• Physics 9702 Doubts | Help Page 7

Question 40: [Mechanics > Dynamics > Forces and motion]

One horse pulls, with a force of X N, a cart of mass 800kg along a horizontal road at constant speed. Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8ms^-2. Find the time it would take two horses to increase the speed of the cart from 2ms^-1 to 5ms^-1, given that each horse pulls with a force of X N, and that the resistance to motion has the same constant value at all times.

Solution 40:

From the question, resistance to motion (frictional force) is constant at all times.

Consider

“One horse pulls, with a force of X N, a cart of mass 800kg along a horizontal road at constant speed.”

With a force of X N, the cart moves with constant velocity. That is, acceleration is zero. The resultant force is thus zero. Let the force of friction be F_R.

X – F_R = ma = 0

So, F_R = X N   (Frictional force = X N)

“Three horses, each pulling with a force of X N, give the cart an acceleration of 0.8ms^-2”

Resultant force = 3X – F_R = 3X – X = 2X N

This causes the cart of mass of 800kg to have an acceleration of 0.8ms^-2.

2X = ma = 800(0.8)

Force X = 800(0.8) / 2 = 320 N

When 2 horses are pulling,

Resultant force = 2X – F_R = 2X – X = X = 320 N

320 = 800a

Acceleration = 320 / 800 = 0.4ms^-2

Initial velocity, u = 2ms^-1 and Final velocity, v = 5ms^-1

v = u + at

Time, t = (v – u) / a = (5 – 2) / 0.4 = 7.5s

Question 41: [Waves > Progressive & Stationary]

(a) Transverse progressive wave travels along stretched string from left to right. Shape of part of string at particular instant is shown.

Frequency of wave is 15 Hz. For this wave, use Fig to determine

(i) Amplitude

(ii) Phase difference between points P and Q on string

(iii) Speed of wave

(b) Period of vibration of wave is T. Wave moves forward from position shown in Fig for a time 0.25 T. On Fig, sketch new position of wave:

(c) Another stretched string is used to form stationary wave. Part of this wave, at particular instant, is shown.

Points on string are at their maximum displacement.

(i) State phase difference between particles labelled X and Y:

(ii) Explain following terms used to describe stationary waves on string: Antinode, Node

(iii) State number of antinodes shown for this wave

(iv) Period of vibration of wave is τ. On Fig, sketch stationary wave 0.25 τ after instant shown

Reference: Past Exam Paper – June 2011 Paper 23 Q6 & Specimen Paper 2016 Paper 2 Q7

Solution 41:

(a)

(i)

{Amplitude is the maximum displacement}

Amplitude = 7.6mm

(ii) Phase difference = 180^o / π rad

(iii) Speed of wave, v = f λ = 15 x 0.8 = 12ms^-1

(b)

The sketch should be correct with the peak moved to the right. The curve should be moved by the correct phase angle / time period of 0.25T

{For time = period T, the wave travels a distance equal to the wavelength, λ. So, for a time = 0.25T = T/4, the distance moved by each point on the wave is λ / 4 = 80 / 4 = 20cm.}

(c)

(i)

{For a stationary wave, points in the same segment are in phase, while points in adjacent segments are in anti-phase (180^o). This is because, apart from nodes, the points of a stationary wave can only move upwards or downwards. Here, point X and Y will both move downwards since the position displayed is that where the points are at their maximum displacement}

Phase difference = zero (rad)

(ii)

Antinode: maximum amplitude

Node: zero amplitude / displacement

(iii)

Number of antinodes = 3

(iv)

The sketch contains a horizontal line through the central section of the wave

In the time = period of vibration τ, the new position of the points of the stationary wave will be the same as displayed. [Consider the antinode between X and Y. In 1 period, the point would have to move downwards to the horizontal line, then downwards to the maximum amplitude. Then, the point moves upwards to horizontal line again and finally to upwards to its original position. So in a time = τ, a point travels a distance equal to 4 x amplitude.] So, for time = τ /2, the new position would be the reflection about the horizontal line. For time = 0.25τ = τ /4, the new position be at the horizontal line}

 

Question 42: [Waves > Phase difference]

How to calculate the phase difference between 2 points on a wave?

Solution 42:

Consider 2 points on the wave which are separated by a distance equal to the wavelength, λ of the wave. These 2 points would be in phase (they move the same way). So, a separation of λ corresponds to phase difference of 2π rad (they are in phase).

Therefore, the phase difference between 2 points on a wave can be related to the separation apart.

If point A and B on a wave of wavelength λ are separated by a distance Δx, the phase difference, ϕ between them is given by

ϕ = (Δx / λ) 2π rad

Alternatively, if the start of the wave [of wavelength, λ] (which is at equilibrium position) is taken as the reference, then the phase of point A which is at a distance x₁ from the point of origin is given by

ϕ₁ = (x₁ / λ) 2π rad

Similarly, the phase of a point B which at a distance x₂ from the origin is given by

ϕ₂ = (x₂ / λ) 2π rad

Then the phase difference between the 2 points is

Δϕ = ϕ₂ – ϕ₁ = {(x₂ / λ) 2π} – {x₁ / λ) 2π} = (Δx / λ) 2π rad

To directly obtain the phase difference in degrees, use 360^o instead 2π rad.

Question 43: [Current of Electricity > Variable resistor]

(a) Variable resistor is used to control current in circuit, as shown. 

Variable resistor is connected in series with 12 V power supply of negligible internal resistance, ammeter and 6.0 Ω resistor. Resistance R of variable resistor can be varied between 0 and 12 Ω.

(i) Maximum possible current in circuit is 2.0 A. Calculate minimum possible current:

(ii) On Fig, sketch variation with R of current I₁ in circuit:

(b) Variable resistor in (a) now connected as potential divider, as shown. Calculate maximum possible and minimum possible current I₂ in ammeter:

(c)

(i) Sketch I-V characteristic of filament lamp:

(ii) Resistor of resistance 6.0 Ω is replaced with filament lamp in circuits of Fig. 5.1 and Fig. 5.3. Give an advantage of using circuit of Fig. 5.3, compared to circuit of Fig 5.1, when using circuits to vary brightness of the filament lamp:

Reference: Past Exam Paper – June 2011 Paper 21 Q5

Solution 43:

Go to
A variable resistor is used to control the current in a circuit, as shown in Fig. 5.1.

Question 44:   [Measurements > Uncertainty]

Measurements made for sample of metal wire are shown.

Quantity          Measurement               Uncertainty

Length             1750mm                      ±3mm

Diameter         0.38mm                       ±0.01mm

Resistance       7.5Ω                            ±0.2Ω

(a) State appropriate instruments used to make each of these measurements.

(i) Length

(ii) Diameter

(iii) Resistance:

(b)

(i) Show that resistivity of the metal is calculated to be 4.86 × 10^–7 Ω m:

(ii) Calculate uncertainty in resistivity:

(iii) Use answers in (b) to express resistivity with its uncertainty to appropriate number of significant figures:

Reference: Past Exam Paper – June 2011 Paper 21 Q1

Solution 44:

(a)

(i) Length: metre rule / tape

(ii) Diameter: micrometer (screw gauge) / digital caliper

(iii) Resistance: an ammeter and a voltmeter / ohmmeter / multimeter on ‘ohm’ setting

(b)

(i)

Resistivity ρ of metal = RA / L = [7.5 x π(0.38x10^-3)² / 4] / 1.75 = 4.86x10^-7Ωm

(ii)

{ρ = RA / L. So, (Δρ /ρ) x 100% = [(ΔR / R) + (ΔA / A) + (ΔL / L)] x 100%}

% Uncertainty in R (= ΔR/R x 100%) = [0.2 / 7.5] x 100% = 2.7%

And % Uncertainty in L (= ΔL/L x 100%) = [3 / 1750] x 100% = 0.17%

{A = πr² = πd² / 4. So, (ΔA / A) x 100% = 2(Δd / d) x 100%}

% Uncertainty in A (= 2(Δd / d) x 100%) = 2[0.01 / 0.38] x 100% = 5.3%

Total uncertainty (= 2.7 + 0.17 + 5.3) = 8.17%

{(Δρ /ρ) x 100% = 8.17%. So, Δρ = (8.17 / 100) x ρ}

Uncertainty (= (8.17/100) x 4.86x10^-7) = 0.397x10^-7Ωm

(iii)

Resistivity ρ of metal = (4.9 × 10^–7 ± 0.4 × 10^–7) Ω m