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PENGGAL 2

960/2 STPM 2015

JABATAN PENDIDIKAN NEGERI KELANTAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

MARKING SCHEME

PHYSICS 2 (960/2)

Module 2

2

Jawapan : Bahagian A

1 D 9 C

2 A 10 C

3 C 11 C

4 A 12 C

5 D 13 D

6 B 14 C

7 B 15 A

8 A

16 An electron is released from rest in a uniform electric field strength of 7.5x104 V m-1. The distance between the points XY is 0.45 m.

- - - - - - - -

X

0.45 m

Y

+ + + + + + + +

(a) Find the potential difference between points X and Y.

= 7.5 x 104

V= 3.38 x104 V

[2 marks]

(b) Find the change in potential energy of the electron when it moves from point X to point Y.

U = q V

= (1.6 x 10-19

)( 3.38 x104)

= 5.4 x 10-15 J

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[2 marks]

(c) If the speed of the electron at X is 2.0 x 106 m s-1, find the speed of the electron at Y. Let Vy be speed of electron at y

K.E at y = K.E at x + Gain in K.E

( )

( )

[3 marks]

17 An electric circuit consisting of five resistors and three batteries with negligible internal resistance is shown below.

(a) Calculate the currents I1, I2 and I3.

[6 marks]

I2 = I1 +I3 I3 = I2 - I1

Loop (1): E =(IR) 8.0 + 4.0 = I1(2.0)+ I2(4.0) +I1(2.0)

12.0 = 4.0 I1 + 4.0 I2

3.0 = I1 + I2 .(1)

Loop(2) : E =(IR) 4.0 +10.0 = I2 (4.0)+ I3(3.0) + I3(3.0)

14.0 = 4 I2 + 6 I3

14.0 = 4 I2 + 6 (I2 - I3)

14.0 = 10I2 - 6I1 7.0 = -3I1 + 5I2 (2)

(1)x5: 15.0 = 5I1 + 5I2 (3) (2)-(3): -8.0 = -8 I1 I1 = 1.0 A

R1= 2.0 A R3= 3.0

I3 I1

R5= 4.0

R2= 2.0 R4= 3.0

E1= 8.0V

E2= 4.0V

E3= 10.0V

B

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Substitute I1 = 1.0 A into (1),

3.0 = 1.0 + I2

I2 = 2.0A

Substitute I1 = 1.0 A and I2 = 2.0A into

I3 = I2 - I1 = 2.0 -1.0

I3 = 1.0A

(b) Determine the potential difference VAB. [2 marks]

VAB = E2 - I2 R5

= 4.0 (2.0)(4.0) = -4.0 V

Section C

18 (a) (i) Faraday's law states that the induced electromotive force is directly proportional to the rate of change of magnetic flux.

Lenz's law states that the direction of the induced current is such that it tends to oppose the change that produced it.

OR

(2 M)

(ii) When an alternating potential difference is connected across a coil, a magnetic field is produced in the coil. // produce magnetic field

- The alternating current in the coil causes the magnetic field produced to be constantly changing.

- By Faradays law, this change of magnetic flux linkage experienced by the coil causes an induced e.m.f. in the coil. //produced induced emf

- By Lenzs law, the direction of the induced e.m.f. (back e.m.f.) is such that it tends to oppose the change that caused its production

(4 marks)

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(b)

19 (a) Definition : The magnetic flux density or magnetic field strength of a magnetic field is 1 tesla (T) if a charge of 1 coulomb moving with a velocity of 1 m s-1 at right angles to the direction of the magnetic field experiences a force of 1 newton.

(2 marks)

(b) The charged particle is stationary

The charged particle is moving in a plane that is parallel to the magnetic field lines.

(2 marks)

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(c) (i) F= (1) = (8.4 x103) (1.6x 10-19) (3.4x107) = 4.6 x 10-14 N ..(1)

(ii)

= 5.0 x 1016 m s-1 ..(1)

(iii)

( )(

)

.(1)

r = 22.9 mm ..(1) (iv)

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(d) (i)

(

)

( ) (ii) If they are electrons,

( )( ) (

)

= 5.3 x 1011

m s-1

>speed of light

Therefore the charged particle causing the aurora cannot be electrons.

(2 marks)

20 (a) (i)

Effective current for an AC current/ sinusoidal current, I = I0 sin t, which obtained by calculating time average of for the quantity I2

rms current is equivalent to DC value

the measured value of AC current (1 m)

(ii) V0 = 240 V, I0 = 0.480 V

= 500

(iii) - phasor diagram: RL

- CORRECT label/ diagram/ must have arrow

(3 marks)

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(iv) Pinst = IV .(1)

= (0.480 sin 5000t) ( (

)) .(1)

= 57.6 sin 10000t

= 57.6 W (1)

Or

( )( )

(3 marks)

(b) (i) XL = 2

= 2 ( )( ..(1)

( )( ) .(1)

( ) .(1)

(ii)

..(1)

( )( ) (1)

Energy dissipated maximum because I max. ..(1)

V

I

OR