# Physics 9702 Doubts | Help Page 1

**Question 1: [Kinematics > Linear motion]**
An aircraft is at rest at one end of
a runway which is 2.2km long. The aircraft accelerates along the runway with an
acceleration of 2.5ms

^{-2}until it reaches its take-off speed of 75 ms^{-1}. Calculate
(i) the time taken to reach take-off
speed

(ii) the distance travelled in this
time

(iii) Just as the aircraft reaches
take-off speed a warning light turns on. The maximum possible deceleration of
the aircraft is 4.0ms

^{-2}and 2.5s elapses before the pilot takes any action, during which times the aircraft continues at its take off speed. Determine whether or not the aircraft can be brought to rest in the remaining length of runway.**Reference:**

*Cambridge International AS and A Level Physics Coursebook – 2*

^{nd}edition – by David Sang, Graham Jones, Gurinder Chadha and Richards Woodside

__Solution 1__:
(i)

Initial speed, u = 0ms

^{-1}(aircraft is at rest)
Acceleration, a = 2.5ms

^{-2}
Final speed, v = 75ms

^{-1}(take-off speed)
a = (v - u) / t

Time taken to reach take-off speed,
t = (v – u) / a = (75 – 0) / 2.5 = 30s

(ii)

Consider the equation for uniformly
accelerated motion:

s = ut + ½at

^{2}
Distance travelled during this time,
s = 0(30) + 0.5(2.5)(30

^{2}) = 1125m__Alternatively__,

Consider the equation for uniformly
accelerated motion:

v

^{2}= u^{2}+ 2as
(75)

^{2}= 0^{2}+ 2(2.5)s
Distance travelled during this time,
s = 75

^{2}/ 2(2.5) = 1125m
{I think the
first method is better since the question asks for the distance travelled
during that

__time__and the second method does not make use of the time}
(iii)

Remaining length of runway when
warning light turns on = 2200 – 1125 = 1075m

Distance travelled during the 2.5s
before the pilot takes any action:

Speed (= distance / time) = 75ms

^{-1}
Distance = Speed x Time = 75 x 2.5 =
187.5m

Remaining length of runway for
deceleration = 1075 – 187.5 = 887.5m

Consider the deceleration to be
maximum (= 4ms

^{-2}) [acceleration, a = – 4ms^{-2}]
Initial speed, u = 75ms

^{-1}and final speed, v = 0ms^{-1}(aircraft brought to rest)
Let’s try to calculate the distance
required to bring the aircraft to rest, s and compare it with the remaining
length of runway.

Consider the equation for uniformly
accelerated motion:

v

^{2}= u^{2}+ 2as
0

^{2}= 75^{2}+ 2(-4)s
s = (75

^{2}) / 8 = 703.125m
Since the distance required to bring
the aircraft to rest (703.125m) is less than the remaining length of runway
(887.5m), the aircraft can be brought to rest in the remaining length of
runway.

__Question 2: [Current of Electricity > Resistance]__**(a)**{Detailed explanations for this part of the question is available as Solution 391 at Physics 9702 Doubts | Help Page 72 -

*http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html*}

**(b)**Some heaters each labelled 230V, 1.0kW. Heaters have constant resistance. Determine total power dissipation for heaters connected as shown in each diagram below.

**Reference:**

*Past Exam Paper - November 2010 Paper 21 Qu6(b)*

__Solution 2:__**(a)**{Detailed explanations for this part of the question is available as Solution 391 at Physics 9702 Doubts | Help Page 72 -

*http://physics-ref.blogspot.com/2015/02/physics-9702-doubts-help-page-72.html*}

**(b)**

{The heaters are each
230V, 1.0kW. This means that when there is a potential difference of 230V
across a heater, there is a power consumption of 1.0kW.}

(i)

{When electrical
components are connected in parallel to a power supply, the same potential
difference (as the emf of the power supply) would occur through each of the
components.

Here, there is a potential
difference (p.d.) of 230V across each heater. So, each of them would have a
power consumption of 1.0kW.}

Total power dissipation (= 1.0 +
1.0) = 2.0kW

**,**

__Alternatively__
Let resistance of a heater = R

Power dissipation = V

^{2}/ R
{For a power supply of
230V, there is a power dissipation of 1.0kW}

The total resistance in the circuit
= [(1/R) + (1/R)]

^{-1}= R/2
Since R is now halved,

Total Power dissipation = V

^{2}/ 0.5R = 2 (V^{2}/ R) = 2(1.0kW) = 2.0kW
(ii)

{When electrical components
are connected in series to a power supply, the total e.m.f of the source is
divided into p.d. across the different components (depending on their
resistance) so that the total p.d. in the circuit is equal to the e.m.f of the
power supply (given there is no loss volts)

Here, since both heaters
have the same resistance, the e.m.f is divided equally into 2 as the p.d.
across each heater. That is, there is a p.d. of 115V across each heater.

Power dissipation across a
heater (= V

^{2}/ R) is proportional to the (p.d.)^{2}across it. So, if the p.d. across a heater is halved, the power dissipation is (= (½)^{2}= ¼) one quarter. Therefore, the power dissipated in each heater in the diagram is (1.0kW) / 4 (= 0.25kW).}
Total power dissipation (= 0.25 +
0.25) = 0.5kW

**,**

__Alternatively__
Total resistance in circuit = R + R
= 2R

Since R is now doubled,

Total power dissipation = V

^{2}/ 2R = 0.5 (V^{2}/ R) = 0.5(1.0kW) = 0.5kW
(iii)

{The first method would be
too time-consuming here. So, consider only the alternative method as the 2 above}

The circuit can be
simplified as a series connection of a power supply and a heater, (both of
which are) connected in series to a parallel combination of 2 heaters.

Total resistance in circuit = R +
[(1/R) + (1/R)]

^{-1}= R + R/2 = 3R / 2
Since total resistance is now 3R /2
(= 1.5R)

Total Power dissipation = V

^{2}/ 1.5R = (2/3) (V^{2}/ R) = (2/3)(1.0kW) = 0.67kW

__Question 3: [Deformation of Solids > Young modulus]__
Wire of length 1.70 m hangs
vertically from fixed point, as shown. Wire has cross-sectional area 5.74 × 10

^{–8}m^{2}and is made of material that has Young modulus of 1.60 × 10^{11}Pa. Load of 25.0 N is hung from wire.
(i) Extension of wire

(ii) Same load is hung from second
wire of same material. This wire is twice the length but same volume as first
wire. Explain how extension of second wire compares with that of first wire

**Reference:**

*Past Exam Paper – June 2011 Paper 21 Q4(b)*

__Solution 3:__
(i)

Young modulus, E = Stress / Strain

Young modulus, E = [F / A] / [e / l] (= Fl / Ae)

Extension, e (= Fl / AE) = (25 x
1.70) / ([5.74x10

^{-8}] x [1.6x10^{11}])
Extension, e = 4.6x10

^{-3}m
(ii)

(Volume = Area x length)

(Since wire is of the same material,
the Young modulus is the same. Also, same load = same F)

The area A becomes A / 2 OR the
stress is doubled. Since the extension, e is proportional to (l / A) OR
Substitute into the full formula {e = Fl / AE}. So, the total extension
increase {for the second wire} is 4e {since l becomes 2(l) and A becomes (A/2)
– both the change in l and A should be considered}

__Question 4: [Current of Electricity > Variable resistor]__
Variable resistor in (a) now connected
as potential divider, as shown. Maximum possible and minimum possible current I

_{2}in ammeter:**Reference:**

*Past Exam Paper – June 2011 Paper 21 Q5(b)*

__Solution 4:__
(The total current from the battery is divided at the
junction connecting the fixed resistor and the variable resistor. I

_{2}is the current flowing through the 6.0Î© resistor. )
{How the circuit works?
Consider the point A and B.

The resistance contributed
by the variable resistor changes as the contact is moved. At A (which is
connected to the +ve terminal of the battery), the resistance of the variable
resistor is zero.

In this second circuit,
the fixed resistor is connected in parallel to the variable resistor while for
the first diagram, the fixed resistor is connected in series with the variable
resistor.

For a parallel connection,
current splits up at a junction. From Ohm’s law, I = V / R. That is, the lower
the resistance, the greater is the current flowing through it. Since the
variable resistor has zero resistance at A, all current would flow through it.
Thus, current I

_{2}through the fixed resistor is zero.
In this case, it’s as if
the fixed resistor has been short-circuited by the variable resistor of zero
resistance.}

Minimum possible current = 0

{The resistance of the
fixed resistor (6.0Î©) is known. Note that by choosing a resistance for the
variable resistor, the circuit becomes a parallel combination of the fixed
resistor and the resistance chosen for the variable resistor.

From Kirchhoff’s law, there
will be the same p.d. across both the fixed resistor and the variable resistor.
This p.d. is equal to the e.m.f. in the circuit (that is 12V).

From Ohm’s law, I = V / R

Current I

_{2}through fixed resistor = 12 / 6.0 = 2.0A.}
Maximum possible current (= 12 / 6)
= 2.0A

__Question 5: [Sound Waves > CRO]__
Loudspeaker produces sound wave of
constant frequency. How cathode-ray oscilloscope (c.r.o.) may be used to
determine frequency:

**Reference:**

*Past Exam Paper – November 2010 Paper 23 Q3*

__Solution 5:__
Below is a diagram for the c.r.o:

A microphone / (terminals of) the
loudspeaker is connected to the Y-plates of the c.r.o.

The c.r.o. is adjusted to produce a
steady wave of 1 (or 2) cycles / wavelengths on the screen.

The length of a cycle / wavelength Î» is measured and the time-base b is noted.

The frequency f is given by f = 1 / Î»b.

Source for 1st diagram: http://jjphysics.pbworks.com/f/10_11_H1_Wave+Motion_notes.pdf

Source for 2nd diagram:

*Cambridge International AS and A Level Physics Coursebook – 2*^{nd}edition – by David Sang, Graham Jones, Gurinder Chadha and Richards Woodside, Page 180

__Question 6: [Kinematics > Linear motion]__
Student takes measurements to
determine value for acceleration of free fall. Some of apparatus used is
illustrated in Fig. Student measures vertical distance d between base of
electromagnet and bench. Time t for iron ball to fall from electromagnet to
bench is also measured. Corresponding values of t

^{2}and d are shown.
(a) Draw line of best fit

(b) Explain why there is a non-zero
intercept on the graph

(c) Determine student’s value for (i)
Diameter of ball (ii)
Acceleration of free fall

**Reference:**

*Past Exam Paper – November 2010 Paper 23 Q4*

__Solution 6:__**(a)**

Line of best fit for points:

An acceptable straight line
(touching every point) should be drawn

**(b)**

Explain why there is a non-zero
intercept on graph:

The distance fallen by the iron ball
is not d. Distance d is the distance fallen by the ball plus the diameter of
the ball

**(c)**

Determine student’s value for

(i)

Diameter of ball:

(The diameter is the value of
y-intercept)

Diameter of ball: (allow) 1.5 ± 0.5cm

(ii)

Acceleration of free fall, g:

Gradient of graph = 4.76 ± 0.1 with evidence that the origin {point
(0,0)} has not been used {since the line intercepts the y-axis at a value
greater than 0}

{A graph of
distance against (time)

^{2}is plotted. Consider the equation ofr uniformly accelerated motion: s = ut + ½at^{2}where d = s and a = g here. Compare with y = mx + c where m is the gradient and c the y-intercept. From equation of motion, gradient of a s-t^{2}graph would be equal to ½g [just like the gradient m is multiplied by x, the gradient should be multiplied by the x-axis quantity (which is t^{2}). So ½ g represents the gradient]}
Gradient of graph = g / 2

So, the acceleration of free fall, g
= 9.5ms

^{-2}

__Question 7: [Units]__
Cylindrical tube rolling down slope
of inclination Î¸ moves distance L in time T. Equation relating these quantities
is L (3+a

^{2}/P) = QT^{2}sinÎ¸ where a is internal radius of tube and P and Q are constants.
Which line gives the correct units
for P and Q?

P Q

A m

^{2}m^{2}s^{-2}
B m

^{2}ms^{-2}
C m

^{2}m^{3}s^{-2}
D m

^{3}ms^{-2}**Reference:**

*Past Exam Paper – November 2011 Paper 11 Q4 & Paper 13 Q5*

__Solution 7:__
{Ans: B.} Expanding the left hand
side gives 3L + a

^{2}L/P.
For equation to be homogeneous (and
hence correct), the units should be the same on both sides of the equation and
the units for 3L = units for a

^{2}L/P.
L has unit m (since it is a distance),
so all the different parts should have an overall unit of m.

Considering the a

^{2}L/P: [m^{2}][m] / [P] = m. So, [P] = [m^{3}] / [m] = m^{2}.
Now for QT

^{2}sinÎ¸: sinÎ¸ has no units. So, [Q][s^{2}] = m giving [Q] = ms^{-2}.
In your answer to question 4(b)(i) why have you given the answer in 2 sf ?

ReplyDeleteit's actually 1 d.p.

Deleteusually we give answers to 3 sf or 1 dp