UMOJA WA WAZAZI TANZANIA.
WARI SECONDARY SCHOOL
PRE-NATIONAL EXAMINATION SERIES
PHYSICS 1 - SERIES 10
INSTRUCTIONS
- This paper consists of section's A and B with a total of ten (10) questions.
- Answer all questions in section A and two (2) questions from section B.
- Section A carries seventy (70) marks and section C carries thirty (30) marks.
- Marks for each question or part thereof are indicated.
- Cellular phones and any unauthorized materials are not allowed in the examination room.
- Mathematical tables or non-programmable calculators may be used.
- Write your examination number on every page of your answer booklet(s).
The following information may be useful:
- Acceleration due to gravity \( g = 9.8m/s^2 \)
- Density of air \( = 1.3kg/m^3 \)
- Density of water \( = 1000kg/m^3 \)
- Stefan-Boltzmann constant \( \sigma = 5.7 \times 10^{-8}W m^{-2} K^{-4} \)
- Electronic charge, \( e = 1.6 \times 10^{-19}C \)
- Radius of earth, \( R_e = 6400km \)
- Resistivity of copper \( 1.72 \times 10^{-8}\Omega m \)
- Resistivity of Aluminium \( 2.63 \times 10^{-8}\Omega m \)
- Velocity of light, \( c = 3 \times 10^8m/s \)
- Molar gas constant, \( R = 8.314 J Mol^{-1}K^{-1} \)
- Mass of earth, \( M_e = 6 \times 10^{24}kg \)
- Average density of the earth's rocks \( = 5.5 \times 10^3kg/m^3 \)
ANSWER ALL QUESTIONS IN THIS SECTION.
| Temperature | Pressure A | Pressure B |
|---|---|---|
| Triple-point of water | 1.250×10\(^5\)Pa | 0.200×10\(^5\)Pa |
| Normal melting point of sulphur | 1.797×10\(^5\)Pa | 0.268×10\(^5\)Pa |
Answer two (2) questions from this section.
UMOJA WA WAZAZI TANZANIA.
WARI SECONDARY SCHOOL
PRE-NATIONAL EXAMINATION SERIES
PHYSICS 1 - SERIES 10
INSTRUCTIONS
- This paper consists of section's A and B with a total of ten (10) questions.
- Answer all questions in section A and two (2) questions from section B.
- Section A carries seventy (70) marks and section C carries thirty (30) marks.
- Marks for each question or part thereof are indicated.
- Cellular phones and any unauthorized materials are not allowed in the examination room.
- Mathematical tables or non-programmable calculators may be used.
- Write your examination number on every page of your answer booklet(s).
The following information may be useful:
- Acceleration due to gravity \( g = 9.8m/s^2 \)
- Density of air \( = 1.3kg/m^3 \)
- Density of water \( = 1000kg/m^3 \)
- Stefan-Boltzmann constant \( \sigma = 5.7 \times 10^{-8}W m^{-2} K^{-4} \)
- Electronic charge, \( e = 1.6 \times 10^{-19}C \)
- Radius of earth, \( R_e = 6400km \)
- Resistivity of copper \( 1.72 \times 10^{-8}\Omega m \)
- Resistivity of Aluminium \( 2.63 \times 10^{-8}\Omega m \)
- Velocity of light, \( c = 3 \times 10^8m/s \)
- Molar gas constant, \( R = 8.314 J Mol^{-1}K^{-1} \)
- Mass of earth, \( M_e = 6 \times 10^{24}kg \)
- Average density of the earth's rocks \( = 5.5 \times 10^3kg/m^3 \)
ANSWER ALL QUESTIONS IN THIS SECTION.
In experimental physics, errors are always maximized to determine the worst-case scenario or maximum possible error in measurements. This approach ensures that:
- The true value lies within the calculated error bounds with high confidence
- It provides a safety margin in engineering and scientific applications
- It helps identify the limiting factors in experimental precision
- It ensures reliable and conservative estimates in critical applications
Repeating measurements and taking the arithmetic mean minimizes random errors because:
- Random errors follow a normal distribution (Gaussian distribution)
- Positive and negative errors tend to cancel each other out
- The standard error of the mean decreases as 1/√n, where n is the number of measurements
- This process increases the reliability and precision of the final result
Precision refers to the reproducibility or consistency of measurements (how close measurements are to each other).
Accuracy refers to how close a measurement is to the true or accepted value.
Relationship:
- A precise experiment may not be accurate if there are systematic errors
- An accurate experiment must have both high precision and proper calibration
- High precision is necessary but not sufficient for high accuracy
- Good experimental design aims for both high precision and high accuracy
Let's analyze the dimensions:
Moment of inertia I = [ML²]
Force F = [MLT⁻²]
Velocity V = [LT⁻¹]
Work W = [ML²T⁻²]
Length L = [L]
Now, \( x = \frac{I F^2}{W L} \)
Dimensions: \( [x] = \frac{[ML²] × [MLT⁻²]²}{[ML²T⁻²] × [L]} = \frac{[ML²] × [M²L²T⁻⁴]}{[ML³T⁻²]} \)
\( [x] = \frac{[M³L⁴T⁻⁴]}{[ML³T⁻²]} = [M²LT⁻²] \)
This corresponds to dimensions of Energy × Mass or specifically, it could represent a quantity like momentum squared or related to action in physics.
More precisely, M²LT⁻² = (MLT⁻¹)² / M, which is momentum squared divided by mass.
From \( T = 2\pi \sqrt{\frac{L}{g}} \), we get \( g = \frac{4\pi^2 L}{T^2} \)
Fractional error: \( \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \)
Given: L = 10 cm = 0.1 m
Least count of meter rule = 0.1 cm = 0.001 m
So \( \frac{\Delta L}{L} = \frac{0.001}{0.1} = 0.01 \) (1%)
Time for 100 oscillations = 100 × 0.5 = 50 seconds
Least count of stopwatch = 0.1 seconds
Error in time for 100 oscillations = 0.1 seconds
Error in one period \( \Delta T = \frac{0.1}{100} = 0.001 \) seconds
\( \frac{\Delta T}{T} = \frac{0.001}{0.5} = 0.002 \) (0.2%)
So \( \frac{\Delta g}{g} = 0.01 + 2(0.002) = 0.01 + 0.004 = 0.014 \) (1.4%)
Precision in determination of g = 1.4%
Carpet can be cleaned by beating it with a stick due to inertia. When the carpet is beaten suddenly:
- The carpet moves quickly due to the force applied
- The dust particles tend to remain at rest due to their inertia
- This relative motion causes the dust to separate from the carpet
- The dust then falls away due to gravity
This happens due to the conservation of horizontal momentum and Newton's first law of motion:
- The ball shares the bus's horizontal velocity when thrown
- While in air, no horizontal force acts on the ball (neglecting air resistance)
- The ball continues moving horizontally with the same velocity as the bus
- Therefore, it maintains its position relative to the thrower and returns to the same point
The bicycle undergoes greater damage due to:
- Conservation of momentum: Both experience equal but opposite forces (Newton's third law)
- Acceleration: The bicycle has much less mass, so it experiences greater acceleration (a = F/m)
- Kinetic energy: The change in kinetic energy causes deformation and damage
- Structural strength: The truck is designed to withstand greater impacts
According to Newton's second law (F = ma), for the same force, the smaller mass (bicycle) experiences much greater acceleration and therefore greater damage.
No, a rocket in flight is not an example of projectile motion because:
- Projectiles move under the influence of gravity only (after initial impulse)
- Rockets have their own propulsion system and can accelerate during flight
- Rockets can change their trajectory and speed during flight
- Projectiles follow a parabolic path, while rockets can follow various trajectories
Yes, the bullet will hit the monkey (assuming no air resistance and perfect aim). This can be explained by:
- Both the bullet and monkey experience the same gravitational acceleration (g)
- The bullet has two components of motion: horizontal (constant velocity) and vertical (free fall)
- The monkey only has vertical motion (free fall)
- In the vertical direction, both fall the same distance in the same time: h = ½gt²
- The horizontal motion of the bullet brings it to the point directly below the monkey's original position at the same time the monkey falls to that point
This demonstrates the independence of horizontal and vertical motions in projectile motion.
Yes, the height of jump matters in long jumping because:
Factors determining the span of jump:
- Take-off speed: Higher speed gives longer horizontal distance
- Take-off angle: Optimal angle is about 45° for maximum range
- Height of jump: Higher jump increases time of flight, allowing more horizontal distance
- Air resistance: Affects the trajectory
- Technique: Body position during flight and landing
The range formula for projectile motion is: \( R = \frac{v^2 \sin 2\theta}{g} \), where higher take-off gives more time of flight.
No, rockets do not need to achieve escape velocity (11.2 km/s) initially because:
- Rockets have continuous propulsion and can accelerate gradually
- They work against gravity continuously rather than needing one initial push
- The escape velocity calculation assumes no additional propulsion after launch
- Rockets can escape Earth's gravity with much lower initial speeds by burning fuel continuously
Escape velocity from Earth: \( v_e = \sqrt{\frac{2GM}{R}} \)
Orbital speed: \( v_o = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \)
Orbital radius = R (Earth's radius)
Total energy in orbit = Kinetic + Potential = \( \frac{1}{2}mv_o^2 - \frac{GMm}{R} \)
\( = \frac{1}{2}m\left(\frac{GM}{2R}\right) - \frac{GMm}{R} = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R} \)
When stopped, kinetic energy = 0, so total energy = potential energy = \( -\frac{GMm}{R} \)
Energy is conserved during fall.
At Earth's surface, total energy = \( \frac{1}{2}mv^2 - \frac{GMm}{R} \)
Equating: \( -\frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \)
This gives \( \frac{1}{2}mv^2 = 0 \) ⇒ v = 0, which is incorrect.
Let's recalculate properly:
When stopped at orbital radius R, total energy = \( -\frac{GMm}{R} \)
At Earth's surface (radius R), total energy = \( \frac{1}{2}mv^2 - \frac{GMm}{R} \)
By conservation: \( -\frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \)
This gives v = 0, which suggests an error in the approach.
Actually, when stopped at orbital radius R, the satellite will fall directly to Earth.
Using conservation of energy:
Initial energy (when stopped) = \( -\frac{GMm}{R} \)
Final energy (at surface) = \( \frac{1}{2}mv^2 - \frac{GMm}{R} \)
Equating: \( -\frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \)
This gives \( \frac{1}{2}mv^2 = 0 \) ⇒ v = 0
This suggests the satellite would have zero speed when it reaches the surface, which is not physically reasonable.
The correct approach: The orbital speed is \( v_o = \sqrt{\frac{GM}{R}} \) for low Earth orbit, not \( \sqrt{\frac{GM}{2R}} \).
If \( v_o = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \), then:
Initial total energy = \( \frac{1}{2}mv_o^2 - \frac{GMm}{R} = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R} \)
When stopped, kinetic energy = 0, so total energy = \( -\frac{GMm}{R} \)
At surface: \( -\frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \) ⇒ v = 0
This is still giving v = 0. There seems to be a conceptual issue with the problem statement.
In reality, if a satellite is stopped in orbit, it would fall to Earth and hit with substantial speed.
Let's assume the orbital radius is r, not R (Earth's radius).
For a circular orbit: \( v_o = \sqrt{\frac{GM}{r}} \)
If \( v_o = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}} \), then \( \sqrt{\frac{GM}{r}} = \frac{1}{2}\sqrt{\frac{2GM}{R}} \)
Squaring: \( \frac{GM}{r} = \frac{1}{4} \cdot \frac{2GM}{R} = \frac{GM}{2R} \) ⇒ r = 2R
So the satellite is at height R above Earth's surface.
Initial energy when stopped = \( -\frac{GMm}{2R} \)
Final energy at surface = \( \frac{1}{2}mv^2 - \frac{GMm}{R} \)
By conservation: \( -\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \)
\( \frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} \)
\( v^2 = \frac{GM}{R} \)
But \( v_e = \sqrt{\frac{2GM}{R}} \), so \( v = \frac{v_e}{\sqrt{2}} = \frac{11.2}{\sqrt{2}} = 7.92 \) km/s
Impact speed = 7.92 km/s
There is no loss in mechanical energy because:
- The friction in rolling without slipping is static friction
- Static friction does no work as there is no relative motion at the point of contact
- The mechanical energy is conserved, with potential energy converting to both translational and rotational kinetic energy
- The friction force provides the torque for rotation without dissipating energy
The role of friction in rolling without slipping is to:
- Provide the necessary torque for rotation
- Prevent slipping at the point of contact
- Convert some translational motion into rotational motion
- Ensure pure rolling motion occurs
Using conservation of energy:
Initial potential energy = mgh
Final energy = Translational KE + Rotational KE
\( mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)
For rolling without slipping: \( \omega = \frac{v}{r} \)
For solid cylinder: \( I = \frac{1}{2}mr^2 \)
So: \( mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 \)
\( mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 \)
\( gh = \frac{3}{4}v^2 \)
\( v^2 = \frac{4}{3}gh \)
\( v = \sqrt{\frac{4}{3}gh} \)
Where:
- m = mass of cylinder
- g = acceleration due to gravity
- h = height of incline
- v = translational velocity at bottom
- I = moment of inertia
- ω = angular velocity
- r = radius of cylinder
Amplitude Modulation (AM):
- The amplitude of the carrier wave is varied in proportion to the message signal
- Frequency and phase remain constant
- More susceptible to noise and interference
- Simpler circuitry
- Used in medium wave and short wave broadcasting
Frequency Modulation (FM):
- The frequency of the carrier wave is varied in proportion to the message signal
- Amplitude remains constant
- Less susceptible to noise and interference
- More complex circuitry
- Used in VHF radio, television audio, and satellite communication
Three advantages of FM over AM:
- Better noise immunity: FM is less affected by amplitude variations caused by noise
- Constant amplitude: Allows more efficient power amplification
- Higher fidelity: Can accommodate wider bandwidth for better sound quality
- Less interference: Adjacent FM stations cause less interference
- Capture effect: Stronger signal captures the receiver, reducing interference
The three basic elements of a communication system are:
- Transmitter:
- Converts the message signal into a form suitable for transmission
- Includes components like modulator, amplifier, and antenna
- Processes the information signal and impresses it on a carrier wave
- Communication Channel:
- The medium through which the signal travels from transmitter to receiver
- Can be wired (coaxial cable, optical fiber) or wireless (free space)
- Introduces attenuation, distortion, and noise to the signal
- Receiver:
- Receives the transmitted signal from the channel
- Extracts the original message signal from the carrier
- Includes components like antenna, demodulator, and amplifier
Carrier frequency f_c = 2 MHz = 2000 kHz
Modulating frequency f_m = 4 kHz
Lower side band frequency = f_c - f_m = 2000 - 4 = 1996 kHz
Upper side band frequency = f_c + f_m = 2000 + 4 = 2004 kHz
Bandwidth = 2 × f_m = 2 × 4 = 8 kHz
Modulation factor m = 55% = 0.55
Carrier amplitude A_c = 70 V
Side band amplitude = \( \frac{m A_c}{2} = \frac{0.55 × 70}{2} = \frac{38.5}{2} = 19.25 \) V
Amplitude of both upper and lower side bands = 19.25 V
