4

Click here to load reader

Mass of average person ~ 60 kg -1 KE = ½ m v 2 = ½ (60) (1 ...a-leveltuition.com/.../2011/12/2011-CJC-PH-H2-P1-Prelim-ans.pdf1 CJC Prelim Paper 1 Solutions 1 B Mass of average person

Embed Size (px)

Citation preview

Page 1: Mass of average person ~ 60 kg -1 KE = ½ m v 2 = ½ (60) (1 ...a-leveltuition.com/.../2011/12/2011-CJC-PH-H2-P1-Prelim-ans.pdf1 CJC Prelim Paper 1 Solutions 1 B Mass of average person

1 CJC Prelim Paper 1 Solutions

1 B

Mass of average person ~ 60 kg Normal walking speed ~ 1 – 2 m s-1

KE = ½ m v2 = ½ (60) (1.5)2 = 68 J

2 C

( ) kgM

M

M

M

G

FrM

Moon

Moon

Moon

Moon

Moon

22

22

222

108.02.7

10883.0

7.1

1.02

699.1

004.0

10361.7)1(

×±=∴

×=∆

+=

×==

3 D 4 B

14.57

)5.3(120

2

2

2

2

22

=

−−=

−=

+=

s

s

asu

asuv

5 A

6 B s

F

dpdt

dt

dpF 5.0

100

)25(2===⇒=

7 B

mk

Fx

kxF

31035.7)6000000(2

)81.9(9000

2

−×===

=

8 D Change in momentum = 116)4(8

2

1 −= kgms

Thus change in vel = 16 and hence final speed = 16 + 4 = 20 ms-1 9 B Solution:

Loss of GPE = Gain in KE of the car mg (∆h) = ∆KE Since the same car is used for the 2 scenarios, the final KE of the car at the end of the slope depends only on the height of release.

10 C Solution: a = - ω2r ω = √(120/0.030) = 63.25 rads-1

v = r ω = 0.050(63.25) = 3.16 ms-1

11 A Solution: Taking centripetal motion, T – mg cos θ = mac T = mac + mg cos θ

a

t

Page 2: Mass of average person ~ 60 kg -1 KE = ½ m v 2 = ½ (60) (1 ...a-leveltuition.com/.../2011/12/2011-CJC-PH-H2-P1-Prelim-ans.pdf1 CJC Prelim Paper 1 Solutions 1 B Mass of average person

2 12 B Solution:

Originally, E = (1/2)mωωωω2A

2 With the frequency doubled and the amplitude halved,

E’ = (1/2)m(2ωωωω)2(A/2)2 = E

13 D Solution:

a = - ωωωω2x

The acceleration is in anti-phase with the displacement. 14

D During light damping, the amplitude of the oscillation decreases with time while the period of the oscillation stays constant.

15 C

Solution: Distance between two adjacent nodes is half a wavelength. Therefore 5 nodes will be equivalent to 2 wavelength between them.

16

C

Solution:

πππλ

nv

nv

Hencendifferencepath

21400

2

1000

4.1;2 =⇒==

280114001680

=⇒=− vvv

m s-1

17

C

Solution: v = fλ = λ/T in a time 0.75 T, the wave would have traveled v(0.75T) = 0.75 λ

Therefore it point X would be at the lowest displacement (-A) and at 0 ms-1

. 18

B

Let the final temperature be T

TT

T

TT

TT

2.308

5.4574814099400

12600000420005.37481499400

)300)(4200(10)400)(2499(5.1

=

=

−=−

−=−

19

A

204.1

302

3

−=∆

−=∆

T

TNk

20 B )( WQU −+=∆

in first case, Q =0 in second case, ∆U = 0, thus Q is positive and heat is gained.

21 C From Fig. 1.1, When current = 5 mA, p.d. = 0.8 V (voltage across diode) Voltage across 50 Ω resistor = (5 mA)(50 Ω) = 0.25 V Total voltage across supply = 0.8 + 0.25 = 1.05 V

22 B

23 A Let r be resistance of R and S; then 0.5r is the resistance of R and S. When bulb S blows, total circuit resistance increases from r/3 to r/2 total circuit current decreases from Io to (2/3)Io . Since current through each bulb unchanged, bulb brightness unchanged.

24 B Vo = 400/2 = 200 V Voltage across the diode:

Page 3: Mass of average person ~ 60 kg -1 KE = ½ m v 2 = ½ (60) (1 ...a-leveltuition.com/.../2011/12/2011-CJC-PH-H2-P1-Prelim-ans.pdf1 CJC Prelim Paper 1 Solutions 1 B Mass of average person

3

Vrms across diode = √ (Vo

2/4) = √ 2002 / 4 = 100 V 25 C For sinusoidal a.c., W = ½ Po = Vo

2/(2R) For the square wave a.c.,

Vrms = √ (Vo2/2) = Vo/√2

<P> = Vrms2 / R = (Vo/√2 )2 / R = W

26 D Ans: D

Induced E is directly proportional to dB/dt, which is proportional to dI/dt. dI/dt is the gradient of the given graph. Consider one cycle: For the first ¼ of the cycle, gradient becomes steeper, hence induced E increases. For the second ¼ of the cycle, gradient becomes gentler, hence induced E decreases. For the third ¼ of the cycle, gradient becomes steeper but is now negative, hence induced E increases in magnitude but reverse in direction. For the fourth ¼ of the cycle, gradient becomes gentler, hence induced E decreases in magnitude.

27 C The liquid level changes due to the magnetic force acting. The magnetic force balances the extra height of liquid. Hence, Weight of extra height of liquid = magnetic force ρVg = BIL (900)(1.2 x 10-4 h)(9.81) = (0.20)(5.0)(√ 1.2 x 10-4) h = 0.0103 m = 1.03 cm

28 B F = BQv sinΦ Where Φ: angle between B and v, which in this case us 90 Hence, F = BQv

29 B This is a typical example of SHM since g α r and g is always directed towards the centre of the Earth (the equilibrium position where g = 0). If we drop a stone through the hole it would accelerate and increase its velocity of fall until the acceleration reaches zero as it passes through the Earth's center. The velocity on the other hand, starts at zero and reaches a maximum as it goes through the Earth's center after which deceleration sets in bringing the stone to a stop as it reaches the other pole. Time taken to reach opposite pole = half a period (T/2) amax = gsurface = 9.81 m s-2 For SHM, amax = ω2xo = ω2 RE 9.81 = (2π/T)2 (6.4 x 106) T/2 = 2500 s (2 s.f.)

30 C Direction of E doesn’t change, only magnitude changes. E is greatest close to the charge and minimum at midpoint.

31 B Electron experiences a constant force opposite to the electric field lines.

32 A Force on electron, F = qE (to the left)

Page 4: Mass of average person ~ 60 kg -1 KE = ½ m v 2 = ½ (60) (1 ...a-leveltuition.com/.../2011/12/2011-CJC-PH-H2-P1-Prelim-ans.pdf1 CJC Prelim Paper 1 Solutions 1 B Mass of average person

4 Work done by electric force on electron = - Fx When the electron moves to the right, it gains EPE. The amount of EPE gained is equal to the work done by the electric force (Fx), where x is the distance moved in the direction of the field lines.

33 D

34 A 3 possible transitions as a photon can go from -1.51 -> - 3.39 -> -13.6 -1.51 -> -13.6

35 A Even though electrons from the Valence band require a lot less energy to get into the dopant hole energy level, it still requires some energy, and at absolute zero, there is not even that amount of energy to spare.

36 C E = (-1.5 – (-3.5)) x 1.6 x 10-19 = 3.2 x 10-19

E = hf = hc / λλλλ λλλλ = hc / E = 6.63 x 10-34 x 3 x 108 / 3.2 x 10-19 = 6 x 10-7 m visible light 37 B A is incorrect : ons occurs for frequency lower than the threshold frequency

C is incorrect: The velocity of the emitted electrons is dependent on the photo energy and work function of metal surface D is incorrect: The number of electrons emitted per second is dependent of the intensity of the incident radiation

38 B The two peaks are the ones that are dependent. The wavelengths depend on the energy of the electron and how much it loses in each interaction with an atom

39 B Energy released = rest mass energy of reactants – rest mass energy of products ml c

2 + mp c2 – 2( mh c2) 40 B

Ao = 360-20= 340 min -1

A = 190 – 20 = 170 min -1

t = 1 half life = 5700