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Wednesday, January 27, 2010

Hwa Chong Institution Physics Prelim 2009 Paper 1 2 3 Answers

HCI Prelim Ans 1 2 3

Hwa Chong Institution Physics Prelim 2009 Paper 1

Hci Physics Prelim 09-p1

List of key Physics Definitions

List of key Physics Definitions

Tutorials -Electromagnetism

Physics - Tutorials - Electromagnetism

Tutorials - Electrical Field - No worked Solutuons

Physics - Tutorials - Electrical Field

Tutorials - DC Circuits

Physics - Tutorials - DC Circuits
Physics - Tutorials - DC Circuits

Tutorials - Current of Electricity

Physics- Tutorials - Current of Electricity

Physics A level H2 Paper 2 2008

A Level H2 Paper 2

Victoria Junior College Prelim Physics 2009 Paper 3 Answers

VJC-phy-pap3 answers

Jurong Junior College Physics Prelim 2009

JJC-phy-prelim02

Catholic Junior College Physics Prelim Paper 1 Answers

CJC Prelim Paper 1 Ans

Tips for handling A level Physics Paper 1


Catholic Junior College Physics Prelim Paper 1, 2 and 3

CJC Phy Prelim 1-2-3

Specimen Paper for H2 Physics 9745 Paper 1

Specimen Paper 1 Phy

Superposition Lecture Notes

Superposition Lecture Notes

Waves and Oscillations Revision

Waves & Oscillations Revision

Monday, January 4, 2010

NJC Physics prelim paper 2009

NJC_2009_phy_paper2

Anglo-Chinese JC Physics prelim 2009 paper 3

ACJC-2009-phy-paper3

ACJC-2009-phy-paper3-ans

Newtonian's birthday!

This site has been founded to commemorate Sir Issac Newton's Birthday on 4 January 1643.


2-Kinematics Structured Questions with Solutions

2-Kinematics Structured Questions with Solutions

1.) A shell is fired from a hill 400 m above the ground at an angle 30° to the horizontal with a speed of 45.0 m s-1. How far from the hill and at what angle will the shell hit the ground? [2]



2.)

Is it possible for a body to be at rest and yet be accelerated? State an example if yes. [2]

Answer:
Yes. An example will be when an object is thrown vertically upwards and reaches the peak of its motion. The velocity will be momentarily zero while acceleration is the gravitational acceleration.

Sunday, January 3, 2010

2-Kinematics MCQ with Solutions

2-Kinematics MCQ(with Solutions)

1.) At time t = 0 s, a body has starts from rest and moves with constant acceleration. Which graph best represents how s, the displacement of the body, varies with time t?



Answer: E
The parachutist experience free-fall initially, implying the only force acting on him or her is gravitational force. The acceleration will therefore be constant at 9.81 m s-2. The opening of the parachute causes a significant air resistance and hence, causes a resultant force upwards instead of down. Hence, acceleration is now in the opposite direction. The object slows down and hence, air resistance reduces gradually to arrive at terminal velocity when acceleration is zero.


An object is projected at an angle to the horizontal in a gravitational field and it follows a parabolic path, PQRST. These points are the positions of the object after successive equal time intervals. T being the highest point reached. Assuming negligible air resistance, the displacements PQ, QR, RS and ST.

A are equal.
B decrease at a constant rate.
C have equal horizontal components.
D increase at a constant rate.
E have equal vertical components.

Answer: C

The horizontal speed remains constant and hence, the horizontal components are equal.

10

When a rifle is fired horizontally at a target P on a screen at a range of 25 m, the bullet strikes the screen at a point 5.0 mm below P. The screen is now moved to a distance of 75 m and the rifle again fired horizontally at P in its new position.



Answer: C
Initial velocity is zero since object starts from rest. Displacement-time graph should have zero gradient at time t = 0 s.

2 A particle starts from rest and moves in a straight line. Its motion is represented by the acceleration–time graph shown below. At which point is its velocity maximum?



Answer: B

Area under acceleration-time graph represents change in velocity. The particle experiences an increase in velocity in the negative direction continuously to point B. From point B onwards, the velocity decreases until acceleration is zero again, as acceleration and velocity is now in opposite direction (i.e. acceleration positive, velocity negative). Hence, the velocity decreases. The velocity increases again to point C but, the increase is less than the decrease before that, hence overall, the velocity is still lesser than at B.

3.)
An aeroplane, flying in a straight line at a constant height of 400 m with a speed of 300 m s-1, drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so. Taking g as 10 m s-2 and ignoring air resistance, which one of the following gives the values of t and d?
t d
A 1.3 s 0.40 km
B 8.9 s 2.7 km
C 8.9 s 4.0 km
D 40 s 3.0 km
E 40 s 120 km

Answer: B



4.) A body dropped from a tower is timed to take (3.0±0.3) s to fall to the ground. If the acceleration of free fall is taken as 10 m s-2, the calculated height of the tower should be quoted as

A (15.0±0.2) m
B (15±0.1) m
C (45.0±0.2) m
D (45±5) m
E (45±9) m



5.)
A lunar landing module is descending to the Moon’s surface at a steady velocity of 25.0 m s-1. At a height of 40 m, a small object falls from its landing gear.
Taking the Moon’s gravitational acceleration as 1.60 m s-2, at what speed does the object strikes the Moon?

A 22.3 m s-1 B 27.4 m s-1 C 497 m s-1 D 753 m s-1

Answer: B




Take note as u,s and a are all in the same direction, they should have the same sign.

6.)

A parachutist steps from an aircraft, falls freely for 2 s and then opens his parachute. Which graph best represents how his vertical acceleration a, varies with time t during the first 5 s?




Answer: E
The parachutist experience free-fall initially, implying the only force acting on him or her is gravitational force. The acceleration will therefore be constant at 9.81 m s-2. The opening of the parachute causes a significant air resistance and hence, causes a resultant force upwards instead of down. Hence, acceleration is now in the opposite direction. The object slows down and hence, air resistance reduces gradually to arrive at terminal velocity when acceleration is zero.

7.)
When a rifle is fired horizontally at a target P on a screen at a range of 25 m, the bullet strikes the screen at a point 5.0 mm below P. The screen is now moved to a distance of 75 m and the rifle again fired horizontally at P in its new position.



Assuming that air resistance may be neglected, what is the new distance below P at which the screen would now be struck?

A B 15.0 mm C 20.0 mm D 25.0 mm E 45.0 mm

Answer: E
For horizontal direction,
Speed remains constant. Hence, if time needed to travel 25 m is t, time needed to travel 75 m will be 3t.

3-4-Forces and Dynamics Questions Structured Questions (with Solutions)

3-4-Forces and Dynamics Questions Structured Questions (with Solutions)

1.) A uniform plank of weight 70.0 N is 2000 mm long. It rests on a support that is 700 mm from end F. A load of 50.0 N is placed at end E of the plank.



At what distance from F must a 180 N weight be placed in order to balance the plank?



2.) A jet of water of cross-sectional area of 3.0 cm2 is directed normally against a wall. If the speed of the jet of water is 30 m s-1, what is the average force exerted on the water, assuming (unrealistically) that the water is stopped by the wall ( Density of water = 1000 kg m-3.) [2]



3.) An ice cube of mass 40.0 g floats on the surface of a strong brine solution of volume 250.0 cm3 inside a measuring cylinder. Calculate the level of liquid in the measuring cylinder
(i) before and [2]
(ii) after the ice is melted. [2]
(Assume density of ice = 900 kg m-3, density of water = 1000 kg m-3 and density of brine = 1100 kg m-3)

Answer:
(i) Floating ice displaces 40 g of brine since upthrust equals weight of ice.

Volume displaced = mass/ density = 40.0/1.100 = 36.4 cm3 [1]

Level on the measuring cylinder = 286.4 cm3 [1]

(ii) 40.0 g of ice forms 40.0 g of water when all of it is melted.

Volume of 40.0 g of water = 40.0/1.000 = 40.0 cm3 [1]
Level on the measuring cylinder rises to 290.0 cm3 [1]

3-4-Forces and Dynamics Questions MCQ (with Solutions)

Forces and Dynamics

1 A constant mass undergoes negative uniform acceleration when the resultant force acting on it

A increases uniformly and is in the opposite direction of the mass.
B is constant and opposite to the direction of motion of the mass.
C is constant and in the direction of motion of the mass.
D is constant and in the negative direction.
E decreases uniformly and is in the opposite direction of the motion of mass.

Answer is D.
For uniform acceleration, the resultant force must be constant. Negative acceleration just represents the acceleration is in the negative direction and does not imply decrease nor movement opposite to the direction of motion of the mass.

2 Which of the following statements is untrue about Newton’s Third Law of Motion?

A It involves forces which are equal in magnitude and opposite in direction.
B It involves forces which act on different bodies.
C The force of attraction between an electron and a proton in a hydrogen atom is an example of the forces referred to in Newton’s law of motion.
D The two forces are equal and opposite so that the bodies are in equilibrium.

Ans: D
The forces acts on different bodies and hence, they will not keep the bodies in equilibrium.

3 An object is dropped from a high height and is acted upon by two forces: a constant gravitational force and air resistance which is directly proportional to its velocity.
Which one of the following statements about the subsequent motion of the particle is true?


A Its acceleration increases from zero to a maximum and then decreases.
B Its acceleration increases from zero to a maximum.
C Its velocity increases from zero to a maximum and then decreases.
D Its velocity increases from zero to a maximum.

Ans: D
The object will experience a decreasing resultant force which drops to zero and remains at zero. Hence, the velocity will increase from zero to terminal velocity, which is also the maximum velocity it will attain. The acceleration will decrease from maximum to zero.

4.) A helicopter of mass 2.0 x103 kg rises vertically with a constant speed of 35 m s-1. Taking the acceleration free fall as 10 m s-2, what is the resultant force that is acting on the helicopter?

A zero
B 3.0 x 104 N downwards
C 4.5 x 104 N upwards
D 7.5 x 104 N upwards
E 10.5 x 104 N upwards

Answer is A.
Since speed is constant, resultant force is zero.

5.) A 20 g object traveling to the right at 6.0 m s-1 collides head on with a 30 g object traveling to the left at 7.0 m s-1. What is the loss in kinetic energy for the collision if it was perfectly inelastic?

A 0.01 J
B 1.01 J
C 0.01 kJ
D 1.01 kJ

Ans: B
Loss in kinetic energy = Total final kinetic energy – total initial kinetic energy
Total initial kinetic energy = 0.5[(0.020)(6.0)2+(0.030)(7.0)2]=1.095 J
By conservation of linear momentum and based on the fact that the collision is perfectly inelastic,
m1 u1+m2 u2= (m1+ m2)v
20(6.0)-30(7.0)=(20+30)v
v=-1.8 m s-1
Total final kinetic energy = 0.5(0.050)(1.8)2= 0.081 J
Loss of kinetic energy = 1.014 J

6.)

A particle X (of mass 8 units) and a particle Y (of mass 2 units) move directly towards each other, collide and then separate. If vx is the change of velocity of X and vY is the change of velocity of Y, what is the magnitude of the ratio vx/vY?

A 1/4 B 1/sq root 2 C 1/2 D 2 E 4

Answer is A.
Applying Conservation of Linear momentum,
mX uX+mY uY =mX vX+mY vY
mX vX- mX uX=-( mY vY-mY uY)
mX vx= mYvY
vx/vY= mY /mX= 2/8 =1/4

7.)
A tennis player serves in a tennis match. The ball has a mass of 0.2 kg and was travelling at 40 ms-1 towards the racket just before being struck horizontally by the racket. Figure below shows the magnitude of the force exerted on the ball by the racket as a function of time.

What is the speed of the ball immediately after being struck?



A 10 ms-1.
B 40 ms-1.
C 50 ms-1.
D 90 ms-1.

Answer: A




Area under graph = change in momentum of ball
200 x 50 x 10-3 = m(v – u)
v= (10/0.2) -40 = 10 ms-1. (take final direction of ball as positive)

8.)
An object immersed in a liquid in a tank, experiences upthrust.

What is the physical reason for this upthrust?

A The density of the body differs from that of the liquid.
B The density of the liquid increases with depth.
C The pressure in the liquid increases with depth.
D the value of g in the liquid increases with depth.

Answer C.

9)
A uniform beam of length L is hinged at X and is supported by a tension rod as shown below.



The force exerted by the wall on the beam is in the direction

A XW B XZ C XY D YX

Answer is C
For equilibrium, the line of action of the three forces must pass through a single point. This is only possible if the force acts in the direction XY. It cannot be YX because there will be no translational equilibrium in the horizontal direction.

10)
An object of mass 3.9 kg is on a horizontal frictionless surface. It is held in equilibrium by three strings with different tensions, as shown below. The diagram is not drawn to scale.



What is the value of the tension, T?

A 3.5 N B 9.8 N C 10.7 N D 26.4 N

Answer is C.

For translational equilibrium in the horizontal direction,
T = 14.0 sin 35° + 15.2 sin 10° = 10.7 N

8-10-Waves and Oscillations MCQ

1 The ionosphere contains free electrons. What is the amplitude of oscillation of the oscillation of these electrons when subject to a 200 kHz electromagnetic wave in which the oscillations of electric field have amplitude 5 ´ 103 V m 1 ?

A 3.2 ´ 1015 m B 4.0 ´ 109 m C 2.5 ´ 108 m D 5.6 ´ 104 m

Force on electron F = eE = (1.6 ´ 1019) (5 ´ 103)

Since Fnet = F = ma (m = 9.1 ´ 1031 kg)

F = mw2x (w = 2pf)

Hence x = eE / (m4p2f2) = 5.6 ´ 104 m

Ans: D

2. A suspension bridge is to be built across a valley where it is known that the wind can gust at 5 s intervals. It is estimated that the speed of transverse waves along the span of the bridge would be 400 ms1. The danger of resonant motions in the bridge at its fundamental frequency would be greatest if the span had a length of

A 2000 m B 1000 m C 400 m D 80 m

If l is the length of the suspension bridge and l is the wavelength associated with the speed of at which the wind is blowing, then

I = l/2 = 0.5 (v/f) = 0.5 (400) )5) = 1000 m

Ans: B

3.) A mass hanging from a light helical spring produces an equilibrium extension of 0.1 m. The mass is pulled vertically downwards by a distance of 0.02 m and then released. The angular frequency of the simple harmonic motion is 10 rad s m1.The equation relating the displacement x of the mass from its equilibrium position and the time t after release is

A x/m = 0.02 sin [10(t/s)]

B x/m = 0.03 sin [10(t/s)]

C x/m = 0.02 cos [10(t/s)]

D x/m = 0.03 cos [10(t/s)]

Amplitude of simple harmonic motion, xo­ = 0.02 m since this is the distance from the equilibrium position. The motion results in a cosine graph since it starts from the maximum displaced position at t = 0.

Ans: C

4.) Values of the acceleration a of a particle moving in simple harmonic motion as a function of its displacement x are given in the table below.

a / mm s2

16

8

0

8

16

x / mm

4

2

0

2

4

The period of the motion is

A 1/p s B 2/p s C p / 2 s D p s

Angular frequency, w2 = a/x = 16/(4) = 4 or w = 2 rad /s

Hence, period of the motion , T = 2p/w = p s

Ans: C

5 In which group below do all three quantities remain constant when a particle moves in simple harmonic motion?

A

acceleration

force

total energy

B

force

total energy

amplitude

C

total energy

amplitude

angular frequency

D

amplitude

angular frequency

acceleration

In simple harmonic motion, the energy is conserved and hence the total energy (sum of KE and PE) remains constant. The equation that describes the motion in terms of displacement x is given by

a = w2x

Amplitude of the SHM is the maximum displacement of the motion and hence remains unchanged.

Ans: C

6.) A particle P performs simple harmonic motion of period 8 s about a fixed point O in the horizontal direction. The maximum displacement of P from O is 5 m. If P is initially at O and moving to the left, the 17 s later P will be

A moving towards O.

B moving with increasing speed.

C moving with increasing acceleration.

D at a distance of 2.5 m to the left of O.

P passes by O at 0, 8 and 16 s. At t = 17 s, it will be moving leftward away from O. Since a µ x, acceleration increases as displacement from O increases.

Ans: C


7 A point mass moves with simple harmonic motion. Which of the following statements is false?

A

The maximum kinetic energy is dependent on the frequency of the oscillation.

B

The time taken for the system to change from maximum kinetic energy to maximum potential energy is a quarter of the period of the oscillation.

C

An oscillation system with larger amplitude will have a greater maximum velocity

D

An oscillation system with larger amplitude will have longer period.

Ans: D

8.

During the “live” telecast of Campus mega hotstar, a television viewer in Pasir Ris hears the sound picked up by a microphone directly in front of a contestant. This viewer is seated 2.00 m from the television set.

A “live” audience is in the recording studio located 10.0 m from the microphone. The “live” audience and the television viewer hear the same sound at a time difference of 0.025 s.

If the distance between the “live” audience and the television viewer is 15.0 km and assuming that the electrical signal is transmitted via land cables at the speed of light, determine the speed of sound.

A 311 ms-1 B 319 ms-1 C 330 ms-1 D 343 ms-1

t1 = time taken for sound travel from mic to live viewer

t2 = time taken for sound travel from mic to television

t3 = time taken for sound travel from television to home viewer

Assuming that home viewer hears the sound first,

t1 -(t2 + t3) = 0.025

10/vs -(15´103/c + 2/vs) = 0.025 where (c=3.0´108 ms-1)

vs = 319 ms-1

Ans: B


9)

A small loudspeaker at P generates a sound wave of frequency 660 Hz. If the speed of sound in air is 330 ms-1, the phase difference between the air vibrations at Q and R, 0.25 m apart, is

A zero B 2p radians C p/2 radians D p radians

v=fl

l=330/660=0.50 m

Df=o.25/0.50 ´ 2p = p rad

Ans: D


10)

The intensity of a wave is proportional to the square of the amplitude of the wave. If two waves of the same frequency and different amplitudes are superimposed at a point in phase the total intensity at that point is proportional to

A the sum of the intensities of the two waves

B the mean value of the intensities of the two waves

C the square of the sum of the two amplitudes

D the square of the mean value of the two amplitudes


Ans: C