RAFFLES INSTITUTION 2010 YEAR 5 TERM 3 COMMON TEST 1 July 2010
Time:
2 hr
H2 PHYSICS RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION
Section A INSTRUCTIONS TO CANDIDATES
There are 2 sections in this paper. Section A consists of 15 multiple-choice questions. For each question, four suggested answers are given. You are to choose the most appropriate one and shade your answer on the OMR form. You are advised not to spend more than 30 minutes on Section A. Section B consists of 9 structured questions. You are to write your answers in the spaces provided. Attempt all questions in Sections A and B.
There are 7 printed pages, inclusive of the cover page and data & formulae page, in this booklet.
2 DATA AND FORMULAE Data speed of light in free space,
c = 3.00 x 108 m s−1
permeability of free space,
µ0 = 4 π x 10−7 H m−1
permittivity of free space,
ε0 = 8.85 x 10−12 F m −1
elementary charge,
e = 1.60 x 10−19 C
the Planck constant,
h = 6.63 x 10−34 J s
unified atomic mass constant ,
u = 1.66 x 10−27 kg
rest mass of electron,
me = 9.11 x 10−31 kg
rest mass of proton,
mp = 1.67 x 10−27 kg
molar gas constant,
R = 8.31 J K−1 mol−1
the Avogadro constant,
NA = 6.02 x 1023 mol−1
the Boltzmann constant,
k = 1.38 x 10−23 J K−1
gravitational constant,
G = 6.67 x 10−11 N m2 kg−2
acceleration of free fall,
g = 9.81 m s−2
Formulae uniformly accelerated motion,
s = ut + 21 at 2 v 2 = u 2 + 2as
work done on/by a gas,
W = p∆V
hydrostatic pressure,
p = ρgh
gravitational potential,
φ=−
displacement of particle in s.h.m.,
x = x0 sin ωt
velocity of particle in s.h.m.,
v = v 0 cos ωt
Gm r
= ±ω x02 − x 2
mean kinetic energy of a molecule
3
of an ideal gas,
E=
resistors in series,
R = R1 + R2 + . . . .
resistors in parallel,
1/R = 1/R1 + 1/R2 + . . . .
electric potential,
V = Q/4πε0r
alternating current/voltage,
x = x0 sinωt
transmission coefficient,
T = exp(−2kd) where k =
radioactive decay,
x = x0 exp(−λt)
decay constant,
λ=
2
kT
0.693 t1 2
8π 2 m(U − E ) h2
3
Section A (30 marks) Which of the following expressions could be correct for the speed v of waves travelling along a string in terms of T the tension in the string, m the mass of the string and L the length of the string?
1
L T
A
2
3
B
T m
C
D
TLm
TL m
How many nanoseconds are there in 1.0 hour?
A
6.0 × 107
B
6.0 × 1010
C
3.6 × 1012
D
3.6 × 1015
A car, initially at rest, moves along a straight line. The graph below shows the variation with time t, of the car’s acceleration a.
A a
B
D t
C
At which point, A, B, C or D, is the velocity of the car maximum?
4
A ball of mass m is thrown from a point P with an initial velocity u at an angle 60o to the horizontal. In the absence of air resistance, it follows a parabolic path before landing back on the ground at point Q. P and Q are at the same height. What is the magnitude of the change in momentum between P and Q?
A 0 B C
mu 3mu
D 2mu
4
5
A block of mass m1 is resting on top of another block of mass m2. The blocks are connected by a light rope, strung through a frictionless pulley. A force F = 0.750 N is applied on m2 as shown. frictionless pulley
m1 m2
F
If m1 and m2 are 200 g and 800 g respectively, and friction between m2 and the ground is 0.150 N while friction between the two blocks is 0.050 N, find the acceleration of m2.
6
A
0.50 m s−2
B
0.60 m s−2
C
0.70 m s−2
D
0.75 m s−2
A car of mass 1200 kg was travelling at 30 m s−1 when a traffic light ahead turned amber. The driver immediately applied his brakes such that the stopping force F increased to a maximum and decreased steadily till the car came to a halt, as shown in the following diagram. Stopping Force, F Fmax
0
10
Find the magnitude of the maximum deceleration of the car.
A
1.5 m s−2
B
3.0 m s−2
C
1.8 x 103 m s−2
D
3.6 x 103 m s−2
20
Time / s
5
7.
A uniform cantilever beam of weight W is hinged on one end and held by a wire on the other, carrying a load of similar weight W as shown. The tension in the wire is T.
C D
E
T B
A W W
What is the direction of the force exerted by the wall on the beam?
A
8
AB
B
C
AC
AD
D
AE
The spring system shown is composed of two springs, X and Y, of negligible mass. X has a spring constant of 60 N m −1 and Y has a spring constant of 15 N m−1.
X
Y 3.0 N load
The overall extension produced when a load of 3.0 N is supported by the system is
A 40 mm B 150 mm C 225 mm D 250 mm
6
9
The diagram shows two bodies, X of mass 1.0 kg and Y of mass 2.0 kg, connected by a light inelastic cord passing over a light, free-running pulley. X starts from rest and when moving on the horizontal surface, experiences a constant frictional force of 3.0 N. 2.0 m 1.0 kg X
Y 2.0 kg
What is the kinetic energy of Y when X has travelled 2.0 m along the surface?
A
22.2 J
B
26.2 J
C
30.2 J
D
33.2 J
10 A car of mass 2800 kg is moving at a constant speed of 22.0 m s−1 and the total resistive force acting on it is 341 N. If the car is using fuel at the rate of 2.00 × 10−3 litres per second and one litre of fuel contains 34.8 MJ of chemical energy, what is the efficiency of the car engine in moving the car? A
1.95 %
B
2.16 %
C
10.8 %
D
97.4 %
11 Which statement about internal energy is correct? A
The internal energy of a system can be increased without transfer of energy by heating.
B
The internal energy of a system depends only on its temperature.
C
When the internal energy of a system is increased, its temperature always rises.
D
When two systems have the same internal energy, they must be at the same temperature.
7
12
A mixture of two monatomic ideal gases X and Y is in thermal equilibrium. The mass of a molecule of gas Y is four times the mass of a molecule of gas X. The mean translational kinetic energy of the molecules of Y is 8.0 x 10−21 J. What is the mean translational kinetic energy of the molecules of X?
A
3.2 x 10−20 J
B
1.6 x 10−20 J
C
2.0 x 10−21 J
D
8.0 x 10−21 J
13 A car travels in uniform circular motion. Which of the following correctly describes the linear velocity, angular velocity and centripetal acceleration of the car? linear velocity
angular velocity
centripetal acceleration
A
constant
constant
constant
B
constant
constant
varying
C
varying
constant
varying
D
varying
varying
constant
14 The maximum speed for a car to go around a corner without skidding is 18 m s−1 when the road is dry. The maximum frictional force between the road surface and the wheels of the car is halved when the road is wet. What is the maximum speed for the car to go round the corner without skidding when the road is wet?
A
4.5 m s−1
B
6.4 m s−1
C
9.0 m s−1
D
12.7 m s−1
15 A marble of mass m is released from rest at the rim of a smooth semi-spherical bowl of radius R. What is the normal force acting on the marble when it passes the bottom of the bowl? Take the acceleration of free fall to be g. A
mg
B
mgR
C
3mg
D
3mgR
END OF SECTION A
1
Name:
(
)
CT Group: 11SO
RAFFLES INSTITUTION 2010 YEAR 5 TERM 3 COMMON TEST 1 July 2010
H2 PHYSICS RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION RAFFLES INSTITUTION
Section B INSTRUCTIONS TO CANDIDATES Write your name, index number and CT Group. Write your answers to Section B in the spaces provided in this booklet.
For Examiner’s Use Section A
Section B
MCQ 16 17 18 19 20 21 22 23 24
Total
There are 15 printed pages, inclusive of the cover page, in this booklet.
/ / / / / / / / / / /
30 8 9 8 7 7 8 8 7 8 100
2
Section B (70 marks) Answer all questions in this section.
16
Fig. 16.1 shows the set-up of the apparatus used to determine g, the acceleration of free fall.
Electromagnet X
A steel ball X is held by an electromagnet and positioned directly above a hinged metal plate P that is initially closed. When switch S is quickly switched from A to B, the electromagnet is switched off and the ball is released. At the same time, the circuit operating a clock C is completed and the clock starts. When the ball hits plate P, the plate swings open, causing the clock circuit to be broken and stopping the clock.
(a)
A
S
H B
C P Fig. 16.1
The above procedure is repeated four times and the following set of results were recorded: Distance H from P to bottom of ball = 0.982 m Times recorded in each of four runs: 0.47 s, 0.44 s, 0.48 s, 0.46 s. The distance H is measured using a metre rule. (i)
Express the value of t, the mean time of fall, together with its uncertainty to the correct number of significant figures. [1]
(ii)
Using the values given above, calculate the acceleration of free fall using the formula H = ½ g t 2 and express its value together with its uncertainty to the correct number of significant figures. [3]
3
(b)
The experiment was repeated for different values of H and the results plotted on a graph. Fig. 16.2 shows the graph obtained.
H x x x x x x
t
0 Fig. 16.2
The intercept on the t-axis is due to the presence of a systematic error. (i)
Explain what is meant by a systematic error.
[1]
(ii)
State the feature of the graph that shows the presence of random errors.
[1]
(iii)
Explain why drawing a best fit line through the points obtained from a number of readings reduces random errors. [1]
(iv)
A student wrongly attributed the presence of the intercept on the t-axis to air resistance. On Fig. 16.2 above, sketch the graph that would be obtained if air resistance was not negligible.
[1]
4
17
In the absence of air resistance, a rubber band is launched from a rubber band launcher at point O with a speed u at an angle θ to the horizontal. Point A is the maximum horizontal distance that can be reached by the rubber band. The path of the rubber band is shown in Fig. 17.
u O
θ
A 4.00 m Fig. 17
(a)
State the angle θ that will give the maximum range of the rubber band.
[1]
(b)
The maximum horizontal distance OA is 4.00 m. Show that u = 6.26 m s−1.
[3]
(c)
Calculate the maximum height that can be reached by the rubber band.
[2]
(d)
In reality, air resistance cannot be neglected. State and explain whether the time of flight for the upward motion is longer, shorter or equal to the time of flight for the downward motion, in the presence of air resistance. [3]
5
18
(a)
(b)
(i)
State Newton’s Second Law of Motion.
[2]
(ii)
Show how the equation F = ma is derived from Newton’s Second Law.
[1]
(i)
Distinguish between the mass and weight of a body.
[2]
(ii)
In a lift with upward acceleration a, a spring balance indicates a box to have a mass of 10.0 kg. When the lift has a downward acceleration 2a, the spring balance indicates 7.0 kg. Find the value of the acceleration a and the actual mass of the box. [3]
6
19
(a)
State the principle of conservation of linear momentum.
(b)
Fig. 19.1 shows two persons A and B, each standing on a cart placed on a frictionless surface, facing each other. Initially, both persons are stationary and person A is holding a ball of mass 2.0 kg.
[1]
Person A then throws the ball towards person B with a horizontal velocity of 11.6 m s−1. The mass of person A and B are 70.0 kg and 75.0 kg, respectively. Assume that the mass of each cart is negligible.
B 75.0 kg
A 70.0 kg
Fig. 19.1 (i) Calculate the velocity of person A just after the ball leaves his hand.
[2]
(ii) Show that person B moves to the right with a speed of 0.301 m s−1 at the instant he receives the ball.
[1]
7
(c)
During a company’s team building session, two persons were made to stand on a large stationary cart placed in the middle of an ice-skating rink, as shown in Fig. 19.2. A large number of 2.0 kg balls were placed on the floor of the rink. The two persons were then tasked to move the cart from the middle of the skating rink to the side, without stepping off the cart. This was achieved when person A continuously picked up a ball from his side and threw it towards person B, who caught it, and put it down on the floor of the rink. A B
Figure 19.2
By describing the motion of the cart from the instant person A throws a ball to the instant person B catches it, explain why they were able to complete the task. [3]
8
20
(a)
Define upthrust.
[1]
A student wanted to use Archimedes Principle to find the mass of a tennis ball. He first placed a square block of dimension 0.150 m x 0.150 m x 0.060 m in a pail of water (as shown in Fig. 20.) and measured the depth of immersion h of the block in water. The densities of the block and water are 560 kg m−3 and 1000 kg m−3 respectively.
0.150 m
0.150 m 0.060 m
h
Fig. 20 (b)
Calculate the value of h.
(c)
When the tennis ball was placed on the block, the block was submerged to a depth of 0.053 m. The radius of tennis ball is 0.067 m.
[3]
(i)
Calculate the mass of the ball.
[1]
(ii)
Hence find the density of the ball.
[2]
9
21
(a)
State the conditions for a system of coplanar forces to be in equilibrium.
[2]
(b)
Fig. 21 shows a crane being used to lift a load of girders, each of which has a mass of 500 kg.
A
cable E
B
jib
Q 4.0 m
P
F
30°
30o
R
girders
cab 9.0 m
Fig. 21 The jib of the crane has a mass of 2500 kg and the cab has a mass of 20 000 kg. The centres of mass of the jib and the cab are at their mid-points E and F respectively. The hook and the cable have negligible mass. (i)
When a single girder is lifted with a constant speed, calculate 1.
the tension in the cable AB,
[1]
10
2.
(ii)
the corresponding tension in the cable PQR.
[2]
For the jib in the position shown in Fig. 21, determine 1.
the tension in AB which will just topple the crane.
[2]
2.
the maximum number of girders which can be lifted without the crane toppling over.
[1]
11
22
A light elastic string of natural length 0.450 m and force constant of 24.5 N m−1 is attached to point X at one end and a sphere of 0.600 kg at its other end, as shown in Fig. 22. When the sphere is released from rest at 0.250 m below X, the sphere reaches its maximum speed at a distance of 0.690 m below X. X retort stand with boss and clamp
elastic string
0.250 m m sphere
bench top
clamp
Fig. 22 (a) Find the extension of the string when the sphere reaches its maximum speed.
[1]
(b) Calculate the elastic potential energy of the elastic string when the sphere reaches its maximum speed.
[2]
(c) Determine the maximum kinetic energy of the sphere.
[2]
12 (d) Discuss the energy changes which take place as the sphere falls from rest till it reaches its maximum speed. You need to refer to the different forms of energy involved.
23
[3]
An ice-cube at −25.0 oC is heated until it becomes steam at 100 oC. Fig. 23 shows the variation with energy Q added to the system of the temperature θ of the system.
θ / oC 100 75 50 25 0 -25 Q/J Y
62.7 Fig. 23
(a) Explain briefly why the temperature remained constant for two sections of the graph.
[2]
13
(b)
The specific heat capacity of ice is 2090 J kg−1 K−1, specific heat capacity of water is 4200 J kg−1 K−1, specific latent heat of fusion of ice is 3.33 x 105 J kg−1 and specific latent heat of vaporization of water is 2.26 x 106 J kg−1. Calculate
(i)
the mass of the ice cube,
(ii)
the value of Y on the graph, the point when all the water has completely changed to steam. [3]
[2]
14 24
A sample of a monatomic ideal gas starts at A (see Fig. 24) with a volume of 5.0 x 10−3 m3, pressure of 1.0 x 105 Pa and temperature of 300 K. It is heated at constant volume to B where the pressure is 3.0 x 105 Pa. It is then allowed to expand at constant temperature to C at which the pressure is 1.0 x 105 Pa. Finally, it is compressed to its original state at constant pressure. p /105 Pa 3.0
B
2.0
1.0
0
C
A
5.0
10
VC
V / 10−3 m3
Fig. 24 (a)
Calculate (i)
the amount, in mol, of gas in the sample,
[1]
(ii)
the temperature at B,
[1]
(iii) the volume VC of the gas at C,
[1]
(iv) the internal energy of the gas
[2]
1.
at A,
2.
at B.
15
(b)
Complete the table below for each section of the cyclic process A → B → C → A. Heat supplied
Work done
to gas
on gas
Increase in internal energy
/J
/J
/J
A→B B→C
-1650
C→A
END OF SECTION B
[3]
Year 5 Physics H2 Common Test – MCQ Answers
Raffles Institution
2010
Question Key D 1 C 2 B 3 C 4 A 5 B 6 C 7 D 8 A 9 C 10 A 11 D 12 C 13 D 14 C 15
1 (For Internal Use Only)
Year 5 Physics H2 Common Test – MCQ Answers 1
Raffles Institution
2010
Option A: 1
1
1 2 1 2 L m 2 = = = Base units of kg s -2 T kg m s-2 kg s
Option B: 1
1 1 T kg m s-2 2 -2 2 2 = = = m s m s-1 ( ) m kg
Base units of
Option C:
(
Base units of TLm = ( kg m s-2 ) ( m )( kg)
)
1 2
1
= ( kg2 m2 s-2 ) 2 = kg m s-1
Option D: 1
-2 2 1 TL ( kg m s ) m = ( m2 s-2 ) 2 = m s-1 Base units of = m kg
Answer: D
2
1.0 hour = 3600 seconds = 3600 x 109 nanoseconds = 3.6 x 1012 nanoseconds Answer: C
3
The variation with time t of the car’s velocity v is given by the area under the a-t graph. v increases from 0 between t = 0 and point B, where a is positive. Subsequently, v decreases over the time interval between point B and D, where a is negative. Hence the velocity of the car at point B is maximum. Answer: B
4
pi (of magnitude mu) pf 60
o
pf - pi 60o
60o 60o
pf (of magnitude mu)
-pi
The magnitude of the change in momentum (pf - pi) is given by 3 2 ( mu sin 60o ) = 2 mu = 3mu 2 Answer: C
2 (For Internal Use Only)
Year 5 Physics H2 Common Test – MCQ Answers
5
Isolate m1,
Raffles Institution
2010
a T m1 0.050 N
Isolate m2, T
a
0.050 N
m2
F
0.150 N Apply Newton's 2nd law to m1 , T − 0.050 = 0.200a .......... (1) Apply Newton's 2nd law to m1 , F − 0.150 − 0.050 − T = 0.800a 0.750 − 0.150 − 0.050 − T = 0.800a ..........(2) Adding (1) and (2), we obtain a = 0.50 m s-2
Answer: A
6
impulse acting on the car = change in momentum of the car area under the F -t graph = change in momentum of the car Fmax (20 s) = (1200 kg)(30 m s-1 ) 2 Fmax = 3600 N Magnitude of maximum deceleration of the car Fmax = = 3.0 m s-2 1200 kg
Answer: B
7
Since the weights of the beam and load are the same, their combined force 2W acts downwards along the mid-line between their lines of action and passes D. It meets the line of action of tension T at D. The direction of the force exerted by the wall on the beam must be along AD to meet at this common point. Answer: C
3 (For Internal Use Only)
Year 5 Physics H2 Common Test – MCQ Answers
8
Raffles Institution
2010
The tension in both springs is 3.0 N. Assume the springs obey Hooke's Law, i.e. tension = (spring constant)(extension), 3.0 N = 0.050 m 60 N m-1 3.0 N the extension of spring Y = = 0.20 m 15 N m-1 Overall extension = 0.050 + 0.20 = 0.25 m = 250 mm the extension of spring X =
Answer: D
9
Conservation of energy: increase in total Ek of X and Y + work done by X against frictional force = decrease in gravitational Ep of Y ⇒ increase in total Ek of X and Y + (3.0 N)(2.0 m) = (2.0 kg)(9.81 m s-2 )(2.0 m) ⇒ increase in total Ek of X and Y = 33.24 J ∴ increase in Ek of Y mass of Y = ( 33.24 J) total mass of X and Y 2.0 = ( 33.24 J) = 22.2 J 3.0 Since the system starts from rest, the kinetic energy of Y is 22.2 J. Answer: A
10
efficiency of the car engine in moving the car power output = power input Fv = rate of usage of chemical energy =
( 341 N) ( 22.0 m s-1 )
( 34.8 x 10
6
J litre-1 )( 2.00 x 10-3 litre s-1 )
= 0.108 = 10.8% Answer: C
4 (For Internal Use Only)
Year 5 Physics H2 Common Test – MCQ Answers
Raffles Institution
2010
11 Answer: A The internal energy of a system can be increased with external work done on the system. The internal energy is the sum of a random distribution of kinetic and potential energies associated with the molecules of the system. The potential energy of the molecules of the system does not depend on the temperature. Hence the other options are incorrect. If the system is an ideal gas, option B is incorrect because its internal energy also depends on the number of gas molecules. The internal energy of an ideal gas can be increased by an increase in the number of molecules, without a rise in temperature. Hence option C is incorrect. When two systems of ideal gases have the same internal energy, one can be at a higher temperature if it has a smaller number of gas molecules. Hence option D is incorrect.
12
1 3 m < c 2 >= kT 2 2 The two monatomic gases X and Y are in thermal equilibirum (i.e. at the same T ) ⇒ mean translational Ek of molecules of X = mean translational Ek of molecules of Y
mean translational Ek of monatomic molecules =
⇒ mean translational Ek of molecules of X = 8.0 x 10-21 J Answer: D
13 Answer: C The car has varying linear velocity - its direction changes constantly with time such that it is always tangential to its circular path. The car has varying centripetal acceleration - its direction changes constantly with time such that it is always pointing towards the centre of its circular motion. The car has a constant angular velocity because the change of its angular displacement per second is uniform.
5 (For Internal Use Only)
Year 5 Physics H2 Common Test – MCQ Answers
14
Raffles Institution
2010
The frictional force f between the road surface and the tires of the car provides the centripetal force for the car to go around a corner If fmax is the maximum frictional force for the car (of mass m ) to go round a circle of radius r , and v max its maximum speed without skidding, we have v max 2 r ∝ v max 2
fmax = m ⇒ fmax
1
( fmax ) 2 (v max ) wet wet ∴ = f v ( max )dry ( max )dry 1
(v max )wet
( fmax ) 2 wet = (v max )dry f ( max )dry
1
12 = (18 m s-1 ) 2 = 12.7 m s-1
Answer: D
15
Let N be the normal force acting on the marble at the bottom of the bowl. Apply Newton's 2nd law to the marble when it is at the bottom of bowl: v2 .......... (1) R where v is speed of the marble there.
N − mg = m
Conservation of mechanical energy: 1 mv 2 2 or v 2 = 2gR .......... (2) mgR =
Substitute (2) into (1): N =m
v2 + mg = 2mg + mg = 3mg R
Answer: C
6 (For Internal Use Only)
1
Suggested Solutions to 2010 Year 5 Common Test Section A
1 2 3 4 5
D C B C A
6 7 8 9 10
B C D A C
11 12 13 14 15
A D C D C
Section B 16
(a)
(i)
Mean value of t =
0.47 + 0.44 + 0.48 + 0.46 = 0.46 s 4
(0.46 ± 0.02) s (ii)
g=
2H 2 × 0.982 = = 9.28 m s −2 t2 0.462
∆g ∆H ∆t = +2 g H t 0.001 0.02 = + 2 0.982 0.46 = 0.08797 ∆g = 0.08797 × 9.28 = 0.8 m s −2 (1 s.f.)
g = (9.3 ± 0.8) m s−2 (b)
(i)
Systematic errors produce results that differ from the true value by a fixed positive amount (or negative amount).
(ii)
Scatter in the points about the best fit line.
(iii)
Drawing a best fit line through many points is a form of averaging the results obtained from a number of trials and finding an average minimizes random errors.
Either curve (iv)
Hx x x x x x x
0
t
2
COMMENTS: (a)(i) Many students did not score this mark because they fail to consider the fluctuation in the readings. The uncertainty should be half the range of the readings. (a)(ii) Some students did not include the fractional uncertainty of H in their calculations and were not awarded full credit. Error carried forward (ECF) from (a)(i) was allowed. Part (b) was generally poorly done. For parts (i) – (iii), many did not answer the crux of the questions and did not obtain the one mark allotted. (b)(i) Many students did not mention that the error is fixed in magnitude and will cause all readings to be higher (or lower) than the true value. (b)(ii) Many answers did not make any reference to the scatter around the best fit line but vaguely refer to a true or mean value. Also, answers such as “Not all the points lie on the straight line” seem to suggest there are anomalous results and are not awarded the mark. Answers such as “the points do not follow (or form) a straight line” is not true of the data shown. (b)(iii) Many regurgitated the following and are deemed not to have answered the crux of the question: “Random errors tend to cancel out and the residual error spread out over all the readings.” Other common mistakes are: 1. The best fit line averages the random errors or is the average of the random errors. 2. The best fit line cancels out the points on both sides of the line. 3. It is the average of a reading. (Vague as to what ‘it’ and ‘a reading’ refer to.) (b)(iv) Many students did not draw the graph on Fig. 16.2 and was not awarded the mark even when the shape of the curve is acceptable because it was not possible to judge if the initial gradient of the curve drawn is the same as that of the given line.
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17
(a)
45o
(b)
1 2 ax t " 2 4.00 = u(cos 45o )t − − − − − − − − − (1) 1 " s y = u y t + ay t 2 " 2 1 0 = u(sin 45o )t − gt 2 2 2 t = u(sin 45o ) − − − − − − − − − − − (2) g (2)int o(1) 2 4.00 = u(cos 45o )[ u(sin 45o )] g " sx = ux t +
u = 6.26 m s−1
(c)
" v y 2 = u y 2 + 2ay sy " 0 = (6.26 sin 45o )2 − 2(9.81)sy sy = 0.999m
(d)
In reality, air resistance cannot be neglected. State and explain whether the time of flight for the upward motion is longer, shorter or equal to the time of flight for the downward motion, in the presence of air resistance.
The time of flight for the upwards motion is shorter than the downwards motion. On the upward journey, the vertical component of the drag force and the gravitational force of the Earth both acts in the same direction (downwards), while on the downward journey, the vertical component of the drag force and the gravitational force of the Earth acts in the opposite direction. Hence the net retarding deceleration on the object on the way up is larger than the net acceleration on the way down, at the same vertical position. As the vertical distance travelled is the same, this means that the average speed for the upwards journey is larger than the average speed for the downwards journey. Thus the time taken for upward journey is less than the time taken for downward journey. Alternative approach
In the presence of air resistance, work will be done against air resistance throughout the whole journey. At the same vertical height, the rubber band will have greater kinetic energy on the upward journey than the downward journey, since more work is lost to air resistance on the downward journey. This means that the average speed for the upwards journey is larger than the average speed for the downwards journey. Thus the time taken for upward journey is less than the time taken for downward journey.
4
COMMENTS:
(b) All workings must be shown, and all steps clearly explained. Marks will be deducted if one u 2 sin 2θ 2u sinθ or t = without deriving them. g g (c) Several students argued that the maximum height will be reached at angle θ = 90o and subsequently calculated the maximum height from this. Note that you cannot arbitrarily change the values once it is set in the question! used the formulas s =
(d) No marks will be awarded if one states the answer without explanation, or if the explanation is incorrect. Many students correctly explained that the deceleration for the upward journey is larger than the acceleration for the downward journey, and concluded that the time taken is hence shorter on the upward journey. It must be noted that you can only make this conclusion after comparing the average speed for both journeys.
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(a)
(i) Newton’s 2nd law: the rate of change of linear momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of the force. (ii)
F=
d ( mv ) dv =m = ma for a body of constant mass, where F is in Newtons, dt dt
N (b)
(i) Mass is a measure of a body’s inertia, whereas weight is the gravitational force exerted by the Earth on it. Mass is constant and does not depend on the location of the body whereas weight can vary with location since it depends on the gravitational field strength. (ii)
N1 − mg = ma ∴ N1 = m(g + a ) = 10.0g ------------ (1) mg − N2 = 2ma ∴ N2 = mg − 2ma = m(g − 2a ) = 7.0g -------------- (2)
g + a 10 1 = , ….. solving, get a = g = 1.1 m s −2 g − 2a 7 9 Substitute back into either eqn, get m = 9.0 kg Eqn (1)/(2) , get
COMMENTS:
On the whole the question was poorly done. Not many could write the statement of Newton’s second law correctly. For the derivation of F = ma , many students did not state that its for a body of constant mass. For part (b) (i) – its not correct to say that mass is the ‘amount of matter’, or number of molecules in the body. The physics definition for mass is ‘inertia’ of the body. Weight is the gravitational force on the surface of the earth. Part (b) (ii) was also badly done because students did not understand that a reading of 10 kg on the spring balance means an apparent weight of 10g N which is equal to (10 x 9.81) N.
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(a)
The principle of conservation of momentum states that when bodies in a system interact, the total momentum remains constant, provided no net external force acts on the system.
(b)
(i)
By the principle of conservation of momentum, Total initial momentum of person & ball = Final momentum of person + Final momentum of ball 0 = mAvA + mballvball 0 = (70.0) vA + (2.0) (11.6) vA = 0.331 m s-1 to the left [direction must be clearly stated/indicated in the answer]
(ii)
Initial momentum of person B + initial momentum of ball = Total final momentum of person & ball 0 + mballvball = (mb +mball) v (2.0)(11.6) = (75.0 + 2.0) v v = 0.301 m s−1 to the right (shown)
(c)
Each time person A throws the ball towards B, the cart will move to the left. Each time person B catches the ball, the cart will stop moving. This is because the total momentum of the system must be zero, just as it was before person A threw the ball. Eventually, the cart would move towards the left side of the skating rink.
COMMENTS:
(a) ‘isolated system’ in place of ‘no net external force’ was accepted. It must be noted that in a system, there must be at least 2 bodies/objects interacting. (b) (i) Velocity is a vector quantity so students are expected to state the direction in their answers. Simple, clear answers such as ‘to the left’ or ‘to the right’ are appreciated. (c) Many students did not realise the 2 persons were now on the same cart. Hence, the cart and the 2 persons should be considered as one system. Answers were often not clear, indicating that either person A or person B moves but without mentioning motion of the cart. There were a lot of references to the answers calculated in part (b). However, this is incorrect as the total mass of the system has changed.
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20
(a)
It is the vertical upward force exerted on a body by a fluid when it is fully or partially submerged in the fluid. COMMENTS: Mark is not awarded if students miss out any of the underlined words. Many students did not write ‘force exerted by fluid’. It is important to recognise that the force comes from the fluid due to pressure difference.
(b)
For the block to be floating, Upthrust = Weight of Block
0.15 × 0.15 × h × 1000 = 0.15 × 0.15 × 0.06 × 560 h = 0.0336 m (c)
(i)
Mass of ball = additional vol. of water displaced x density of water 2
( 0.150 ) ( 0.053 − 0.0194 )(1000 ) = 0.437 kg COMMENTS:
Many students gave weight instead of mass. (ii)
ρ=
m m 0.4365 = = = 346 kg m-3 3 3 4 4 V 3π r 3 π ( 0.067 )
COMMENTS: Many students do not know the formula for volume of a sphere!!! There are also several students who used rounded answers from part (b) and (c)(i) to find the density of the ball. The final answer became quite different from the actual answer. Please remember to use more s.f. in your intermediate workings. 21
(a)
1. The vector sum of the forces is zero. 2. The vector sum of the torques due to the forces is zero.
(b)
(i)
1.
T = mg = 500 × 9.81 = 4.91 × 103 N
2.
Considering forces acting on the hook and ignoring its weight, 2T sin 30 = 4.91× 103 T = 4.91× 103 N
(ii)
1.
Take moment about the right wheel, 20000 × g × 2 = 2500 × g × 4.5 + T × 9 T = 3.13 × 104 N
2.
N × 500 × g = 3.13 × 104 N = 6.4 Hence, the maximum number of girders is 6. (Largest integer less than 6.4).
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22
(a) x = 0.969 – 0.450 = 0.240 m (b) Elastic potential energy = ½ kx2 = ½ (24.5)(0.690−0.450)2 = 0.706 J (c) KEinitial + GPEinitial + EPEinitial = KEmax + GPEfinal + EPEfinal 0 + [0.600×9.81×(0.690-0.25)] + 0 = KEmax + 0 + 0.7056 KEmax = 1.88 J
(d) When the sphere is released, its gravitational potential energy decreases and its kinetic energy increases. After it reaches 0.450 m below X, its gravitational potential energy continues to decrease and its kinetic energy and the elastic potential energy of the string increases.
COMMENTS: (c)
Many students calculated the maximum velocity instead of the required maximum kinetic energy.
(d)
Symbols or abbreviations should not be used until they have been explained or defined in the answer given.
23
(a) The two sections represent ice melting and water boiling. The temperature does not increase as the heat energy is used to break the bonds between the molecules. (b)
(i)
Q = mc ∆θ 62.7 = m(2090)(25) m = 0.0012 kg
(ii)
62.7 + 0.0012(3.33 × 105 ) + 0.0012(4200)(100 − 0) + 0.0012(2.26 × 106 ) = 3680J Value of Y = 3680
24
(a)
(i)
the amount, in mol, of gas in the sample, pV = nRT 1.0 x 105 x 5.0 x 10−3 = n x 8.31 x 300 n = 0.2006 = 0.20
(ii)
pBVB = nRTB 3.0 x 105 x 5.0 x 10−3 = 0.2006 x 8.31 x TB TB = 900 K
(iii) the volume VC of the gas at C, pCVC = nRTC 1.0 x 105 x VC = 0.2006 x 8.31 x 900
8 VC = 0.015 m3 also accept VC = 15 as correct value on the horizontal axis.
(iv) 1.
at A, UA = 3/2 NkT = 3/2 (0.2006 x 6.02 x 1023)(1.38 x 10-23)(300) = 750 J
2.
at B. UB = 3/2 (0.2006 x 6.02 x 1023)(1.38 x 10-23)(900) = 2250 J
(b) Heat supplied
Work done
to gas
on gas
Increase in internal energy
/J
/J
/J
A→B
1500
0
1500
B→C
1650
- 1650
0
C→A
- 2500
1000
- 1500
