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 FR.
X – FR = ma = 0
So, FR = 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 – FR
= 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 – FR
= 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 = 180o
/ 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 (180o). 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 360o 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:
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
