PHYSICS FIZIK

1ACE AHEAD Physics First Term Second Edition© Oxford Fajar Sdn. Bhd. 2018

MODEL PAPER SET 2

STPM960/1

Instructions:

Answer all questions in Section A. Marks will not be deducted for wrong answers. For each question, four suggested answers are given. Choose the correct answer and circle the answer.

Answer all questions in Section B. Write your answers in the spaces provided.

Answer any two questions in Section C. All essential working should be shown. For numerical answers, unit should be quoted wherever appropriate. Begin each answer on a fresh sheet of paper and arrange your answers in numerical order.

Arahan:

Jawab semua soalan dalam Bahagian A. Markah tidak akan ditolak bagi jawapan yang salah. Bagi setiap soalan, empat cadangan jawapan diberikan. Pilih jawapan yang betul dan buat bulatan pada jawapan tersebut.

Jawab semua soalan dalam Bahagian B. Tulis jawapan anda di ruang yang disediakan.

Jawab mana-mana dua soalan dalam Bahagian C. Semua jalan kerja yang sesuai hendaklah ditunjukkan. Bagi jawapan berangka, unit hendaklah dinyatakan di mana-mana yang sesuai. Mulakan setiap jawapan pada helaian kertas jawapan yang baharu dan susun jawapan anda mengikut tertib berangka.

PAPER 1 (KERTAS 1)One and a half hours

(Satu jam setengah)

PHYSICS FIZIK

MODEL PAPER SET 2

2 ACE AHEAD Physics First Term Second Edition© Oxford Fajar Sdn. Bhd. 2018

MODEL PAPER SET 2

Section A

[15 marks]

Answer all questions in this section.

1 The retarding force F on a sphere of radius R moving with a speed of v in the streamline flow of a viscous fluid is given by

F = CvR

where C is a constant. The dimension of C is A M–1 T2 L2 B M T–2 L–2 C M T–1 L–1 D M–1 T L–1

2 A ball is projected horizontally with a speed of 30 m s–1 from the top of a cliff on the surface of the Earth. What will be its speed 3 s later if there is no air resistance?A 42.03 m s–1 B 50.21 m s–1 C 61.38 m s–1 D 70.04 m s–1

3 A body starts from rest at time t = 0 and moves with constant acceleration. Which graph best represents how displacement, s of the body varies with time, t?A s

t t t tOOOO

s s sBs

t t t tOOOO

s s sCs

t t t tOOOO

s s s Ds

t t t tOOOO

s s s

4 Two masses of 6 kg and 8 kg have the same kinetic energy. What is the ratio of the momentum of the 6 kg mass to the momentum of the 8 kg mass?

A 2

3 B 2

3 C 3

2 D 3

2

5 The Mars planet moves in a circular orbit with a constant speed. The force acting on the planet isA zeroB parallel with the direction of its motionC in the direction toward the centre of the orbitD in the direction away from the centre of the orbit

6 A mass m experiences a constant frictional force F when it moves on a rough plane inclined at an angle of a to the horizontal. Another mass M is attached to it by a light inelastic string over a smooth pulley. Mass m moves up the inclined plane when mass M falls through a vertical distance of h.

F

mM

h�

How much heat is generated by friction in this process?A Mgh sin a B Mgh sin a – Fh C mgh D Fh

'13STPM


MODEL PAPER SET 2

7 A raindrop of mass m is falling vertically through the air with a steady speed v. It experiences a retarding force cv due to the air, where c is a constant. The acceleration of free fall is g. Which expression is the kinetic energy of the raindrop?

A mgc

B mg2

2c2 C m3g2

2c2 D m

3g2c2

2

8 Which of the following is true about the linear momentum and energy of a body which moves in a circle with constant angular velocity?

Linear momentum Energy

A constant decreasing

B changes constant

C constant constant

D changes increasing

9 A pendulum bob of mass m is hung from a light string. The bob then moves in a horizontal circle of radius r and the string makes an angle u with the vertical. If the angular velocity of the bob is and g is the acceleration due to gravity, which of the following is equal to tan u?

A r2

g B g

r2 C gr2 D

rg2

10 The escape velocity from the surface of the Earth is 1.1 × 104 m s–1. What is the escape velocity from a point at a height of 0.3 R from the Earth’s surface?(R = radius of the Earth)A 1.0 × 104 m s–1 B 9.9 × 103 m s–1 C 9.6 × 103 m s–1 D 8.1 × 103 m s–1

11 A body of mass m moves up a smooth inclined plane at an angle of b to the horizontal with uniform acceleration a. What is the power generated by the body when its velocity is v?A mav sin b + mgvB mavg sin bC (mav + mgv) sin bD mav + mgv sin b

12 A rubber band is stretched by an increasing force and then the force is slowly decreased. It is found that from a particular stress, the strain during unloading is always higher than that during loading. Which of the following graphs best represents the stretching and un-stretching of the rubber band?A

O

stress stress

strain strain strain strain

stress stress

O O O

B

O

stress stress


stress stress

O O O

C

O

stress stress


stress stress

O O O

D

O

stress stress


stress stress

O O O


MODEL PAPER SET 2

13 At what temperature will the rms speed of nitrogen molecules be half their rms speed at 280 K?A 50 K B 70 K C 100 K D 200 K

14 A fixed mass of an ideal gas undergoes the changes represented by PQRP in the graph below.

Q

O

RP

pressure

volume

Which of the following describe the changes?

PQ QR RP

A isothermal expansion adiabatic compression

compression at constant pressure

B isothermal expansion adiabatic compression

pressure reduction at constant volume

C adiabatic compression

isothermal expansion compression at constant pressure

D adiabatic compression

isothermal expansion pressure reduction at constant volume

15 The energy emitted by a black body of surface area A at a temperature T in a time t is directly proportional toA AT 4t B A2T 4t C A2T 2t D ATt 4

Section B

[15 marks]

Answer all questions in this section.

16 A block P of mass 2 kg slides with uniform speed down a plane inclined at an angle of 10° to the horizontal.

10°

P

(a) Show on a sketch, the forces acting on the block. [2 marks](b) What is the kinetic frictional force between the block and the plane? [2 marks](c) The block is now given an impulse so that it moves up the plane with an initial velocity

of 1.0 m s–1. (i) What is the distance travelled by the block before it stops? [2 marks] (ii) Discuss whether the block will slide down the plane again. [2 marks]


MODEL PAPER SET 2

17 A man is at a position of latitude 70° N on the Earth. Find his(a) angular velocity in radian per second, [2 marks](b) linear speed, [3 marks](c) acceleration [2 marks] due to the rotation of the Earth about its axis. (1 day = 8.6 × 104 s, radius of the Earth = 6.4 × 106 m)

Section C

[15 marks]

Answer any two questions in this section.

18 (a) (i) Define work and state the relationship between work and energy. [2 marks] (ii) Using Newton’s Second Law of motion, show that the work done on an object is

equal to the change in its kinetic energy. [3 marks](b) A motor pulling a load of mass 100.0 kg at rest

with a constant force of 1900.0 N is shown in the diagram.

pulley

load

table

motor

The frictional force on the pulley is 30.0 N and the mass of the pulley is negligible. Calculate (i) the acceleration of the load, [3 marks] (ii) the increase in kinetic energy of the load in 1.0 s, [2 marks] (iii) the work done by the motor in the first second. [3 marks]

Explain the difference between your answers in (b)(ii) and (b)(iii). [2 marks]

19 (a) State the conditions for the equilibrium of a rigid body. [2 marks](b) A spherical steel ball of mass 1.0 kg is supported by two cords fixed to a ceiling and wall is

shown in the diagram below.ceiling

cord

spherical steel ball

wall1.0 kg

50.0°

The angle between the cord and the ceiling is 50.0° while the other cord is attached horizontally to the steel ball.

(i) Sketch a triangle of forces representing the forces in equilibrium. [1 mark] (ii) Determine the tensions in the two cords. [4 marks] (iii) If the breaking tension of both cords is 40 N, calculate the maximum mass of the steel

ball which can be supported by the cords. [2 marks](c) A uniform ladder of weight 120 N is leaned against a smooth wall, and is placed on a

rough floor as shown in the diagram below.


MODEL PAPER SET 2

smooth wall

rough floor

�

The coefficient of friction between the floor and the ladder is 0.35. (i) Sketch a diagram showing all the forces acting on the ladder. [1 mark] (ii) Determine the angle u when the ladder is about to slide.

[5 marks]

20 (a) Give the definition of (i) the root mean square speed of a gas molecule, [1 mark] (ii) the degree of freedom of a diatomic gas, [1 mark]

(iii) the principle of equipartition of an energy. [1 mark](b) A tank of volume 6.00 × 10–2 m3 contains 1.5 moles of oxygen at a pressure of

3.12 × 105 Pa. Assuming that the gas behave as an ideal gas, calculate (i) the temperature of the gas, [2 marks] (ii) the root mean square speed of the gas molecules, [2 marks] (iii) the internal energy of the gas. [3 marks] (molecular mass of oxygen is 32 u)

(c) The gas in (b) is compressed to one quarter of its initial volume by an adiabatic change. (i) State the practical conditions for the adiabatic change. [2 marks] (ii) Determine the final pressure of the gas. [3 marks]


MODEL PAPER SET 2

Suggested Answers

Section A

1 C F = cvR c = FvR

[c] = – [F]

[v][R] = MLT–2

LT–1 × L = MT–1L–1

A vx = 30 m s–1

vy = u

y + at = 0 – 9.81(3) = –29.43 m s–1

v = 302 + (–29.43)2 = 42.03 m s–1

3 B s = ut + 12

at2 = (0)t + 12

gt2

s = 12

gt2 The graph is a parabola.

4 D12

(6)v21 = 1

2 (8)v2

2

v21

v22

= 43

v1

v2

= 2

3

m1v

1

m2v

2

= (m1

m2)(v

1

v2)

= ( 36/28/4/) 2/

3

= 32

5 C The centripetal force which is provided by the gravitational force acting on the Mars planet is perpendicular to the motion of the planet towards the centre of the orbit.

6 D: Heat dissipated = Work done against friction = Fh 7 C: F = ma

cv = mg

v = mgc

KE = 12

mv2 = 12

m ( mgc )2

= m3g2

2c2

8 B: Linear momentum changes because its direction changes. Energy is conserved.

9 A:

θT

mg

r

ω

T sin θ = mrw2 (1)T cos θ = mg (2)

(1)(2)

, tan θ = mrw2

mg =

rw2

g

10 C: v = 2GM

R = 1.1 × 104 m s–1

v′: = 2GM

R + 0.3R =

11.3

(1.1 × 104) = 9.6 × 103 m s–1

11 D P = Fv = (ma + mg sin θ)v 12 A


MODEL PAPER SET 2

13 B: vrms

2 ∝ TT

2

T1

= (vrms2

vrms1

)2 = ( 1

2

1 )2

T2 =

14 T

1 =

14

(280) = 70 K

14 C: Gradient of PQ > gradient QR ∴ PQ is adiabatic compression and QR is isothermal expansion. RP is compression of gas at

constant pressure.

15 A: Energy = Power × Time = sAT 4 × t = sAT 4· t ∴ Energy ∝ AT 4t

Section B 16 (a)

θθ

mg

NFfr

(b) Ffr = mg sin θ = (2)(9.81) sin 10° = 3.41 N

(c) (i) ma = –mg sin θ – Ffr

a = –g sin θ – Ffr

v2 = u2 + 2as v2 = u2 + 2(–g sin θ – F

fr)s

s = v2 – u2

–2g sin θ – Ffr

= 02 – 1.02

–2(9.81) sin 10° – 3.41 = 0.147 (ii) No, because static friction > kinetic friction = mg sin θ

17 (a) w = 2p

8.6 × 104 = 7.31 × 10–5

(b) v = rw = (6.4 × 106 cos 70°)(7.3 × 10–5) = 159.8 m s–1

(c) a = vw = (159.8)(7.31 × 10–5) = 0.012 m s–2

Section C 18 (a) (i) Work is the product of force in the direction of the displacement with the displacement of

the object.Work done on/by an object causes a change in its kinetic energy. (ii) W = F.s = mas

= ma × v2 – u2

2a = 1

2m × (v2 – u2)

= 12

mv2 – 12

mu2

(b) (i) Fnet

= ma

1900 – 30 – 1000 = 100a a = 8.7 m s–2

(ii) u = 0, t = 1 s, a = 8.7 m s–2

v = u + at = 0 + (8.7)(1) = 8.7 m s–1

12

mv2 – 12

mu2 = 12

m(v2 – u2) = 12

(100.0)(8.72 – 02)

= 3784.5 J in 1 second


MODEL PAPER SET 2

(iii) P = mgh

t =

mg(12

at2)t

= mgat2

= (100)(9.81)(8.7)(1)

2 = 4267.35 J in 1 second (iv) The difference of 4267.35 – 3784.5 = 482.85 J in second is dissipated as heat due to friction

from the pulley.

19 (a) No resultant force on the rigid body. No resultant torque on the rigid body.(b) (i)

mg

T2

T1

50°

(ii) sin 50° = mgT

1

⇒ T1 =

mgsin 50°

= (1.0)(9.81)

sin 50° = 12.81 N

cos 50° = T

2

T1

⇒ T2 = T

1 cos 50° = 12.81 cos 50°

= 8.23 N

(iii) T1 40

mg

sin 50° 40

m 40 sin 50°

g =

40 sin 50°9.81

= 3.12 kg

Maximum mass = 3.12 kg

(c) (i)

Ffr

Nw

Nf

120 N90 – θ

θ

d

p

(ii) x: Ffr = N

w ⇒ m

sN

f = N

w (1)

y: Nf = 120 N (2)

Torque about P = Nw d cos θ – 120 ( d

2 sin θ) = 0 (3)

From (1), Nw = (0.35)(120) = 42 N (4)

(4) into (3), (42) d cos θ – 60 d sin θ = 0 42 d cos θ = 60 d sin θ tan θ = 0.7 θ = 35°

20 (a) (i) The square root of the mean of the squares of the speeds of gas molecules. (ii) Number of independent ways in which a molecule can absorb, release or store energy. (iii) When a system is in thermodynamic equilibrium,

the average energy per molecule is 12

kT for each degree of freedom.(b) (i) PV = nRT

T = PVnR

= (3.12 × 105)(6.00 × 10–2)

(1.5)(8.31) = 1.5 × 103 K


MODEL PAPER SET 2

(ii) M = 1 kg100032 mol

= 0.032 kg mol–1

vrms

= 3RTM

= 3(8.31)(1.5 × 103)

0.032 = 1081 m s–1

(iii) U = f2

nRT = 52

(1.5)(8.31)(1081) = 3.37 × 104 J

(c) (i) Adiabatic change must be carried out very quickly. The wall of the container must be thick and insulated.

(ii) P1V

1g = P

2V

2g

P2 = (V1

V2)g

P1

= ( V14

V)1.40 × (3.12 × 105) = 2.17 × 106 Pa

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PHYSICS FIZIK