝑡) = 𝑦(𝑥, 0), as required for a travelling wave. Yes, there are nodes, which occur where 𝑦(𝑥, 𝑡) = 0 for all times, 𝑡. The nodes therefore satisfy sin 𝑘𝑥 = 0, which leads to 𝜋 2𝜋 ,… 𝑥 = 0, ± , ± 𝑘 𝑘 Since 𝑘 = 2𝜋/𝜆, we see that the nodes are separated by 𝜆/2.

(iv)

In general, the transverse speed is 𝑣𝑡 = 𝜕𝑦/𝜕𝑡. The maximum possible transverse speed is 2𝐴𝜔.

(c) (i)

Proof required. There are many waves to show this. Here’s one way: ⇒

𝑦 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡) + 𝐵 sin(𝑘𝑥 + 𝜔𝑡)

𝑦 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡) + (𝐵 − 𝐴) sin(𝑘𝑥 + 𝜔𝑡) + 𝐴 sin(𝑘𝑥 + 𝜔𝑡)

⇒ 𝑦 = 𝐴[sin(𝑘𝑥 − 𝜔𝑡) + sin(𝑘𝑥 + 𝜔𝑡)] + (𝐵 − 𝐴) sin(𝑘𝑥 + 𝜔𝑡) ⇒ 𝑦 = 2𝐴 sin 𝑘𝑥 cos 𝜔𝑡 + (𝐵 − 𝐴) sin(𝑘𝑥 + 𝜔𝑡).

(ii)

Proof required. Given that 𝑦 = 𝑦1 + 𝑦2 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡) + 𝐵 sin(𝑘𝑥 + 𝜔𝑡),

clearly the largest possible displacement 𝑦 will occur when 𝑦1 = 𝐴 and 𝑦2 = 𝐵. This follows from the principle of superposition applied to the two travelling waves. It is impossible for the sine functions to have values larger than one. However, we still need to prove that values of 𝑥 and 𝑡 exist such that it is possible to have 𝑦1 = 𝐴, 𝑦2 = 𝐵 simultaneously at some point(s) on the string. There are many possible ways to do this. One way is to consider what is happening at time 𝑡 = 0. PHYC10001 Physics 1: Advanced

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We have 𝑦(𝑥, 0) = 𝑦1 + 𝑦2 = 𝐴 sin 𝑘𝑥 + 𝐵 sin 𝑘𝑥 = (𝐴 + 𝐵) sin 𝑘𝑥,

so at that time the displacement of the string is 𝐴 + 𝐵 everywhere that sin 𝑘𝑥 = 1.

Another way is to consider the position 𝑥 such that 𝑘𝑥 = 𝜋/2. At that position we have 𝜋 𝜋 𝜋 𝑦 ( , 𝑡) = 𝐴 sin ( − 𝜔𝑡) + 𝐵 sin ( + 𝜔𝑡) 2 2 2 = (𝐴 + 𝐵) cos(𝜔𝑡),

showing that the string oscillates with amplitude (𝐴 + 𝐵) at that position.

(iii)

There are no nodes, because this would require that 𝑦(𝑥, 𝑡) = 0 at some 𝑥 for all times, 𝑡, but there are no spatial locations on the modulated wave for which that condition is true. One way to show this is to consider the expression 𝑦(𝑥, 𝑡) = 2𝐴 sin 𝑘𝑥 cos 𝜔𝑡 + (𝐵 − 𝐴) sin(𝑘𝑥 + 𝜔𝑡). As previously demonstrated (in part (b)(iii)), the first term is zero at all times only where sin 𝑘𝑥 = 0, i.e. where 𝑘𝑥 = 𝑛𝜋, 𝑛 = 0, ±1, ±2, …. The 𝑥 values derived this way are the only possible candidate positions for nodes. But suppose the condition 𝑘𝑥 = 𝑛𝜋 holds. Then we still need the second term in the expression for 𝑦(𝑥, 𝑡) to be zero at all times. That is, we need sin(𝑘𝑥 + 𝜔𝑡) = sin(𝑛𝜋 + 𝜔𝑡) = ± sin(𝜔𝑡) = 0.

Obviously, this is not true for all values of 𝑡, so we conclude there are no nodes. 6. (a)

Parallel rays focus behind Alberto’s retina. The cornea-retinal distance is fixed, so forming a focused image of an object 25 cm from the eye requires the eye to adjust to a shorter focal length than is required to focus on an object at 45 cm. Thus, Alberto’s eye is unable to converge light rays as well as a ‘normal’ eye. Alberto’s resting eye would have a focal length that is longer than the 20 mm distance between the cornea and the retina, meaning that parallel rays would focus behind his retina.

(b)

The image should be virtual. Each lens needs to form an image on the same side of the lens as the object.

(c)

Positive focal length. To focus parallel rays on Alberto’s retina, the rays need to be converged more than what Alberto’s unaided eye can achieve, so he needs converging lenses.

(d)

The required lens power is 𝑃 =

PHYC10001 Physics 1: Advanced

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𝑓

= +1.8 dioptre. Use the thin lens equation to calculate 𝑓.

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(e)

Only two of the three construction rays are really needed to locate the image. Notice that the object is closer to the lens than the focus. 7. (a)

Explanation required. For example:

The spot arises from light waves diffracting around the edges of the disc or sphere. Light that is at equal radial distances from the centre of the disc or sphere as it passes the disc/sphere travels equal distances to a point on a screen behind the object, in the centre of the geometric shadow. Therefore, if that light on all sides is in phase as it passes the disc, it will still be in phase at the centre of the screen (this can be achieved using a coherent light source, or a point source at a large distance from the sphere/disc). Therefore, the light interferes constructively at that point, and we see bright spot.

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(b) (i)

(ii)

𝜆2 = 346 nm. This in in the ultraviolet range, so perhaps “colour” isn’t the best word to use here. (Recall that visible light has 400 nm < 𝜆 < 700 nm, approximately.) First find the angle to the 4th order blue spot (22.9°), then use the given wavelength to calculate the grating spacing. We find 𝑑 = 4.45 × 10−6 m.

The number of rulings per unit distance is 1/𝑑 = 2.25 × 105 m−1 , which is 225 lines per millimetre.

(Note: using the small angle approximation in this problem results in a final answer of 244 lines per millimetre, an error of about 8%. As we see here, for diffraction gratings the diffraction angles often aren’t particularly ‘small’, so it’s usually best not to make the approximation.) (iii)

The distance between adjacent maxima on the screen would increase. A larger number of lines per millimetre means the grating spacing, 𝑑, is smaller, and the angles to the various maxima increase.

(iv)

Relevant points here that you might mention include: ▪

The interference maxima for the 2-slit arrangement would be found at the same locations on the screen as before (at the same angles).

▪

The 2-slit maxima would be dimmer than the grating maxima.

▪

The 2-slit fringes would be broader than the grating fringes (i.e. the intensity would not drop off so rapidly on either side of each maximum).

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8. (a) (i)

The laws of physics take the same forms in all inertial reference frames. The speed of light in vacuum has the same value in all inertial reference frames.

(ii)

The proper distance between two objects is the distance measured between those objects by an observer who is at rest with respect to both objects. Equivalently, the rest length of a single object is the length measured by an observer who is at rest with respect to the object. A proper time interval is the time interval between two events measured in a frame in which both events occur at the same place (i.e. same spatial coordinates).

(b) (i)

(ii)

𝐿𝐼𝑆𝑆 =

𝐿𝐴𝐵 , 𝛾

where the Lorentz factor is 𝛾 =

1

2

√1−(𝑣 ) 𝑐

.

Each spaceship was at a distance of 𝐿𝐴𝐵 /2 from the ISS when the signal was sent. That is, the ISS was exactly halfway between A and B when the signal was sent. According to A and B, the signal travelled at the same speed, 𝑐, in both directions from the ISS. Since the signal was received by both spaceships simultaneously, it must have travelled an equal distance from the ISS to each spaceship (𝑠 = 𝑐𝑡).

(iii)

Yes, it is correct that the two spaceships were at equal distances from the ISS when the signal was sent, as observed in the ISS frame. From part (ii), above, we know that the spaceships were at equal distances from the ISS, as measured in the spaceship frame, when the signal was sent. If we imagine a point C, half way between A and B in the spaceship frame (i.e. it has a fixed 𝑥′ coordinate), that is the position of the ISS at time 𝑡 ′ = 0. Then the distance between C and A is a proper distance, as is the distance between C and B, and these distances are equal in the spaceship frame. The ISS sees both distances contracted by a factor of 𝛾, and therefore concludes that both spaceships were at equal distances from the ISS when the signal was sent.

(iv)

The signal was not received simultaneously, according to observers on the ISS. When the signal was sent, A and B were at equal distances from the ISS. But A was moving towards the ISS, while B was moving away from it. Since the speed of light is the same (𝑐) in both directions, according to the ISS, the light would have had to travel less distance to reach A than to reach B. Therefore, spaceship A received the signal before spaceship B.

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(v)

Here is the Minkowski diagram:

𝑐𝑡

A B ISS

0

𝐿𝐼𝑆𝑆 /2

𝐿𝐼𝑆𝑆

𝑥

Important features of this graph include: •

• •

The spaceship worldlines, marked A and B, are parallel to one another, with a gradient that is greater than the gradient of the light signal worldlines (because the spaceship...