stpm trials 2009 physics answer scheme (johor)w

8/12/2019 STPM Trials 2009 Physics Answer Scheme (Johor)w

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(b)(i)

SUGGESTED ANSWERS AND MARKING SCHEMEJOHOR STPM PHYSICS TRIAL EXAMINATION 2009PAPER 2SECTION ASuggested nswerThe Pr inciple of Conservation of Linear Momentum states that the totallinear momentum of a system is constant ( or conserved ) if there is noexternal force acting on the system .co nservation of linear momentum2,Ox 10 x 500 = 1.0 v) + 2 0x 10 x 100v 0.80 m 5.1

(b)(ii) Loss in K.E wooden block=work done against constant or averagefrictional force

2, (0)(b)

3 (o)(i)(ii)

. .(1.0)(0.8) ' = FR 0.20)2> R = 1 6 N

image is realv =0.25(20) =5.0 em1 1 1 1 1-=- - = - -J U v 20 5.0J=4 0cm

1 1 1- = n - 1 - - / T T2I 2- = (1.65 - 1-4.0 r

r l = T = 5 2cm1 1f= - =25HzT 0.04 50 f iangular frequency : w m 157=160 radsI d - G, 50 20 10-amp Itu e , 0- - , = ( )2= . x mIV 157

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(b)

4. (a)(b)

x x1020 \ _ \ \ .U 0.1220 . .. . .. ......... . .......................... .

rittleshow no plastic deformation before t breaksE= FLxA

60xO.24= -....:..:.,:.:...:..::3x104 x8x10 7

=6 .0xI0 IO Nm2

Us

5. (a) A multiplier is a resistor of very high resistance connected in series to ag l v n o m t ~ r

(b) The function of the multiplier is to reduce the incoming current down tothe maxirnum current that can be carried y the galvanometer.

c)

6. (a)

ORThe function of the multiplier is to convert the galvanometer into avoltmeter.

sdi {

V=3.0VJ= I

ft= V = 3.0 = 0.02

r Rm 30 Rm30 +R = 3.0 = 150 => R = 12000.02The multiplier is connected in series to the galvanometer.to reduce the distortion of the outputor to reduce the gain

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9 (a)

(b)

SECTION BCentripetal force ::: force that causes a body to move in a ci rcle, and itsdirection is always towards the centre of the circle .The statement is false .If the two forces balance each other, the resultant force;:: O. Accordingto Newtons first law of motion, the body will move in a straight line with

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constant velocity. That is the body will not perform circular motion. 1

(e)

(d)(i)

( i)

(e)(i)

(ii)

R

l lmg

Resultant force towards the centreof circle ;:: mg - RmvHence mg - R;::

For maximum v, R::: 0v::lrg ;F81 9 I ,,-'Newtons Law of gravitation states that the force between two massesis directly proportional to the product of the masses, and is inverselyproportional to the square of the distance between them.

The gravitational force of attraction is F=

F G Mm ., .F gR X

R 9 81x150; 4; 368 N

F = mr a

150

M = gR2; r = 2R

368 ; 150 x 2 x 6.38 x 10' ,'

G Mmr

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(iii)

10,(0)

(b)(i)

weL i I)(iii)

(c)(i)

2Period T = iJ= 1.435 x 10' s.

1 1 J 1K E ~ m v ~ m i J )2 2K.E ~ 1 5 0 ) 2 x 6.3 8x 10 ) (4.38 x 10-) ~ 2.3 4 x 10 ' J2

Heat and electric conduction in metals are both caused by the manyfree electrons that moves with high mobility.There are no free electrons in thermal and electric insulators.Thus thermal and electric insulators are poor conductors.Steady state is' achieved when the temperatures at all points along themetal rod are stable and not changing.IITemperature gradient is the difference in temperature-per unit lengthalong a conductor. ~ J,..uJI x .Qll t dC/ , kA 1Current, f - r - A ) ~ \ o(;t ~~ . ..L Kv l ' -Cl p ~where: 1 is the current in the tlowinl ln the conductor,p is the resistivity of the material conductor,

V is the potential gradient along the conductor.Irate of heat flo w in P=rate of heat flow in Q V 111 .. ;.,4kA 8-0) =kA(100-8) V, l q ; i, I I

t- ) p .,-1' : IDO - f

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ii) At H is first minimum where m=1 1(m- I )DA 1 1 ) 1.50) 600 xW )

y d = 1.0x 10-3 11

12.(a) a The stationary negative charged particle will move in the direction ofthe gravitational field. 0 eJ '11

The stationary negative cha rged particle will move oPPosite in directionto an electric field and will not move in a magnetic field. 1

(b)

(e) i)ii) _

d) i)

Upward electric force =downward gravitational forceeE mg

charge on Ne ion = 1 .6 x 10.19 C. M 20 x 10-3mass of Ne lon, m = 3 3 3x 26 kN, 6.0 10

Path of e ions

ORE

Path of Ne ions

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d)

e)

f)

14. a)

b) I)

Intensityline spectrum

continuous spectrum

wavelength

When an electron collides with a target atom, the electron willdecelerateand is stopped. The loss of aU the kinetic energy E of the electron in asing le collision with the atom means that the ray emitted has

. h f hemaxImum paton energy 0

heeV= ,Nn

A = helIua 20xlO J x1.60xlO - 9

= 6 .22 x 10 'm

A

Half-life : the time taken for the number of radioactive atoms in asample to decay to half of its initial number.

decay constant =ddl

Ra -----+ Rn + ; e(ii) Mass defect /: m = 226 .025402u - 222.017570 + 4.002603)u

= 0.005229 uTotal K.E =0.005229 x 931= 4 .88 MeV

c) I) 4 a + 9Se .to C + X 0

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ii)

iii)d)

alpha particle ; nucleus of helium: Neutron

neutron is not charged and does cause ionization

= 8qvm = q r

vAssumption: Ions of the element and ions of C 12 have the samecharge then m IX: rHence mass number A :: rA = 26.2 x 12 = 1422.4The element is nitrogen N

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