Download physics 8866/02 - A Level Tuition

NANYANG JUNIOR COLLEGE
JC 1 PROMOTIONAL EXAMINATION
Higher 1
CANDIDATE
NAME
CLASS
TUTOR’S
NAME
PHYSICS
8866/02
Paper 2 Structured Questions
5 October 2012
1 hour 30 minutes
Candidates answer on the Question Paper.
No Additional Materials are required
READ THESE INSTRUCTIONS FIRST
Write your name and class on all the work you hand in.
Write in dark blue or black pen on both sides of the paper.
You may use a soft pencil for any diagrams, graphs or rough working.
Do not use staples, paper clips, highlighters, glue or correction fluid.
Answer all questions.
The number of marks is given in brackets [ ] at the end of each question or
part question.
For Examiner’s Use
1[8]
2[8]
3[8]
4[16]
5[20]
Total
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This document consists of 13 printed pages.
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DATA AND FORMULAE
Data
8
–1
speed of light in free space
c =
3.00 x 10 m s
permeability of free space
µo =
4π x 10 H m
permittivity of free space
εo =
8.85 x 10
–7
–1
–12
–1
Fm
–9
–1
(1/(36π)) x 10 F m
–19
elementary charge
e =
1.60 x 10
the Planck constant
h =
6.63 x 10
unified atomic mass constant
u =
1.66 x 10
rest mass of electron
me =
9.11 x 10
rest mass of proton
mp =
1.67 x 10
gravitational constant
G =
6.67 x 10
acceleration of free fall
g =
9.81 m s
s =
ut + ½ at2
C
–34
Js
–27
kg
–31
kg
–27
kg
–11
–2
Formulae
uniformly accelerated motion
work done on/by a gas
hydrostatic pressure
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
v2 =
u2 + 2as
W =
pΔV
p =
ρgh
2
N m kg
–2
3
1)(a)(i) Explain how random error may arise when using a micrometer screw gauge to
measure the diameter of a pencil.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
………………………………………………………………………………………………[1]
Suggested answers:
Non-uniformity of the pencil
Parallax error when reading the micrometer screw gauge
(ii) Suggest how this error can be minimised.
………………………………………………………………………………………………..
…………………………………………………………………………………………………
…………………………………………………………………………………………………
………………………………………………………………………………………………[2]
This random error can be reduced by taking a large number of individual measurements
and calculating the average. The average value will comprise a minimized random error
(b) To determine the volume of glass in a length of glass tubing, an experimenter obtains the
following measurements:
length L of tubing:
400 ± 5 mm
external diameter D of tubing:
15.0 ± 0.2 mm
internal diameter d of tubing:
10.0 ± 0.2 mm
The volume V of the glass is calculated using the formula
V =
(i)
1
π L( D 2 − d 2 )
4
Show that the absolute uncertainty for ∆( D 2 − d 2 ) is 10 mm2.
[2]
Solution:
∆(D2 − d 2 ) = ∆D2 + ∆d 2
 ∆D 
2
2
∆D2 =  2
 × D = 2(15.0)(0.2) = 6mm
D


2
2
∆d = 4mm
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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(ii)
Hence calculate the percentage uncertainty in the value of V using these readings.
Solution:
∆V ∆L ∆( D 2 − d 2 )
5
10
=
+
=
+ 2
= 0.09 = 9%
2
2
V
L
400 15 − 10 2
(D − d )
Percentage uncertainty = ……………………………. [2]
2
A motorcycle helmet consists of a hard outer shell which is lined with a plastic foam material
which squashes relatively slowly when the helmet strikes a hard surface such as the road or
a car.
(a) Explain why the process of squashing 'relatively slowly' is helpful in protecting a
motorcyclist wearing the helmet.
………………………………………………………………………………………………………
Squashing
'relatively slowly' increases the time for the motorcyclist head to come to
a stop.
………………………………………………………………………………………………………
This longer time reduces the impact force which is change in momentum
………………………………………………………………………………………………………
divided
by time for the momentum change. A smaller force will cause less injury to
………………………………………………………………………………………………………..
the
motorcyclist.
……………………………………………………………………………………………………[2]
(b) In one safety test for a motorcycle helmet, the helmet is loaded with a dummy head and
dropped so that it reaches a speed of 6.4 m s-1 before hitting a flat metal block. The total
mass of the dummy head and helmet in one test is 5.0 kg and it rebounds at a speed of
3.1 m s-1. The time of contact between the helmet and the metal block is approximately
4.0 ms.
(i) Calculate the magnitude of the change in momentum of the helmet.
Change in momentum = mv - mu = (5.0 x( - 3.1)) - (5.0 x( 6.4))
= 5.0 x (- 9.5 ) = - 47.5 kg m s-1
Magnitude of the change in momentum = 47.5 kg m s-1
change in momentum = ……………………… kg m s-1 [2]
(ii) The helmet 'passes' the safety test if the instrumentation connected to the dummy head
registers an average deceleration of less than 300 g where g is the gravitational
acceleration. Deduce with justification whether or not this helmet passes the test.
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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Decelerating force = Change in momentum / time = 47.5 / 4 x 10-3
= 11880 N = m a
Decelerating acceleration a = F / m = 11880 / 5.0 = 2375 m s-2
= 242 g < 300 g . It passed the test!!
[2]
(iii) The efficiency of the plastic foam material is defined as the percentage of the kinetic energy
that is dissipated during the test. Calculate the efficiency of the material.
Efficiency = (KE dissipated / initial KE) x 100
= [( ½ m v2 - ½ m u2 )/ ½ m u2 ] x 100 = [(v2 - u2 )/ u2 ] x 100
= [ | ( 3.12 - 6.42 ) | / 6.42 ] x 100 = (31.35 / 40.96) x 100
= 77 %
efficiency = …………………… % [2]
3
A ball of mass 2.0 kg is dropped from a height 1.5 m above a spring with spring constant
25 N m -1. The subsequent positions of the ball and spring are as shown in Fig 3.
1.5 m
x
At start of
release of
ball
At start of
compression
of spring
y
At equilibrium
compression of
spring
At the lowest
position
Fig. 3
(a) Show that the compression of the spring when the ball is in equilibrium is 0.78 m.
[2]
mg = kx
mg
x=
k
2.0(9.81)
=
25
= 0.78 m
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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(b) By applying the principle of conservation of energy, calculate the speed of the ball
when it is at its equilibrium position.
[4]
Loss in GPE = Gain in KE + Gain IN EPE
1
1
mv 2 + kx 2
2
2
1
1
mv 2 = mg ( h + x ) − kx 2
2
2
= 37.1
mg ( h + x ) =
v = 6.1 m s-1
(c) Calculate y the compression of the spring when the ball is at its lowest position.
↓ GPE =↑ EPE
1
mg (h + y ) = ky 2
2
2mg (h + y ) = ky 2
ky 2 − 2mgy − 2mgh = 0
y = −0.94 m (reject) or
y = 2.5 m
(d) State an assumption made in your calculation.
No air resistance.
………………………………………………………………………………………………..
…………………………………………………………………………………………………
………………………………………………………………………………………………[1]
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
[2]
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4
In an attempt to verify the relationship d = ½ a t 2, a student sets up the apparatus as shown
in Fig 4.1 where a is the linear acceleration.
d
Fig 4.1
(a) Define acceleration.
Acceleration is the rate of change of velocity (with respect to time).
[1]
The time taken for the car to move from rest through a distance d is t. The values of d and t
obtained are shown in Table 4.2.
d / cm
t 2 / s2
t/s
1
100
3.1
9.6
2
80
2.9
8.4
3
60
2.5
6.0
4
40
2.2
4.8
5
20
1.4
2.0
Table 4.2
(b) Complete the table by filling in the missing t2 values.
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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(c) On the grid below, plot the 2 missing points for the graph of d vs t2 using the values
in Table 4.2 and draw a best fit straight line.
[3]
d / cm
100
80
60
40
20
t2/s2
0
(d) i)
2
4
6
Fig 4.3
8
10
Calculate the gradient of the best fit straight line obtained in Fig 4.3.
The coordinates are (0.6,0), ( 8.0, 78 )
Gradient of the graph = (78 - 0) / (8.0-0.6) = 10.5
gradient of the graph = ………………….….[2]
(ii) By making use of the relationship d = ½ a t2 and your answer in (d)(i), deduce the
acceleration of the toy car as it travels down the slope.
For d = ½ a t2
Gradient = ½ a = 10.5
a = 2(10.5) = 21.0 cm s-2 = 0.210 m s-2
acceleration = ………………….….m s-2 [2]
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(e)
The values of t and d are measured using a stopwatch and a metre rule respectively.
For each of the following types of experimental error, describe the feature of the
graph in Fig 4.3 that indicates its presence and suggest a possible source of that
error in the experiment.
i) random error
Feature : The readings are scattered above and below the best fit line.
Possible source: The reaction time of the student is not consistent, resulting in
readings that is an underestimate or overestimate from the true value.
[2]
ii)
systematic error
Feature: The graph did not cut the origin.
Possible source: The ruler is placed incorrectly such the measured distance d is
always an overestimate from the true value.
[2]
(f)
The experiment is repeated with another toy car of identical shape and size but made
of denser material. Deduce and explain how the acceleration of this toy car will differ,
if at all, from the answer in (d)(ii).
The acceleration will remain constant since the denser material will increase the
mass, but acceleration is independent of the acceleration, a larger mass will
experience a larger force but the same acceleration.
[2]
5
(a)
(i)
Define the moment of a force.
The
moment of a force about a point is the product of the force and
…………………………………………………………………………………………..…
the perpendicular distance of from the line of action of the force to the
point.
……………………………………………………………………………………… [1]
(ii)
State the two conditions necessary for a body to be in static equilibrium.
Resultant force acting on a body is zero.
…………………………………………………………………………………………..…
Resultant torque acting on a body is zero.
…………………………………………………………………………………………..…
…………………………………………………………………………………………..…
………………………………………………………………………………………
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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Fig 5.1 shows a disc and spring system which is mounted on a wall. It is used to move
two cargo boxes of equal weight towards each other on a rough floor by applying a force
at the wrench to turn the disc.
disc
Box A
wrench
Box B
1.5 m
Rough floor
Fig 5.1
(b)
Fig 5.2 shows two of the forces acting on the box A when it is about to slide.
(i)
Complete the free body diagram by drawing two additional forces acting on Box A
on Fig 2.2. Label the forces clearly.
[2]
2.0 m
Normal
reaction force
Box A
x
0.80 m
Fig 5.2
Weight of Box A
CARGO man
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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(ii)
Box A has a mass of 500 kg. The maximum static frictional force between Box A
and the rough floor is 2200 N. Calculate the tension in the spring when Box A
begins to slide.
ΣF = 0
T–f=0
When the block is about to slide, f = max limiting value.
T = 2 200 N.
minimum tension =
(iii)
N [1]
Calculate x, the separation between the weight of Box A and the normal reaction
force.
ΣT = 0
2 200(1.5) – 500(9.81)(x) = 0
x = 0.67 m.
separation, x =
(iv)
m [2]
Box B is identical to Box A except that it is attached to the spring and disc system at
a higher point. Deduce and explain whether the separation between the weight of
Box B and its normal reaction force is smaller, equal or greater than x.
…………………………………………………………………………………………..…
Separation will be greater than x since point of connection is higher
resulting in a greater torque.
…………………………………………………………………………………………..…
…………………………………………………………………………………………..…
………………………………………………………………………………………
(v)
[2]
The system may not be able to move the boxes keeping it upright if it is attached to
the boxes too high above the ground. By considering the rotational equilibrium of
the boxes, suggest why this is so.
…………………………………………………………………………………………..…
If the
connection point is too high, the perpendicular distance between T
and f will be large. This will results in a greater toppling moment/torque.
The …………………………………………………………………………………………..…
block may topple before it start to slide.
………………………………………………………………………………………
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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(c)
The disc is free to rotate about its axle which is normal to its plane and passes through
the centre C of the disc, as shown in Fig 5.3.
disc
20 cm
C
40 cm
wrench
θ
Fig 5.3
F
A wrench of length 40 cm is attached to the disc. When a force F is applied at right
angles to the lever at its end, equal forces are produced in the two springs.
As the wrench is turned from its initial position, the tension in each spring increases from
zero. When the wrench has been turned through an angle of θ, the boxes begin to slide.
(i)
The spring constant of each spring is 2.0 × 104 N m-1. Using your answer in (b)(ii),
calculate the extension in each spring when the boxes begin to slide.
F = ke
2200 = 2000 x e
e = 0.11 m
extension =
(ii)
Hence calculate the angle θ.
e=rθ
(0.11)= 0.1(θ)
θ = 1.1 rad
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
m [1]
13
angle, θ =
(iii)
rad [2]
Calculate the torque produce by the forces in the springs.
T = 2200( 0.20) = 440 Nm
torque produced =
(iv)
N m[1]
Hence calculate the magnitude of F, the force applied to the wrench.
F(0.4) – 440 = 0
F = 1100 N
magnitude of F =
(d)
(i)
N [1]
As the boxes move towards each other at a slow constant speed, the kinetic
frictional force acting on each cargo is 1500 N. By using the principle of
conservation of energy, calculate the energy required to turn the disc by one
revolution to maintain the motion of the boxes.
Energy supplied = work done against friction
=2 x 1500 x( 2πr)
= 1880 J
energy required =
(ii)
J [2]
The system is modified to have a longer wrench so that the applied force
required to move the boxes at the same speed as in (i) may be reduced.
Deduce with explanation how this modification affects, if at all, the energy
required to move the boxes by the same distance.
…………………………………………………………………………………………..…
Energy required will remain the same since the work done against
………………………………………………………………………………………
friction
is the same.
2012 NYJC JC1 Promo Exam – 8866 H1 Physics
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