trialstpm2019YUHUApaper1

SMJK Yu Hua, Kajang, Selangor
STPM Trial Examination
Physics Paper 1
October 2018

Form : 6BS1 Prepared by : Pn Rosmaya Mokhtar
Date :23th October 2018 (Tuesday)
Time : 11:30 am – 1:00 pm (1 ½ h) Verified by : Pn. Dayang Setiah
(Penolong Kanan Tingkatan 6)
Number of printed pages :

Section A. (15 marks)
Answer all questions.

1. A car moves with a constant speed of 40 ms-1 towards north, and then moves with a constant speed of 30 ms-1

towards east. What is the change in velocity of the car ?

A. 10 ms-1 at 307o B. 10 ms-1 at 217o C. 50 ms-1 at 143o D. 50 ms-1 at 053o

2. An athlete doing the long jump takes off at a speed of 10 ms-1 at an angle 30o to the horizontal. How far would

he be able to jump?

A. 5.1 m B. 6.36 m C. 8.8 m D. 12.7 m

3. The acceleration-time graph (a-t) graph of a vehicle travelling along a straight line is shown below.

Which of the following is the displacement-time (s-t) graph of the vehicle?

4. A bullet of mass m moving with a velocity of u penetrates a piece of plasticine of mass M at rest. The bullet

emerges from the plasticine with a velocity v in its initial direction and the plasticine stuck to the bullet has

the same mass as the bullet. The velocity of the plasticine left behind can be expressed as

A. m(u – 2v) B. m(u – 2v) C. m(u – 2v) D. m(u – v)

M – 2m M–m M M–m

5. A spring is placed on a horizontal smooth surface. The spring has an initial length of L and a force constant

of k. One of the ends is tied to a body of mass m and the other end is fixed to a point O. The body is then

rotated about the point O at a constant angular velocity of . The expression for the radius of the circular path

travelled is

A. kL . B. k - 2 C. kL . D. kL – 

k – m2 kL m2 m2

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6. The Palapa satellite orbits around the Earth at a height h from the surface of Earth. If the radius of Earth is r
and the acceleration due to gravity at the surface of the Earth is g, the period of the motion of the satellite is

A. B. C. D.

7. The diagram shows a uniform beam of length L is supported by two vertical cords.

T1 T2

¾L ¼L

What is the ratio T1/T2 of the tensions in the cord ?

A. 2/3 B. ¾ C. 1/3 D. ½

8. A rod of original length L and cross-sectional area A is extended by e. If E is the Young modulus for the rod,

and the change in cross-sectional area can be neglected, the energy per unit volume stored by the rod is

A. Ee B. EeA C. Ee2 D. Ee2A

L2 L 2L2 2L

9. The diagram below shows the expansion process of a fixed mass of an ideal gas.

p/105 Pa

The value of the ratio initial r.m.s speed for the molecules of the gas is
final r.m.s speed

A. B. C. ¾ D. 3/8

10. A container contains a diatomic ideal gas at the absolute temperature T and the average kinetic energy of the

molecules of the gas is E. Another container contains a monoatomic ideal gas at the absolute temperature of

2T. The average kinetic energy, in terms of E for the atoms of the monoatomic gas in the second container is

A. ½ E B. 3/5 E C. 6/5 E D. 2E

11. The specific heat capacity of a gas at constant pressure differs from that of the specific heat capacity at
constant volume because
A. additional heat is needed for the expansion of the gas at constant pressure
B. additional heat is needed to increase the number of degrees of freedom of the gas at constant volume.
C. More work has to be done to overcome the stronger attractive forces between molecules at constant
pressure.
D. the internal energy of the gas at constant pressure is higher than the internal energy of the gas at constant
volume.

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12. The diagram below shows the change in the states of a gas from stage A to stage C through two paths ABC
and ADC. If the change of states follows the path ABC, 70 J of heat is absorbed and 30 J of work is done by
the gas

what is the change in the heat energy, if 40 J of work is done by the gas following path ADC ?

A. No transfer of heat C. 60 J of heat released by the gas

B. 60 J of heat absorbed by the gas D. 80 J of heat absorbed by the gas

13. Two cylinders of equal size are filled with equal amount of ideal diatomic gas at room temperature. Both the

cylinders are fitted with pistons. In cylinder A the piston is free to move, while in cylinder B the piston is

fixed. When same amount of heat is added, to cylinder A raises by 20 K, what will be the rise temperature of

gas in cylinder B?

A. 28 K B. 20 K C. 15 k D. 10K

14. Which of the following graphs best represents the relationship between the power, P radiated by a blackbody
and its thermodynamic temperature, T?

15. A wooden container of surface area 2.0 m2 and wall thickness 3.0 cm contains 30.0 kg of ice at 0oC. The

specific latent heat of fusion of ice is 3.3 x 105 Jkg-1 and thermal conductivity of wood is 0.50 Wm-1K-1. If the

room temperature is 30oC, the time required for all the ice in the container to melt is

A. 8000 s B. 9900 s C. 12500 s D. 15500 s

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Section B. (15 marks)
Answer all questions.

16. A crane cable that is capable of withstanding 22000 N is attached by a hook to a 2000 kg block that is resting
on the ground. The cable initially starts lifting the block at the maximum acceleration that the cable can
withstand for 4 s. It then continues to raise the block at constant velocity for further 2 s. At this time the block
slips off the hook at the end of the cable. Calculate
a) the tension in the cable when the block is moving at constant velocity. [2 mark]
b) the maximum acceleration that the cable can withstand. [2 marks]
c) the maximum height that the block reaches above ground. [4 marks]

17. a) A copper rod and a steel rod, each with a length of 50 cm and a uniform cross-sectional area of 4.0 cm2,
are joined end to end. The free ends of the copper rod and the steel rod are respectively maintained at a
temperature of a 100oC and 0oC whereas the sides of the rods are thermally insulated. When the steady
state is reached, calculate
i) the temperature at the joint [2 marks]
ii) the amount of heat flowing through any cross-sectional area in one minute. [3 marks]

b) Sketch a labeled graph to show the variation of temperature between the ends of the composite rod when
the steady state is reached. [2 marks]
[Thermal conductivity of copper = 380 Wm-1K-1 ; thermal conductivity of steel = 46 Wm-1K-1]

Section C. (30 marks)
Answer any two questions in this section.

18. A bead of mass 15.0 g slides without friction around a loop-the-loop as shown in the diagram below. The
bead is released from a height hand then moves in a vertical circular track of radius 25.0 cm.

(a) At the highest point of the circle, A, the reaction force acting on the bead is almost
equal to zero and the bead is just able to move in a complete circle
(i) Determine the speed of the bead at the highest point, A [3 marks]
(ii) Calculate the minimum height, h the bead must be released so that it can move in a complete circle.
[3 marks]
(iii) Calculate the reaction force acting on the bead at the lowest point of the circle. [3 marks]

(b) If the bead is released at height h = 50.0 cm,
(i) derive an expression for the speed of the bead at the position B in term of the angle θ with the vertical.
[2 marks]
(ii) Sketch a diagram to show the forces acting on the bead at position B. [1 mark]
(iii) Calculate the value of angle θ where the bead will set into a projectile motion. [3 marks]

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19. a) i) State Newton’s Universal Law of Gravitation. [2 marks]
ii) A particle X at a distance r from a particle Y experiences an attractive force F due to particle Y. Sketch
a graph to show the variation of F with r. [1 mark]

b) A two star system consists of star P and star Q respectively of mass 4.0 x 1010 kg and 2.0 x 1010 kg and
separated by a distance of 6.3 x 109 m. The star Pand the star Q move in a circular orbits with a common
centre around the stationary centre of the mass of the system.
i) Explain why the centre of mass of the system is stationary. [1 mark]
ii) Find the position of the centre of mass of the two star system. [2 marks]
iii) Draw a diagram to show the direction of the motion of star P and star Q in their respective orbits at
any instant. [2 marks]
iv) Calculate the speeds of star P and star Q. [3 marks]
v) Calculate the period of star P and period of star Q in their respective orbits. [3 marks]
vi) Discuss the motion of star P and star Q if the mass of star P is very much larger than the mass of star
Q. [1 mark]

20. a) Differentiate between internal energy and thermal energy of a system. [2 marks]
b) A gas which can be assumed to be an ideal gas in a cylinder undergoes a cycle of changes in pressure,
volume and temperature, consisting of three separate processes as shown in the diagram.

The gas is compressed adiabatically (from A to B) and then undergoes an increase in pressure (from B
to C) before it expands back to its initial state (from C to A)

i) Using the values of pressure and volume from the graph above, determine the value of  for the gas

in the cylinder. State whether the gas is monoatomic or diatomic. [3 marks]
ii) If the temperature of the gas in D during an adiabatic compression is 530 K, what is the temperature

of the gas at B? [3 marks]
iii) Calculate the work done on the gas during an adiabatic compression from A to B? [3 marks]
iv) Calculate the temperature of the gas at C. By using the first law of thermodynamics, explain how the

process from B to C can occur. [3 marks]
v) State the physical meaning of the area enclosed by the curve ABCA in the graph above. [1 mark]

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Answers for trialstpm2018paper1 :

1C 6D 11 D
At h : Temperature changes,
2C m(r+h)2 = GmM therefore internal energy
R = u2 sin2/g (r+h)2 changes
= 102 sin (2x30)/9.81 2 = GM ----1
(r+h)3 12 D
= 8.8 m At the surface: Q = U + W
mg = GmM path of ABC :
3B r2 70 = U + 30
GM = gr2 ---------2 U = 40
4B T = 2 --------3 Path of ADC :
 Q = U + 40
= 40 + 40 = 80 J
7D
Taking the moment about the 13 A
centre of the beam Q = nCpT1 -------1
T1( ½L) = T2( ¼L) Q = nCVT2 --------2

8C 14 A
Energy stored per unit volume
= ½ Fe/AL ----1
E = FL/Ae -----2

9A

momentum = momentum

before after

collision collision

mu = 2mv + (M – m)v’

5A 10 C 15 B
F = mr2 Q/t = kA d/dx ----1
E = 5/2 kT ---------1 Q = mL ---------2
kx = mr2 E’ = 3/2 k(2T) ---------2 t = (30.0)(3.3x105)
k(r – L) = mr2 21 ➔ E’ = 6/5 E (0.5)(2.0)(30-0/0.030))

16. a) T = mg = 2000 x 9.8 = 19600 N 6
b) T – mg = ma
a = 22000 – 196000 = 1.2 ms-2
2000
c) at first 4s : v = u + at
= 0 + 1.2(4) = 4.8 ms-1
s1 = ut + ½ at2
= 0 + ½ (1.2)(4)2 = 0.6(4)2 = 9.6 m

at further 2 s : s2 = ut + ½ at2
= 4.8(2) + 0 = 9.6 m

final height = s1 + s2 = 9.6 + 9.6 = 19.2 m

17. a) i) dQ = kAd

dt dx

rate of heat flow per area is constant

k1(100 – T) = k2(T – 0)

T = 100k1 = 100 x 380 = 89.2oC

k2 + k1 426

vi) Q = 380(4x10-4)(100 – 89.2)

60 5 x 10-1

Q = 197 J

b)

18. a) i) at the highest point :
mg + R = mv2/r
R = 0 ➔ mg = mv2/r

v = gr

= 9.81x0.25
= 1.57 ms-1

ii) K = ½ mv2
mg(h – 2r) = ½ mv2
gh – 2gr = ½ gr
2h – 4r = r

h = 5r/2

= 5x0.25

2

= 0.625 m / 62.5 cm

iii) At the lowest point :
mgh = ½ mv2

v = 2gh

= 2(9.81)(0.625)
= 3.50 ms-1

R – mg = mv2/r = 0.882 N
R = mg + mv2/r
= 15x10-3x9.81 + 15x10-3x3.502

0.25
b) i) h = r – r cos 

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mgh = ½ mv2
mg (r – r cos ) = ½ mv2

v2 = 2gr(1 – cos )
v = 2gr(1 – cos )

= 0.5g(1 – cos )

ii)

iii) R + mg cos  = mv2/r
When R = 0 ➔ g cos  = 2g (1 – cos )
cos  = 2/3
 = 48.2o

19. a) i) Newton’s law of gravitation states that the gravitational force of attraction between two particles of
masses M and m separated by a distance r is directly proportional to the product Mm and inversely
proportional to r2.
F = - GMm/r2
ii) .
F

r
b) i) The centre of mass remains at rest because no resultant force acts on the system. There is also no

external torque.
ii) x =  mixi

 mi
= (4.0x1010 x 0) + (2.0x1010 x 6.3x109)

4.0x1010 + 2.0x1010
= 2.1 x 109 m
iii) .

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iv) Gravitational force = centripetal force

GmM = mv2

R2 r

For star P : ➔ 6.67x10-11(mP)(2.0x1010) = mPv2

(6.3x109)2 2.1x109

v = 8.40x10-6 ms-1

For star Q : ➔ 6.67x10-11(mQ)(4.0x1010) = mQv2

(6.3x109)2 (6.3 - 2.1)x109

v = 1.68x10-5 ms-1

v) T = 2r

v
For star P ➔ T = 2(2.1 x 109)

8.40 x 10-6
= 1.57 x 1015 s

For star Q ➔ T = 2(4.2 x 109)
1.68 x 10-5

= 1.57 x 1015 s

vi) The centre of mass will be very close to O, the centre of P.
P will be rotating about its centre while Q will be orbiting with P as its centre of orbit.

20. a) i) Work is done by the gas to increase the separation between the gas molecules when the gas expands.
ii) Internal energy is the sum of kinetic energy and potential energy of the gas molecules.
Thermal energy is the energy transferred between the gas and the surrounding due to temperature
difference.
iii) Q = U + W
Q = heat supplied to gas
U = increase in internal energy
W = work done by gas

b) i) pAVA = pBVB
(VA) = pB
(VB) pA
(10x10-4) = 16x105
(1.4x10-4) 1.0x105
7.1428 = 16

 = log 16

log 7.1428
= 1.41 ➔ the gas is a diatomic gas

ii) TBVB-1 = TDVD-1

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TB = TDVD-1
VB-1

= 530(3.0x10-4)1.41-1
(1.4x10-4)1.41-1

= 530(3.0)0.41
(1.4)0.41

=724 K
iii) Work done = 1 x (pAVA – pBVB)

-1
= 1 x [(1.0x105)(10x10-4) – (16x105)(1.4x10-4)]

(1.41 – 1)
= -302 J
iv) V constant from B to C
pB = pC
TB TC
TC = 20x724.4

16.0
= 905.5 K
Using the first law of thermodynamic
Q = U + W
W = 0, therefore U = Q
Heat needs to be supplied to the gas at constant volume so that W = 0. The heat supplied increases
the pressure and the temperature.
v) ABCA represent the total amount of work done

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