# 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 x

_{1}from the point of origin is given by**ϕ**rad

_{1}= (x_{1}/ λ) 2π
Similarly, the phase of a point B
which at a distance x

_{2}from the origin is given by**ϕ**rad

_{2}= (x_{2}/ λ) 2π
Then the phase difference between
the 2 points is

**Δϕ = ϕ**} – {

_{2}– ϕ_{1}= {(x_{2}/ λ) 2π**x**} =

_{1}/ 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 I

_{1}in circuit:**(b)**Variable resistor in (a) now connected as potential divider, as shown. Calculate maximum possible and minimum possible current I

_{2}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__**(a)**

(i)

{Current is maximum when
the resistance of the variable resistor is minimum. Current is minimum when
resistance of variable resistor is maximum}

Minimum current, I = 12 / (6 + 12) =
0.67A

(ii)

The graph should have a correct
start (0.0, 2.0) and finish points (12.0, 0.67). The curve should have a correct shape with a
decreasing gradient

{Note that the curve
passes through the starting point (0.0, 2.0) and ends at (12.0, 0.67) and that
the curve has a decreasing gradient. Your graph should have the same shape as
above. Some of the other points on which the curve passes have been calculated
below. A better graph than the above can be drawn by calculating more points
(but I don’t think such accuracy is required for these 2 marks. Note also that
the curve should not go beyond R = 12).

To calculate the other
points, R is the resistance chosen for the variable resistor (these values are
between 0 and 12). The value of current I is obtained by I = 12 / (6+R). So,
for example, at R = 6, I = 1.0A. At R = 4, I = 1.2A and at R = 8, I = 12 / 14 =
0.86A. Additionally, at R = 10, I = 12 / 16 = 0.75A and at R = 2, I = 12 / 8 =
1.5A}

**(b)**

This part has previously already
been explained as question 4 at http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-1.html

**(c)**

(i)

The graph is a smooth curve starting
at (0, 0) with a decreasing gradient. The end section should not be horizontal

(ii)

With the circuit of Fig 5.3
(parallel combination), the full range of current / p.d. is possible

OR The currents / p.d. can go down
to zero

OR The brightness (of the lamp)
ranges from off to full brightness

__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^{-7}Ωm
(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 = πr

^{2}= πd^{2}/ 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^{-7}Ωm
(iii)

Resistivity ρ of metal =
(4.9 × 10

^{–7}± 0.4 × 10^{–7}) Ω m
May/june 2013 variant 22

ReplyDeleteQ2 (b)

please show and explain how the graphs are to be drawn

The graphs have been added. More explanations have been included

Deleteoct/nov 2013 variant 23

ReplyDeleteQ2 (b)

please show the steps in calculation

I have added the steps there. Check again

Deletemay/june 2011 variant 21

ReplyDeleteQ1 b (ii)

pls show the steps

check question 44 above

Deleteoct/nov variant 21

ReplyDeleteQ4 c (ii)

which year? if it's already posted, plz specify your doubt at the respective page

Deleteyear 2011

Deleteview question 49 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-8.html

in june2014 variant 21

ReplyDeleteQ1 (c)

why are both the arrows at the left?

shouldn't the acceleration be in the direction of motion?

I have explained this part there. See if you now understand

Deletejune 2014 variant 21

ReplyDeleteQ3 (b)

how have yu decided tht the starting point is (0,100) and ending point is (5,980)?

Explanations for the graph are now included there

Deletenov2013 paper23

ReplyDeleteQ4 (b) (ii)

I saw the paper you've solved but I'm still in doubt

air resistance opposes motion so the acceleration decreases and hence the time taken should increase

details have been added at the page. check it again

DeleteFor solution 41 here, I got why for time = τ, a point travels a distance equal to 4 x amplitude. But I dont understand why for time = τ /2, the new position would be the reflection about the horizontal line and why for time = 0.25τ = τ /4, the new position be at the horizontal line. Please explain :(

ReplyDeletein a time τ/2, a points travels a distance 2 x amplitude. so, if a point is at its highest position, a displacement = 'amplitude' causes it to move to the equilibrium position. And the remaining displacement = 'amplitude' causes it to now be at the lowest position

Deletekindly please solve p2 of o/n 2014 v22 :((

ReplyDeleteQuestion 2 is explained as solution 527 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-102.html

Specify the specific doubts you have and I'll try to answer. I'm not explaining complete papers right now.

Thank you so much for your help, I have a doubt in question 41 shouldn't the graph be like this(link below), corresponding with each point?

ReplyDeleteThe link-

http://postimg.org/image/q2hrljpln/

Like the lower portion of the graph shouldn't be shown as there is no distance before 0 given for original wave?

Thanks,

It is the WAVE on the string that moves forwards. The string will still there. Remember that with a wave, energy is transferred without any movement of the material.

DeleteThe particles on the string are still there, and they move up and down. So, the part before 20cm will still be part of the wave as I drawn above.

For question 41, c (I), does this mean that if one point X (which is above equilibrium), if X was compared with another point that was below the equilibrium position, then they would be said to be out-of-phase? And so, are there only 2 possible phase differences on stationary waves? (i.e. 0 or 180)

ReplyDeleteYes.

DeleteFor solution 41 I don't understand why 10ms is considered t/4 shouldn't it be t/2 because t is 20ms so t/2 should be 10ms?

ReplyDeleteSorry, but I cannot locate where they are having problems. Could you be more specific? Thanks

Deletewhere is it mentioned that t/4 = 10ms???

what is t?