FORM SIX PRE NATIONAL EXAMINATION PHYSICS 1 - SERIES 13 (With Comprehensive Solutions)

PRE NATIONAL EXAMINATION PHYSICS 1 - SERIES 13
UMOJA WA WAZAZI TANZANIA
WARI SECONDARY SCHOOL
PRE-NATIONAL EXAMINATION SERIES
PHYSICS 1 – SERIES 13
131/1
TIME: 2:30 HRS
JANUARY-MAY, 2023

Instructions

  1. This paper consists of section A and B, with a total of ten (10) questions.
  2. Answer all questions from section A and any two (2) questions from section B.
  3. Mathematical table and non-programmable calculators may be used.
  4. Cellular phones are not allowed in the examination room.
  5. Write your examination number in every page of your answer booklet(s).

Useful Information

  • Acceleration due to gravity, \( g = 9.8 m/s^2 \)
  • Density of air = \( 1.29 kg m^{-3} \)
  • Radius of the earth = \( 6.4 \times 10^6 km \)
  • Stefan's constant, \( \delta = 5.67 \times 10^{-8} w m^{-2} k^{-4} \)
  • Speed of light in air, \( c = 3 \times 10^8 m/s \)
  • Specific heat capacity of water = \( 4200J/kg^{-1}k^{-1} \)
  • Pie \( \pi = 3.14 \)
  • Density of metal sphere = \( 8000kg m^{-3} \)
  • Specific heat capacity of metal sphere = \( 400J/kg^{-1}k^{-1} \)
  • Mass of the earth = \( 6.0 \times 10^{24} kg \)
  • Thermal conductivity of wood = \( 0.125w m^2 c^{-1} \)
  • Thermal conductivity of cement = \( 1.5w m^2 c^{-1} \)
  • Universal gravitational constant \( G = 6.7 \times 10^{-11} N m^2 kg^{-2} \)
  • Thermal conductivity of brick = \( 1.0w m^2 c^{-1} \)
  • Density of water = \( 1000kg m^{-3} \)

SECTION A (70 Marks)

Answer all questions from this section

1. (a) (i)
02 marks

Briefly explain two types of errors that are likely to occur whenever an experimental measurement is made.

1. (a) (ii)
03 marks

In an experiment to determine the value of Young's modulus Y of elasticity of steel, a wire of length 325 cm (measured by a metre scale of least count 0.1cm) is loaded by a mass of 2kg and it is found that it stretches by 0.227cm (measured by a micrometer having least count 0.001cm). The diameter of the wire as measured by a screw gauge (least count 0.001cm) is found to be 0.043 cm. Calculate the maximum permissible error in Y.

1. (b) (i)
01 mark

How is a precise physics experiment differ from an accurate one?

1. (b) (ii)
01 mark

The diameter of a steel rod is given as \( (56.47 \pm 0.02)mm \) what does this mean?

1. (c) (i)
01 mark

What is meant by the statement that "an equation is homogenous with respect to its unit"?

1. (c) (ii)
02 marks

The stress S required to fracture a solid can be expressed as \[ S = K \sqrt{\frac{\lambda E}{d}} \] where K is a dimensionless constant, E is a Young's modulus and d is the distance between the planes of the atoms separated by a fracture. If the equation is dimensionally consistent, find the dimensions of the physical quantity \(\lambda\) and suggest the meaning of this quantity.

2. (a)
03 marks

Briefly explain three (3) applications of projectile motion in real life experience.

2. (b)
04 marks

Two paper screens A and B are separated by a distance of 100m. A bullet pierces A and then B. The hole in B is 0.1m below the hole in A. If the bullet is travelling horizontally at the time of hitting the screen A, calculate the velocity of the bullet when it hits the screen A neglecting the resistance of paper and air.

2. (c)
03 marks

A bullet is dropped from a certain height and at the same time another bullet is fired horizontally from the same height. Which one will hit the ground earlier and why?

3. (a) (i)
01 mark

Why do we say that velocity and acceleration of a body executing S.H.M are out of phase?

3. (a) (ii)
03 marks

Show that a body when hanged vertically in a helical spring executes Simple Harmonic Motion (S.H.M.)

3. (b)
03 marks

If the earth was a homogeneous sphere and a straight hole was bored through its centre, so that a body dropped into this hole will execute a S.H.M. Calculate the time period if the radius of the earth is 6400km and g on its surface is 9.8ms\(^{-2}\).

3. (c)
03 marks

Compute the magnitude and direction of total linear acceleration of a particle moving in a circle of radius 0.4m having an instantaneous angular velocity of 2rads\(^{-1}\) and angular acceleration 5rads\(^{-2}\).

Comprehensive Answers

Question 1 Answers

1. (a) (i) Two types of errors in experimental measurement:

Systematic Errors: These are errors that consistently occur in the same direction and magnitude in repeated measurements. They result from flaws in the measurement instrument, incorrect calibration, or observer bias. Examples include zero error in instruments, parallax error, and incorrect calibration of a scale.

Random Errors: These are unpredictable variations in measurements that occur due to unpredictable factors in the measurement process. They can be positive or negative and vary in magnitude. Examples include fluctuations in environmental conditions, observer's inability to judge readings precisely, and minor variations in the measured quantity itself.

1. (a) (ii) Maximum permissible error in Young's modulus:

The formula for Young's modulus is: \( Y = \frac{FL}{A\Delta L} = \frac{4FL}{\pi d^2 \Delta L} \)

Given:

  • Length L = 325 cm, error ΔL = ±0.1 cm
  • Extension ΔL = 0.227 cm, error Δ(ΔL) = ±0.001 cm
  • Diameter d = 0.043 cm, error Δd = ±0.001 cm
  • Mass m = 2 kg (assumed no error)
  • Force F = mg = 2 × 9.8 = 19.6 N

The fractional error in Y:

\[ \frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L} \]

Since F has no error:

\[ \frac{\Delta Y}{Y} = \frac{0.1}{325} + 2 \times \frac{0.001}{0.043} + \frac{0.001}{0.227} \]
\[ = 0.000308 + 0.0465 + 0.00441 = 0.05122 \]

Percentage error = 0.05122 × 100% = 5.12%

1. (b) (i) Difference between precise and accurate experiment:

Precision refers to how close repeated measurements are to each other, indicating the reproducibility of results. A precise experiment gives consistent results when repeated under the same conditions.

Accuracy refers to how close a measured value is to the true or accepted value. An accurate experiment gives results that are very close to the actual value.

An experiment can be precise but not accurate (consistent wrong results) or accurate but not precise (correct on average but with high variability).

1. (b) (ii) Meaning of \( (56.47 \pm 0.02)mm \):

This notation means that the measured diameter of the steel rod is 56.47 mm, with an uncertainty of ±0.02 mm. This indicates that the true diameter lies between 56.45 mm and 56.49 mm. The value 56.47 mm is the best estimate, and 0.02 mm represents the possible error in the measurement.

1. (c) (i) Homogenous equation with respect to units:

An equation is said to be homogenous with respect to its units when all terms in the equation have the same dimensions. This means that each term on both sides of the equation must have identical dimensional formulas. This is a fundamental principle of dimensional analysis used to check the correctness of physical equations.

1. (c) (ii) Dimensions of λ and its meaning:

Given: \( S = K \sqrt{\frac{\lambda E}{d}} \)

Dimensions of stress S = [ML⁻¹T⁻²]

Dimensions of Young's modulus E = [ML⁻¹T⁻²]

Dimensions of distance d = [L]

K is dimensionless.

Squaring both sides: \( S^2 = K^2 \frac{\lambda E}{d} \)

So: \( \lambda = \frac{S^2 d}{K^2 E} \)

Dimensions of λ:

\[ [\lambda] = \frac{[ML^{-1}T^{-2}]^2 [L]}{[ML^{-1}T^{-2}]} = [ML^{-1}T^{-2}] [L] = [MT^{-2}] \]

λ has dimensions of [MT⁻²], which are the dimensions of surface tension (force per unit length).

λ likely represents the surface energy per unit area or fracture toughness of the material, which is the energy required to create new surfaces during fracture.

Question 2 Answers

2. (a) Applications of projectile motion:

  1. Military applications: Calculating trajectories of missiles, artillery shells, and bullets to accurately hit targets.
  2. Sports: Analyzing the motion of balls in games like basketball, football, cricket, and golf to improve performance and strategy.
  3. Space exploration: Planning satellite launches and calculating orbital trajectories for spacecraft.
  4. Engineering: Designing water fountains, fireworks displays, and amusement park rides that involve parabolic motion.

2. (b) Velocity of the bullet:

Given:

  • Distance between screens A and B = 100 m
  • Vertical displacement between holes = 0.1 m
  • Bullet traveling horizontally when hitting screen A

Time taken to travel from A to B:

\[ t = \frac{\text{horizontal distance}}{\text{horizontal velocity}} = \frac{100}{v} \]

Vertical displacement due to gravity:

\[ h = \frac{1}{2}gt^2 \]
\[ 0.1 = \frac{1}{2} \times 9.8 \times \left(\frac{100}{v}\right)^2 \]
\[ 0.1 = 4.9 \times \frac{10000}{v^2} \]
\[ v^2 = \frac{4.9 \times 10000}{0.1} = 490000 \]
\[ v = \sqrt{490000} = 700 \, \text{m/s} \]

The velocity of the bullet when it hits screen A is 700 m/s.

2. (c) Which bullet hits the ground first:

Both bullets will hit the ground at the same time.

Explanation:

The vertical motion of both bullets is independent of their horizontal motion. Both bullets start from the same height with zero initial vertical velocity. The bullet that is fired horizontally has a horizontal component of velocity, but this does not affect its vertical motion. Both bullets experience the same acceleration due to gravity (g) in the vertical direction.

The time taken to fall to the ground depends only on the initial height and the acceleration due to gravity, given by:

\[ t = \sqrt{\frac{2h}{g}} \]

Since both bullets are dropped from the same height and experience the same gravitational acceleration, they will hit the ground simultaneously.

Question 3 Answers

3. (a) (i) Velocity and acceleration out of phase in SHM:

In Simple Harmonic Motion, velocity and acceleration are said to be out of phase because when displacement is maximum, velocity is zero and acceleration is maximum, and when displacement is zero, velocity is maximum and acceleration is zero.

Mathematically:

Displacement: \( x = A\sin(\omega t) \)

Velocity: \( v = A\omega\cos(\omega t) = A\omega\sin(\omega t + \pi/2) \)

Acceleration: \( a = -A\omega^2\sin(\omega t) = A\omega^2\sin(\omega t + \pi) \)

Velocity leads displacement by Ï€/2 (90°), and acceleration leads velocity by Ï€/2 (90°), making acceleration and displacement out of phase by Ï€ (180°).

3. (a) (ii) Body hanged on spring executes SHM:

Consider a mass m attached to a spring of spring constant k.

At equilibrium: mg = kx₀ (where x₀ is the extension at equilibrium)

When displaced by x from equilibrium position, the net restoring force:

\[ F = -k(x_0 + x) + mg = -kx \]

Since mg = kx₀, the equation becomes:

\[ F = -kx \]

According to Newton's second law:

\[ m\frac{d^2x}{dt^2} = -kx \]
\[ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0 \]

This is the standard differential equation for SHM with angular frequency:

\[ \omega = \sqrt{\frac{k}{m}} \]

Hence, the body executes Simple Harmonic Motion.

3. (b) Time period of oscillation through Earth:

For a body inside a uniform sphere, the gravitational force at a distance r from the center is:

\[ F = -\frac{GMm}{R^3}r \]

Where M is mass of Earth, R is radius of Earth, m is mass of body.

This is of the form F = -kr, where \( k = \frac{GMm}{R^3} \)

Time period of SHM:

\[ T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{mR^3}{GMm}} = 2\pi\sqrt{\frac{R^3}{GM}} \]

We know that \( g = \frac{GM}{R^2} \), so \( GM = gR^2 \)

\[ T = 2\pi\sqrt{\frac{R^3}{gR^2}} = 2\pi\sqrt{\frac{R}{g}} \]

Given R = 6400 km = 6.4 × 10⁶ m, g = 9.8 m/s²

\[ T = 2\pi\sqrt{\frac{6.4 \times 10^6}{9.8}} = 2\pi\sqrt{653061.22} = 2\pi \times 808.12 \]
\[ T = 5077.5 \, \text{s} = 84.6 \, \text{minutes} \]

The time period of oscillation is approximately 84.6 minutes.

3. (c) Total linear acceleration in circular motion:

Given:

  • Radius r = 0.4 m
  • Angular velocity ω = 2 rad/s
  • Angular acceleration α = 5 rad/s²

Centripetal acceleration:

\[ a_c = \omega^2 r = (2)^2 \times 0.4 = 4 \times 0.4 = 1.6 \, \text{m/s}^2 \]

Tangential acceleration:

\[ a_t = \alpha r = 5 \times 0.4 = 2.0 \, \text{m/s}^2 \]

Total linear acceleration:

\[ a = \sqrt{a_c^2 + a_t^2} = \sqrt{(1.6)^2 + (2.0)^2} = \sqrt{2.56 + 4.0} = \sqrt{6.56} = 2.56 \, \text{m/s}^2 \]

Direction of acceleration:

\[ \theta = \tan^{-1}\left(\frac{a_t}{a_c}\right) = \tan^{-1}\left(\frac{2.0}{1.6}\right) = \tan^{-1}(1.25) = 51.34^\circ \]

The total linear acceleration is 2.56 m/s² at an angle of 51.34° from the radius vector toward the direction of motion.

WARI SECONDARY SCHOOL - PRE-NATIONAL EXAMINATION SERIES

PHYSICS 1 - SERIES 13 | 131/1

No comments

Post a Comment

© all rights reserved
made with by templateszoo