• Physics 9702 Doubts | Help Page 139

Question 687: [Kinematics]

(a) Drag force D on an object of cross-sectional area A, moving with speed v through a fluid of density ρ, is given by

D = ½ CρAv²

where C is a constant.

Show that C has no unit.

(b) Raindrop falls vertically from rest. Assume that air resistance is negligible.

(i) On Fig.1, sketch a graph to show variation with time t of the velocity v of the raindrop for the first 1.0 s of the motion.

(ii) Calculate velocity of the raindrop after falling 1000m

(c) In practice, air resistance on raindrops is not negligible because there is a drag force. This drag force is given by expression in (a).

(i) State equation relating the forces acting on the raindrop when it is falling at terminal velocity.

(ii) Raindrop has mass 1.4 × 10^–5 kg and cross-sectional area 7.1 × 10^–6 m². Density of the air is 1.2 kg m^–3 and initial velocity of the raindrop is zero. Value of C is 0.60.

1. Show that terminal velocity of the raindrop is about 7ms^-1

2. Raindrop reaches terminal velocity after falling approximately 10m. On Fig.1, sketch variation with time t of velocity v for the raindrop. The sketch should include the first 5s of the motion.

Reference: Past Exam Paper – November 2012 Paper 22 Q1

Solution 687:

(a)

C = 2D / (ρAv²)

Units of force D = kgms^-2     

Units of ρAv² = (kgm^-3) (m²) (ms^-1)² = kgms^-2

So, unit of C = kgms^-2 / kgms^-2 = no units (units cancel each other)

(b)

(i)

{The raindrop falls from rest, so the speed v starts from zero.

Air resistance is negligible: acceleration is constant (= 9.81ms^-2), so gradient (which represents the acceleration) is constant.}

Straight line from (0,0) to (1,9.8) ± half a square

(ii)

{The change in E_K is equal to the change in E_P. OR use the equation for uniformly acceleration motion}

½ mv² = mgh                           or v² = 2as       (u² = 0)

Speed v = (2 × 9.81 × 1000)^1/2 = 140ms^-1      

(c)

(i)

{At terminal velocity, acceleration = 0. So, the weight (downward) is equal to the drag force on the object (which is upward here).}

Weight = drag (D) (+ upthrust)

(ii)

1.

Weight, W = drag (D)

Weight W = mg = 1.4×10^-5 × 9.81     

Drag D = ½ CρAv² = ½ × 0.60 × 1.2 × (7.1×10^-6) × v²

Weight, W = drag (D)

1.4×10^-5 × 9.81 = ½ × 0.60 × 1.2 × 7.1×10^-6 × v²      

Speed v = 7.33ms^-1    

2.

Line from (0,0) correct curvature to a horizontal line at velocity of 7ms^-1

Line reaches 7ms^-1 between 1.5s and 3.5s

{The correct method to find the time at which terminal velocity is reached is to such that the area under the velocity-time graph is 10m. The time t is not known, initial velocity is zero and final velocity is 7m. However, such detailed values are not required here since it is only a sketch.

If the object was falling with constant velocity of 7ms^-1 since t = 0, to travel a distance of 10m, the time of fall would be = distance / speed [since speed = distance / time] = 10 / 7 = 1.4s. But the speed is initially zero and it increases until it becomes 7ms^-1, so the time of fall should definitely be greater than 1.4s. So, the lower range is 1.5s.

The upper limit for the limit is given approximately. Students should be able to estimate some values. We cannot choose a time of fall which is too big, nor one which is too small.

Finally, to remind you again, the best method would be to use the area under the graph. But this is more than required for just a sketch.}

Question 688: [Measurement]

(a) Young modulus of the metal of a wire is 1.8 × 10¹¹ Pa. Wire is extended and the strain produced is 8.2 × 10^–4.

Calculate the stress in GPa.

(b) An electromagnetic wave has frequency 12 THz.

(i) Calculate wavelength in μm.

(ii) State name of the region of the electromagnetic spectrum for this frequency.

(c) Object B is on a horizontal surface. Two forces act on B in this horizontal plane. A vector diagram for these forces is shown to scale in Fig.1.

A force of 7.5 N towards north and a force of 2.5 N from 30° north of east act on B.

Mass of B is 750 g.

(i) On Fig.1, draw an arrow to show approximate direction of the resultant of these two forces.

(ii)

1. Show that magnitude of the resultant force on B is 6.6 N.

2. Calculate magnitude of the acceleration of B produced by this resultant force.

(iii) Determine angle between the direction of the acceleration and the direction of the 7.5 N force.

Reference: Past Exam Paper – November 2014 Paper 22 Q1

Solution 688:

(a)

Stress = Young modulus × strain = (1.8 × 10¹¹) × (8.2 × 10^–4) OR = 1.476 × 10⁸

Stress = 0.15 (0.148) GPa

(b)

(i) Wavelength = (3 × 10⁸) / (12 × 10¹²) = 25 μm

(ii) Infra-red / IR

(c)

(i) The arrow should be drawn up to the left of the 7.5 N force, approximately 5° to 40° to west of north

(ii)

1. A correct vector triangle or working to show that magnitude of resultant force = 6.6 N

A triangle is formed with the 2.5N vector at an angle of 60° to the vertical 7.5N vector. Since the resultant vector is opposite to the 60° angle, the cosine rule can be used.

Let the resultant vector be R.

R² = 7.5² + 2.5² – 2(7.5)(2.5)cos 60°

Resultant vector R = 6.6N

(allow 6.5 to 6.7 N if scale diagram)

2.

{Resultant force F = ma}

Magnitude of acceleration a = 6.6 / 0.75 = 8.8 m s^–2

[scale diagram: (6.5 to 6.7) / 0.75]  [scale diagram: 8.7 – 8.9 m s^–2]

(iii)

{The acceleration is in the same direction as the resultant force. Let the required angle be θ. Consider the triangle formed by the 3 vectors (same triangle as in part (c)(ii)1.). Using sine rule,

sin θ / 2.5 = sin 60 / 6.6

Angle θ = sin^-1 (2.5sin 60 / 6.6) = 19°}

Angle = 19°

[use of scale diagram allow 17° to 21° (a diagram must be seen)]

Question 689: [Waves > Diffraction]

(a) Describe diffraction of monochromatic light as it passes through a diffraction grating.

(b) White light is incident on a diffraction grating, as shown in Fig.1.

The diffraction pattern formed on screen has white light, called zero order, and coloured spectra in other orders.

(i) Describe how principle of superposition is used to explain

1. white light at the zero order,

2. difference in position of red and blue light in the first-order spectrum.

(ii) Light of wavelength 625 nm produces a second-order maximum at an angle of 61.0° to incident direction.

Determine number of lines per metre of the diffraction grating.

(iii) Calculate wavelength of another part of the visible spectrum that gives a maximum for a different order at the same angle as in (ii).

Reference: Past Exam Paper – November 2012 Paper 21 Q4

Solution 689:

(a) The waves pass through the elements / gaps / slits in the grating and spread into geometric shadow

(b)

(i)

1. The displacements add to give a resultant displacement. Each wavelength travels the same path difference or are in phase. Hence, this produces a maximum.

2. To obtain a maximum the path difference must be λ or phase difference 360° / 2π rad. The wavelength λ of red and blue are different, hence the maxima (of red and blue light) are formed at different angles / positions.

(ii)

For diffraction grating: nλ = d sin θ

{Slit separation, d = nλ / sin θ.

Number of lines, N = 1 / d = sin θ / nλ.

For 2^nd order diffraction, n = 2.}

Number of lines, N = sin 61° / (2 × 625 × 10^–9) = 7.0 × 10⁵ m^–1

(iii)

{The path difference nλ = d sin θ is a constant for constructive interference and this constant can be given by 2 × 625 since as given in part (ii), a maxima of order 2 is obtained when the wavelength is 625nm. So, the path difference is 2 × 625.}

nλ = 2 × 625 is a constant (1250)

{Wavelength λ = 2 × 625 / n. We need the wavelength for a different order, so n cannot be 2. Also, the wavelength should be part of the visible spectrum.

We try for different orders (values of n) and see if the wavelength lie in the visible spectrum.}

when n = 1 → wavelength λ = 1250 outside visible range

when n = 3 → wavelength λ = 417 in visible range

when n = 4 → wavelength λ = 312.5 outside visible range

So, wavelength λ (= 417nm) = 420 nm

Question 690: [Thermodynamics]

(a) The equation

pV = constant × T

relates pressure p and volume V of a gas to its kelvin (thermodynamic) temperature T.

State two conditions for equation to be valid.

(b) Gas cylinder contains 4.00 × 10⁴ cm³ of hydrogen at a pressure of 2.50 × 10⁷ Pa and a temperature of 290 K.

Cylinder is to be used to fill balloons. Each balloon, when filled, contains 7.24 × 10³ cm³ of hydrogen at a pressure of 1.85 × 10⁵ Pa and a temperature of 290 K.

Calculate, assuming that hydrogen obeys the equation in (a),

(i) total amount of hydrogen in the cylinder,

(ii) number of balloons that can be filled from the cylinder.

Reference: Past Exam Paper – June 2006 Paper 4 Q2

Solution 690:

(a) Example:

fixed mass/ amount of gas

ideal gas

(b)

(i) Number of moles, n = pV / RT = (2.5 × 10⁷ × 4.00 × 10⁴ x 10^-6) / (8.31 × 290) = 415 mol

{1 cm³ = 10^-6 m³}

(ii)

{From the equation, it can be deduced that

p₁V₁ / T₁ = p₂V₂ / T₂ = constant          for different sets of p, V and T}

At 1.85 × 10⁵ Pa,

{(1.85 × 10⁵) V₁ / 290 = (2.5 × 10⁷) (4.00 × 10⁴) / 290}

Volume of gas = (2.5 × 10⁷ × 4.00 × 10⁴) / (1.85 × 10⁵) = 5.41 × 10⁶ cm³

{1 balloon contains 7.24 × 10³ cm³ of hydrogen. Let N be the number of balloons. As stated in part (a), one condition for the equation to be valid is that the amount of gas in the cylinder should be fixed. So, we should account for the amount of has that remains in the cylinder, which is 4.00 × 10⁴ cm³.}

so, 5.41 × 10⁶ = 4.00 × 10⁴ + (7.24 × 10³) N

Number of balloons, N = 741