confidential960/2Physics2 hours 30 minutes

SEK. MEN. KEB. TINGGI MELAKA(Malacca High School. Estd.1826)Ke Arah Kecemerlangan Pendidikan

Lower Six End of Year Examination2010

PHYSICSPAPER 2 MARKING SCHEME

STRUCTURE AND ESSAY (2 hours and 30 minutes)

Instructions to candidates:

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO. Answer all questions in Section A. Write your answers in the spaces provided. All

working should be shown. For calculati ons, relevant values of constants in the Data

Booklet should be used. For numerical answers, units should be quoted wherever they

are appropriate.

Answer any four questions only in Section B. Write your answers on the answer

sheets. Begin each answer on a fresh sheet of paper and arrange your answers in

numerical order. Tie your answer sheets to this question paper.

A Data Booklet is provided.

Prepared by : Verified by :

___________________ ___________________

Wee Choi Chiang Azmi Bin Sakmis

Ketua Panitia Fizik GPK Tingkatan 6

This question paper consists of 1 3 printed pages.

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Section A : Answer all questions in this section in the question paper.1. Quantum theory states that light consists of discrete packages of energy called photons.

The energy of each photon, E depends on the speed of light in vacuum, c , thewavelength of the light , λ, and the Planck’s constant, h. The value of Planck’s constant is6.63x10-34 J s.(a) Determine the dimensions of Planck’s constant? [2]

Unit of h = J s = Nm s = kg m s-2 m s= kg m2 s-1

[ h ] = M L2 T-1

(b) By using dimensional analysis, derive an expression for E in terms of h, c and λ.[3]

Assume E = k hx cy λz

M L2 T-2 = (M L2 T-1)x (L T-1)y Lz

M L2 T-2 = Mx L2x + y + z T-x-y

M : x = 1T : -x – y = -2 y = -1 + 2 = 1L : 2x +y + z = 2

z = 2-2-1 = -1

hckE or

hcE

(c) Calculate the energy of one photon of a monochromatic light which has a wavelengthof 500 nm.(Assume constant = 1) [1]

JE 199

834

10978.310500

1031063.6

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2. A ball is projected horizontally at 15.0 m s-1 from a point 20.0 m above a horizontalsurface. By taking g = 10.0 m s -2;(a) find the time taken for the ball to reach the horizontal surface. [2]

y = ut + ½ g t2

20 = 0 + ½ (10) t2

t = 2.00 s.

(b) calculate the speed of the ball when it hits the horizontal surface. [3]

Vx = 15.0 m s-1

Vy = u + g t = 0 + 10x2 = 20 m s -1

V2 = Vx2 + Vy

2

V = 25.0 m s-1

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3. The figure below shows a man jumps from boat A to boat B. The two boats ar e stationary initially;they are identical of mass 250 kg. The mass of the man is 50 kg and the speed of his jump is 1 .00ms-1.

Find the velocity of the boat A ( VA ) and boat B ( VB ) after the man jump. [4]

For boat A:MV + mv = 0

250 V + 50x1 = 0

V = - 0.200 m s-1

For Boat B :mv = (m+M)V50x1 = (50+250)V V = 0.167 m s-1

4. A block of mass 20.0 kg is placed on a rough inclined plane at an angle 30o to the horizontal.A man pulls the block with a force F parallel to the inclined plane so that it moves up theplane at a constant speed of 0.400 m s-1. If the coefficient of kinetic friction of the plane is0.500,

(a) calculate the friction acting on the block. [3]

R = mg cosθf = μ R = μ mg cosθ = 0.500x20x9.81x cos30 o

= 84.96 N or 85 N(b) calculate the power of the man. [2]

P = Fv = (mg sinθ + f) v = (20x9.81xsin30o+ 84.96)x0.400 = 73.2 w

300

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5. The diagram shows four horizontal forces acting at a poin t P.

Given that the forces are in equilibrium, calculate the value of T and the size of theangle θ. [4]

NorNTor

TTF

TTF

o

oy

x

110106761.76

028.4tan4.102sin

060sin5080sin60sin42.25cos

0cos60cos5080cos6040

0

0

00

6. (a) Explain what is meant by gravitational field strength. [1]

Gravitational field strength is the gr avitational force per unit mass. 1

(b) The mass of the Moon is 7.50x10 22 kg and its radius 1730 km. Calculate thegravitational field strength on the Moon’s surface. [2]

11

23

2211

2

67.1671.1)101730(

1050.71067.6

kgNorkgN

RGMg

(c) If an athlete can jump a height of 1.50 m on the Earth ’s surface, what is the height he canjump on the Moon’s surface? [2]

Assume work done on Earth = work done on Moon mgh = m g’ h’ 9.81x1.50 = 1.67 h’ h' = 8.81 m

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7. The diagram below shows a toy car which has a flywheel of moment of inertia 0.75 kgm2 attached to the axle of its rear wheels. The flywheel is now accelerated to rotate at2.00 revolution per second and the toy car is allowed to move on a table.

(a) Calculate the kinetic energy of th e flywheel. [3]

J

f

IE

22.59

)2(75.02121

2

2

(b) If the effective decelerating force experienced by the car is 20 N, what is thedistance travelled by the car before it stops? [2]

W=F S59.22 = 20 SS = 2.961 m or 2.96 m or 3.0 m

8. The diagram below shows a disc X of moment of inertia 2.00 kg m 2 rotates freely withan initial angular velocity of 2.00 rad s-1 about a vertical axis passing through its centerand disc Y of moment of inertia 0.500 kg m 2 is held above it. After Y is dropped ontoX, both discs rotates together about the same axis.

(a) Calculate the initial angular momentum of disc X. [2]

L = I w = 2 x 2 = 4.00 kg m s s-1

(b) Calculate the final angular velocity of the system. [2]

L’ = L2.5 w’ = 4.00 w' = 1.60 rad s -1

(c) Calculate the kinetic energy lost by the system. [2]Kinetic energy = ½ I w2

Kinetic energy loss = ½ I w 2 – ½ I’w’2

= ½ (2) (2)2 – ½ (2.5) (1.60)2

= 0.800 J

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Section B: Answer any four questions from this section.

9. (a) State the Newton’s Laws of motion. [3]Newton’s First law of motion: An object will remains stationary or moving in straight line with constant velocity ifthere is no external force acting on it.Second law:The rate of change of momentum is directly proportional to the resultant force andoccurs in the direction of the force.Third law:Action and reaction are equal in magnitude but opposite in direction. 1, 1,1

(b) The diagram below shows a conveyor belt used to transport sand in a factory. When no sand is transported by the belt, a force of 100 N is supplied by the motor to

move the belt at constant speed of 2.00 m s -1.

(i) Calculate the power supplied by the motor to overcome the frictional force.[2]

P = F v= 100x2 = 200 w.

(ii) When sand is loaded on the belt at a rate of 20.0 kg s -1, explain why the motormust supply additional force to maintain the speed of the belt. Calculate theadditional force supplied by the motor. [3]

Additional force is required to in crease the momentum /to accelerate the sand/ to overcome the friction acting on the belt by the sand.

N

vdtdmF

0.40220

(iii) Calculate the additional power supplied by the motor. [2]

Additional power = ΔF v = 40x2

= 80.0 w

(iv) Explain the motion of the sand after it has fallen on the conveyor belt.[2]

The sand will first slide on the conveyor belt. 1 After a short while it will acquire a constant veloci ty same as the belt. 1

2.00 m s-1

sand

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(v) Calculate the energy lost per second due to friction between the conveyor belt andthe sand. [3]

Rate of increase in kinetic energy of the sand

w

vdtdm

40

2202121

2

2

Energy lost per second = 80 – 40 = 40.0 w.

10. (a) (i) Explain why a centripetal force is necessary to keep an object moving round acircle with constant speed. [2]

For an object moving in circle, its direction of motion is changing . 1From Newton’s first law, an object is either stationary or moving in a straight linewith constant velocity if there is no external force. Thus an external force must beacting to change its direction. 1

Or The object experiences an acc eleration since its direction of motion ischanging. 1 From Newton’s Second law, F = ma, there must be a resultant force acting on theobject. 1

(ii) State the direction of the force. [1]Towards the centre of the circle. 1

(b) A motor bike can turn round a bend on a level road. Explain why the bike moving at anexceedingly high speed will easily cause accidents. [2]

The static friction between wheel and the road surface supplies the centripetal force/If the bike is moving at high speed, it requires a very large centripetal force. 1 If the centripetal force is greater than the limiting friction, skidding will occur. 1

(c) A small marble of mass 5 .00 g and negligible size is spinning inside a smoo th conewhose sides slope at an angle of 60o to the horizontal as shown in the Figure below. Themarble is rotated steadily in a horizontal path at a height h = 5.00 cm above the vertex of thecone.

R

30o

mg

1

1

1

8

(i) What is the radius of the circular path of the marble? [1]

cmorcmr 9.2887.260tan5

0 1

(ii) Find the net force acting on the marble. [2]

NRforceNet

NmgR

mgR o

0850.030cos

0981.030sin

81.910530sin

30sin

0

0

3

0

(iii) Hence, determine the speed of the marble and the period of revolution. [3]

sorsv

rT

smv

v

mrForr

mvF

26.0259.0

2701.0

10887.21050850.0

1

2

23

22

(iv) If the path is at a height h = 7 .00 cm above the vertex of the cone instead, what is theincrease in the kinetic energy of the sphere? [3]

cmr 041.460tan7

0

JmvmvKEinIncrease

smv

vr

mvF

422

1

2

23

2

1089.421'

21

8288.0'10041.4

'1050850.0

'

(v) If the sphere in Fig 3.1 slowly loses energy due t o friction, state what happens to the radius of path. [1]

Radius decreases. 1

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11. (a) Define Newton’s law of Universal Gravitation. [1]The gravitational attraction between two point masses is directly proportional to theproduct of the masses but inversely proportional to the square of their separation. 1

(b) Derive a relationship between the period, T of a planet and its orbital radius, r.[3]

32

32

2

22

4

2

rTor

rGM

T

T

rmGMmr

s

s

(c) A binary stars system c onsists of star X of mass 4.00x10 30 kg and star Y of mass1.00x1030 kg separated at distance 4.00x10 11 m apart. Both stars are revolving around thecentre of mass of the system.(i) Explain why both the stars are revolving around their centre of mass. [2]

There is no external force acting on the system. 1Therefore the centre of mass must be stationary. 1

(ii) Calculate the centre of mass of the system from star X. [2]

m

mxm

xi

ii

10

30

113030

1000.8105

1041010104

(iii) Calculate the gravitational force between t he stars. [2]

N

rGMmF

27

211

303011

2

10668.1)104(

1011041067.6

(iv) Calculate the period of each star. [3]

For star X :

yearsorsT

xMF

76.21070.821022.7

10810410668.1

7

8

2103027

2

(v) Discuss the motion of the stars if one of the stars has very large mass compare to theother. [2]

The centre of mass is near the centre of the larger star. 1Therefore the smaller star will revolve around the larger star. 1

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12. (a) Define what is meant by moment of inertia of a rigid body. [1]

Moment of inertia is given by:

2ii rmI

where mi = mass of particle-i , ri = distance of particle from the axis of rotation.OrMoment of inertia of a rigid body about a given axis is the sum of products of masses ofits particles and the square of their respective distances from the axis of rotation.Or Moment of inertia is a measure of a body's resistance to rotational motion.Or Moment of inertia of a body rotating about an axis is the property or tendencyto oppose the change in its state of uniform rotation.

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(b) State two factors which affect the moment of inertia of a rigid body. [2]

1. The mass of the body. 12. The distribution of mass/matter from the axis of rotation/ shape of the body. 1

(c) The diagram below shows a thin bike wheel of radius 0.200 m and mass 2.0 kgspinning at 5.00 revs/second. A block is pushed against the top of the wheel with a forceN of 10 N. The coefficient of kinetic friction between the block and the wheel is 0.900.

(i) Calculate the moment of inertia of the wheel. [1] I = M R2

= 2x0.22 = 0.0800 kg m2 1

(ii) What is the angular momentum of the wheel before the block is applied? [2]

L = I w = 0.0800x2πx5 = 2.513 kg m2 s-1

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(iii) What torque does the block exert on the wheel? [3]

f = μ R = 0.9x10 = 9.00 N

τ = f R = 9x0.2 = 1.80 Nm

(iv) How long does it take for the wheel to stop? [2]

τ = I α- 1.80 = 0.08 α α = - 22.5 rad s-2

ω = ωo + α t0 = 10 π – 22.5 tt = 1.396 s or 1.4 s

(v) Calculate the number of revolution made by the wheel before it stops. [2]

ω2 = ωo2 + 2 α θ

0 = (10π )2 – 2 (22.5) θ θ = 21.93 rad.

Number of revolution

5.349.32

or

(d) If the ring wheel is replaced by a disc wheel of same mass and radius, discuss whether itwill stop in a shorter time interval. [2]

(Moment of inertia of a ring wheel I = MR 2)

A disc wheel has smaller moment of inertia because the mass is distributed closer to axisof rotation. 1Therefore it will stop in a shorter time interval. 1

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13. (a) State the conditions that a rigid body acted by a few coplanar forces are in staticequilibrium. [2]

1. The resultant force acting on the body must be zero. 12. The resultant torque on the body must be zero. 1

(b) The diagram below shows a uniform rod of length L being acted by two forces of equalmagnitude, F but in opposite directions at its ends.

(i) What is the resultant force acting on the rod? [1]

The resultant force = 0 1

(ii) Deduce an expression for the resultant torque ac ting on the rod. [2]The resultant torque = F (L/2) + F (L/2) 1

= FL 1

(iii) Explain whether the rod is in static equilibrium. [2]The rod is not in static equilibrium 1Because the resultant torque is not equal to zero. 1

(c) The diagram below shows a store sign of weight 50.0 N is hung from a uniform bar PQof weight 20.0 N and length 100 cm. The sign is suspended from a point 75.0 cm of theway from the wall. The bar is held up with a cable of length 2.00 m.

L

75 cm P Q

2.00m

θ

13

(i) Calculate the tension T in the cable? [3]

Cos θ = ½ ; θ = 60o

Consider torque about P:20x0.5 + 50x0.75 = T x 1 sin θT = 54.85 N or 55 N.

(ii) Calculate the magnitude of the reaction force at P. [5]

Consider the horizontal force :Rx = T cos 60o

= 54.85 cos60o

= 27.43 NConsider the vertical forces:

Ry + T sin60o = 20 + 50

Ry = 70 – 54.85 sin60o

= 22.50 N

R2 = Rx2 + Ry

2

R = 35.48 N or 35 N

******************************* END *****************************************

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