Exam Questions - Topic Test 5

Question 1

Q1) A student removes the reflective layer from a DVD. She uses the DVD as a transmission diffraction grating. Figure 3 shows light from a laser pointer incident normally on a small section of this diffraction grating. The grooves on this section act as adjacent slits of the transmission diffraction grating. A vertical pattern of bright spots (maxima) is observed on a circular screen behind the disc.

(https: //gyazo.com/61ce466a92ff8d5cb10b6f3fc57b4b37)
(a) Light of wavelength λ travels from each illuminated slit, producing maxima on the screen. State the path difference between light from adjacent slits when this light produces a first-order maximum on the screen.
(b) Explain how light from diffraction grating forms a maximum on the screen

(c) The student has three discs: a Blu-ray disc, a DVD and a CD. She removes the reflective coating from the discs so that they act as transmission diffraction gratings. These diffraction gratings have different slit spacings.
The student also has two laser pointers A and B that emit diffraction colours of visible light. Table 2 and Table 3 show information about the discs and the laser pointers.

(https: //gyazo.com/97ac4da7020d3a556e45c8a1e42d4a87)
(https: //gyazo.com/555060d6abd87eafe020afea583ae408)

Deduce the combination of discs and laser pointer that will produce the greatest possible number of interference maxima.

(d) The students use the CD and laser pointer B as shown in Figure 4. A diffraction pattern is produced on the screen. Laser pointer B and the CD are in fixed positions. The laser beam is horizontal and incident normally on the CD. The height of the screen can be adjusted.
(https: //gyazo.com/fde29619b2fd067b878e5d903136954e)

The screen has a diameter of 30cm and is positioned behind the CD at a fixed horizontal distance of 15cm. The student plans to adjust the height of the screen until she observes the greatest number of spots. The student predicts that using this arrangement, the greatest number of spots on the screen will be 3. Determine whether the student’s prediction is correct.

Answer 1

(a) 1 wavelength
(b) path difference of nλ leads to a phase difference of 0° or 360°, superposition causes constructive interference which forms a maximum
(c) smallest θ through a smaller wavelength and larger slit spacing, so laser pointer A with CD

(d) dsinθ = nλ
(1.6e-6)(sin(45)) = n(6.36e-7)
n = (1.6e-6)(sin(45))/(6.36e-7)
= 1.78

Screen can move up and down, so students can adjust to see more.

Question 2

Q2) Figure 5 shows a spacecraft travelling towards a comet. The spacecraft has an array of blocks designed to capture small dust particles from the comet’s tail.
(https://gyazo.com/1cd0236d7a4f4120c2d7a305860c29fd)
To test the blocks before launch, a spherical dust particle P is fired at a right angle to the surface of a fixed, stationary block. P has a mass of 1.1e-9 kg. It has a speed of 5.9e3m/s when it hits the surface of the block. P comes to rest inside the block

(a) Calculate the work done in bringing P to rest

(b) P travels a distance of 2.9 cm in a straight line inside the block before coming to rest the resultant force on P varies as it penetrates the block.
Calculate the average force acting on P as it is brought to rest.

(c) The block is rectangular with an area of cross-section of 8.0 cm^2 and a thickness of 3.0cm. Figure 6 shows how the density of the block varies with depth up to its maximum thickness.
(https://gyazo.com/2ae1f9871dcfda91db0d87f5ed77af31)
Calculate the mass of the block

(d) In another test, a spherical particle Q is fired at a right angle to the surface of an identical block. Q has the same mass as P and is travelling at the same speed as P when it strikes the surface of the block. Q is made from a less dense material than P.
Compare the distance travelled by Q with that travelled by P as they are brought to rest.

Answer 2

(a) KE = 1/2mv^2 = work done
= 0.19

(b) Wd = Fd
0.019 = f * 0.029
f = 0.66

(c) vol = 2.4e-5
p = v * d
2.4e-5 * (5+50/2) = 6.6e-2 kg

(d) Q would come to rest after a shorter distance travelled than P. This is because p has a small surface area, and hence less work is done for it to travel the same distance as energy is concentrated in a smaller area. Q has a larger surface area, so more energy has to be transported into the metal so as the objects have the same amount of energy, P travels a further distance

Question 3

Q3) Figure 7 Shows an athlete holding a vaulting pole at an angle of 40° to the horizontal.
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Forces D and U are exerted on the pole by the athlete’s right and left hands respectively. U is applied at point U at an angle θ to the vertical. The magnitude of D is 53N and is applied at 90° to the pole at X. The uniform pole is in equilibrium. It weighs 31N.
Figure 8 shows the forces acting on the pole.
(https://gyazo.com/d6b9b5d2fdb7490322bb6ed4981a7ba9)

(a) Determine, using a scale diagram, θ and the magnitude of U.

(b)The athlete now moves the pole to a horizontal position. The pole is held stationary in this position. The athlete’s right hand applies a force S vertically downwards at X as shown in Figure 9. The athlete’s left hand applies a downwards force V at Y
(https://gyazo.com/3dd76c2d04c823450af94ea35800dadc)
Discuss the differences between the magnitudes and directions of force U in Figure 7 and force V applied at Y in Figure 9.

Answer 3

(a) https://gyazo.com/4a01df57f3126228f16252c7953ba72a

(b) moment due to weight increase as the perp distance increases.
Y is vert as S and mg are vertical with no horizontal components
V is greater as it counteracts the forces S and mg

Question 4

Q4) Figure 10 shows a ship leaving a harbour at a constant velocity. The ship moves at the same velocity as a person walking on the harbour wall alongside the ship.

(https: //gyazo.com/e29660649a023e77d254f12dce707713)
(a) The momentum of the ship is approximately 1e7Ns

Estimate the mass of the ship

(b) Figure 11 shows the direction of the thrust exerted by the ship’s propeller as the propeller rotates. The ship’s engines make the propeller rotate. When more water is accelerated, more work is done by the engine.
(https://gyazo.com/7c9611b1a6a457a393f9919992f35fac)
Explain, using Newton’s laws of motion, how the thrust of the propeller on the water the ship to maintain constant momentum.

(c) Figure 12 shows the bottom of the hull with a drag reduction system in operation. Air bubbles are introduced into the water below the hull. This reduces the work done per second against the drag on the hull at any given speed.

However, when the air bubbles reach the propeller they decrease the mass of water being accelerated by the propeller every second. This reduces the thrust produced by the propeller at a given speed of rotation.
(https://gyazo.com/5caf947307c2ba5688892d3a20ea3e4b)
The system enables the ship to save fuel while maintaining the same momentum. Explain why the system delivers this fuel saving. In your answer, consider the effects of the introduction of the system.
• The thrust
• The drag on the hull

Answer 4

(a) momentum = F * t
mass = momentum/velocity
= 3e6kg

(b)
Newton’s First Law states that an object stays at rest and an object in motion stays in motion with the same speed and the same direction unless acted upon by an unbalanced force, because of this, we know that as long as the force remains constant, the momentum of the object will be constant

Newton’s Second Law states that force is equal to mass multiplied by acceleration so as the mass of the object will remain constant, the only things that would change are the force and the acceleration, but as the momentum is constant, as well as the mass, we know that it is not changing in velocity, so there is no acceleration, and the momentum is constant.

Newton’s Third Law states that if object a interacts with object b, object b interacts with the object a with equal force and opposite direction, this means that as the propeller pushes against the water, the water pushes the boat forwards, so as the force which the boat exerts on the water stays constant due to Newton’s second law, the force which the water propels the boat forwards is also constant, leading to the boat having a constant velocity and hence, a constant momentum.

(c) The system delivers fuel-saving by decreasing contact the water has with the hull, this reduces the amount of kinetic energy lost due to drag and increases the efficiency of the kinetic energy from the propeller to have higher efficiency. However, due to the bubbles on the hull of the ship, the thrust of the propeller decreases, while also decreasing the kinetic energy delivered by the propellers, these two systems lead to the ship having the same momentum as before but using less fuel due to the decrease in drag exerted on the ship.