Basic Physics

In this section, you will learn how dimensional analysis helps us create and verify equations, deepening your understanding of the relationships between physical quantities and enhancing your problem-solving skills.

You will also apply basic mathematical concepts, such as trigonometric ratios and the Pythagorean Theorem, to find resultants and break down vectors into components, making it easier to determine the overall effect of multiple forces on systems or structures.

We will explore the concept of floating force (buoyancy) and density to understand why some objects float while others sink. Additionally, you will study how materials behave, focusing on their ability to return to their original shape after being stretched, compressed, or deformed (elasticity), how stiff or resistant they are to stretching (Young’s modulus), and the maximum force or stress they can handle before breaking (breaking stress). These concepts are essential for understanding how materials respond to forces in engineering and everyday applications.

KEY IDEAS

· Dimensional analysis is essential for validating and deriving the relationships between different quantities in physics.

· The understanding of vectors is essential for determining the components of forces as well as determining the overall effect of multiple forces on objects and systems.

· Archimedes’ Principle: When a body is wholly or partially immersed in a fluid, it experiences an upward force (upthrust) which is equal to the weight of fluid displaced.

· Principle of flotation: A floating body displaces its own weight in the fluid in which it floats.

· Elastic deformation is a type of deformation in which the body regains its shape and size after the applied force is removed, while plastic deformation is a type of deformation in which the body does not regain its shape and size after the applied force is removed.

· Hooke’s law: Provided the elastic limit is not exceeded, the extension in an elastic material is proportional to the load or applied force.

· Whilst the stiffness constant of a spring or piece of wire is dependent on its physical dimensions, the Young’s Modulus is not. It is the ratio of the stress (force per unit area) on a material to the stress (extension per unit length) on that material.

Dimensional analysis is the study of the relationship between physical quantities by consideration of their dimensions. In Year One, the concept of dimensions was introduced, focusing on how to determine the dimensions of various physical quantities and how to derive their units based on these dimensions.

Understanding the dimensions of quantities is crucial not only for identifying their respective units but also for validating equations in physics. By using dimensional analysis, it becomes possible to check whether an equation is dimensionally consistent (the dimensions are the same on both sides of the equation), which serves as a preliminary verification step before delving into more complex mathematical proofs.

Additionally, dimensional analysis can aid in deriving relationships between different physical quantities. For instance, if it is assumed that the acceleration of an object is the only factor which determines the resultant force acting on it, we find that the dimensions do not match:

F ∝ a F = ka [MLT⁻²] ≠ [LT⁻²] Furthermore, by analysing the difference between the dimensions on each side of the equation, we can see that resultant force must also be dependent on mass.

Therefore, F ∝ ma F = kma We can find experimentally that k=1, and so, F = ma

Activity 1.1 Recalling the Dimensions

List the 7 fundamental quantities, including their symbols, which you met in Year One 1.

2.

3.

4.

5.

6.

7.

Activity 1.2 Checking Validity Using Dimensions

1. First, you should read through the worked example given below. If you are unclear about any of the steps in the solution, ensure that you ask your teacher or discuss with a peer!

2. Next, try the practice questions which follow independently.

Worked example

The initial and final velocities of a moving object are u and v, respectively. If the acceleration of the object is a and the time of motion is t, check whether the expression, v = u + at is dimensionally correct.

Solution

Step 1: Identify the Physical Quantities and Their Dimensions

a. Final velocity (v): L T⁻¹b. Initial velocity (u): L T⁻¹

c. Acceleration (a): L T⁻²d. Time (t): T

Step 2: Write the Equation in Dimensional Form

[v] = [u] + [a] × [t]

Step 3: Substitute the Dimensions and Simplify

a. Left-hand side:

[v] = L T⁻¹b. Right-hand side: [u] = L T⁻¹, and [a × t] = L T⁻²× T = L T⁻¹Thus, the right-hand side becomes:

[u] + [a × t] = L T⁻¹+ L T⁻¹= 2L T⁻¹But figures are dimensionless constant, therefore [u] + [a × t] = L T⁻¹Step 4: Compare Both Sides and Conclude Both sides of the equation have the same dimensions: L T⁻¹Since both sides of the equation have the same dimensions, the equation is dimensionally consistent.

Practice Problems

Now, using the worked example as a guide, solve the following problems individually or in groups.

Consider the equation of motion given by s = ut + 1/2 a t²(where s is the displacement, u is the initial velocity, a is the acceleration, t is the time). Using dimensional analysis, check whether the given equation is dimensionally correct.

Suppose an equation is proposed: d = v²t (where d is distance, v is velocity, and t is time). Check if this equation is valid by analysing its dimensions.

Activity 1.3 Checking Validity and Establishing Quantities Using

Dimensions

1. First, you should read through the worked example given below. If you are unclear about any of the steps in the solution, ensure that you ask your teacher or discuss with a peer!

2. Next, try the practice questions which follow independently.

Worked example

It is known that the time period depends only on the length l of the pendulum and the acceleration due to gravity g. Using dimensional analysis, derive the formula for the time period T.

Solution

Step 1: Assume a relation between the physical quantities ( T, L, and g) Assume that the time period T is related to L and g as:

T ∝ lᵃgᵇthen introduce a constant of proportionality k T = k lᵃgᵇwhere k is a dimensionless constant, and a and b are unknown powers we need to find using dimensional analysis.

Step 2: Identify the dimensions of physical quantities

a. Period (T): T

b. Length (l): L

c. Gravitational acceleration (g): LT⁻²Step 3: Write dimensions for both sides of the equation and apply your knowledge in indices Using the assumed relation, write down the dimensions of both sides of the equation.

T = Lᵃ( LT⁻²)ᵇT = LᵃLᵇT⁻²ᵇT = Lᵃ+b T⁻²ᵇStep 4: Equate the dimensions of both sides Now, compare the dimensions of both sides of the equation.

a. Left-hand side is [T] = T.

b. Right-hand side is L{a+b} T{−2b}.

For dimensional consistency, the dimensions of both sides must be the same.

Equate the powers of L on both sides of the equation (note: L does not appear on the left-hand side of the equation, and so the power of L = 0 on this side): a + b = 0 Equate the powers of T on both sides of the equation: -2b = 1

Step 5: Solve the system of equations From -2b = 1, we get:

b = − 1_ 2 Substituting b = − 1/2 into a + b = 0:

a − 1_ 2 = 0 = > a = 1_ 2.

Step 6: Write the final equation Now substitute a = 1/2 and b = − 1/2 into the assumed relation:

T = k L{1/2} g {−1/2}.

This simplifies to:

T = k √L_ g Practice Problems Now, using the worked example as a guide, solve the following problems individually or in groups.

Derive the equation for the time taken t for an object to fall from a height h under gravity g using dimensional analysis. Assume that no other factors exist after that time.

For further information and details about the application of dimensional analysis, visit the video link below:

https://www.youtube.com/watch?v=sk9BUMBK6hU

Activity 1.4 Discussion on the importance of Dimensions Objective: To understand the importance of dimensional analysis in everyday life.

Materials needed

1. Textbook or reference material

2. Internet

3. Worksheet What to do

1. Identify a friend or friends to form a small learner group.

2. Work in groups or pairs to research and discuss real-world examples where dimensional analysis is used.

3. Engage in group discussions with the following discussion prompts:

a. Can you think of a time when you have used dimensional analysis in your everyday life, even without realising it?

b. How can dimensional analysis help prevent errors in calculations and measurements?

c. What are the potential consequences of using incorrect units or formulas?

d. Can dimensional analysis be applied to other areas of life besides science and engineering? If so, how?

4. Each group should appoint one member to summarise and record the answers from the discussions.

5. After the discussions, each group should share one key point from their discussion with the class or peers.

See Annex A for the solutions to some activities on dimensional analysis

Vectors are a class of physical quantities that, if we look carefully, we can identify numerous examples of in our day-to-day lives. What makes them special is that they have both magnitude and direction, unlike their counterparts (scalars), which have only magnitude.

For example, when a mango fruit falls from the tree, it travels downward due to its weight.

Figure 1.1: A mango falling under gravity Weight, a type of force, is therefore an example of a vector as it has both a magnitude and a direction (in this case, downward). Other examples of vectors are momentum, velocity, acceleration, etc.

A vector is typically represented with a straight line bearing and an arrow. The length of the line signifies the magnitude of the vector, while the arrow indicates the direction of the vector. If the two vector arrows shown in Figure 1.2 represent the same quantity (e.g. both represent velocities), vector (A) is a larger velocity than vector (B).

(A). A vector acting in a horizontal direction (B). A vector acting in a vertical direction

Figure 1.2: Vectors acting in horizontal and vertical directions

Activity 1.5 Drawing vectors using scale diagrams Using a scale of 1N = 1cm, draw the following force vectors:

1. 5N acting horizontally to the left

2. 7.5N acting horizontally to the right

3. 4N acting at an angle of 35 degrees clockwise from vertically upwards.

Resolving Vectors

Vectors do not always act in the horizontal or vertical directions. When a vector acts at an angle less than 90° to the horizontal, it is said to be inclined. For example, when you kick a football over a wall, the force exerted by your foot on the ball acts in a line inclined to the ground. An inclined vector has both horizontal and vertical components (i.e. a component which acts to move the ball horizontally and a component which lifts the ball into the air vertically).

(A) An inclined vector (B) an inclined vector with both horizontal and vector components shown

Figure 1.3: Inclined vectors Resolving an inclined vector into two components Simple trigonometric functions can be used to find the two components of a vector acting at an angle to the horizontal or at an angle to an inclined plane.

Consider the diagrams in Figure 1.3 (B). If the magnitude of the vector F is 20 N and the angle of incline is 30 degrees, then we can use the sin and cos functions on the calculator to find Fx and Fy :

Hypothenuse = F = 20 N Opposite = Fy Adjacent = Fx θ = 30 degrees sin θ = opp___ hyp Therefore:

opposite = hypothenuse × sin θ = 20 sin 30 = 10 N cos θ = adj___ hyp Therefore:

adjacent = hypothenuse × cos θ = 20 cos 30 = 17.3 N Hence, the vertical component of F, Fy, is 10 N and the horizontal component of F, Fx is 17.3 N.

Finding the resultant of two perpendicular vectors When two vectors act perpendicular to each other at a point, we might want to know the direction and magnitude of the single vector resulting from their action.

For example, if you walk 100 m due East and 50 m due north, what would be the single displacement joining your origin and destination? That is your resultant displacement.

Figure 1.4: The resultant displacement after both horizontal and vertical motion When two forces act on the same object, which can be treated as a single point, simultaneously, we can draw a free-body diagram as shown in Figure 1.5 (A).

The diagram can then be re-drawn, with the vector arrows places top-to-tail, as shown in Figure 1.5 (B).

(A) Two perpendicular vectors F_(y) and Fₓ (B) Fₓ and F_(y) have a resultant F

Figure 1.5: Perpendicular vectors To determine the resultant of any two perpendicular vectors, as in Figure 1.5 (A).

Re-draw the vector arrows top-to-tail as shown in Figure 1.5 (B).

Draw a diagonal from the original meeting point (A) of the vectors to the meeting point (B) of the duplicate vectors.

Applying the Pythagorean theorem, the magnitude of F = √Fₓ ²+ F_(y) ²Determine the direction of the vector as θ = tan⁻¹( F_(y)__ Fₓ)

Activity 1.6 Finding the resultant displacement of an object Use the diagram in Figure 1.4 to determine

1. the resultant displacement

2. the direction of the resultant displacement of the object, which has travelled 100km east and then 50km north.

Finding the resultant of non-perpendicular vectors Sometimes, vectors act at angles less than or greater than 90° to one another. In such cases, we use different methods to determine the magnitude and direction of their resultants.

Look at the vector diagrams below. Imagine two trucks using tow ropes to pull a boulder in the direction of F₁and F₂,respectively. For each situation, a and b, we can determine which direction the boulder will move in and with how much resultant force.

a) θ < 90° between F₁ and F₂ b) θ > 90° between F₁ and F₂

Figure 1.6: Two non-perpendicular vectors To determine the resultant of such vectors

1. Redraw the vector arrows top-to-tail to form a parallelogram (see Figure 1.7 below).

2. Draw a diagonal from their origin to the other vertex of the parallelogram.

This is the resultant vector.

Figure 1.7: Resultant of two non-perpendicular vectors using the parallelogram law of vector addition F is the resultant, making the angle α with the horizontal

3. To calculate the magnitude of the resultant, apply the cosine rule (see more in Annex B – Further Information):

F = √___________________ F₁ ²+ F₂ ²+ 2( F₁ × F₂)cosθ or F = √___________________ F₁ ²+ F₂ ²− 2( F₁ × F₂)cosβ, where β = 180 − θ

4. Calculate the angle of the resultant with respect to the horizontal using the following mathematical rule called the sine rule (see more in Annex B – Further Information):

F_ sinβ = F₁_ sinα

5. Make α the subject is as follows α = sin⁻¹(F₁ sinβ_ F )

Activity 1.7 Resolving vectors

1. Find the horizontal and vertical components of the following vectors:

2. A force of 10.0 N is applied to a ball at an angle of 30° to the horizontal.

Determine the horizontal and vertical components of this force.

3. Looking at the image below of iron filings showing the shape of the magnetic field around a bar magnet, discuss with your peer where the magnetic field has only horizontal components and where it has both horizontal and vertical components. If the equipment is available, use a bar magnet and some iron filings to see if you achieve the same pattern as in the image below!

Note: If you try this for yourself, ensure that the bar magnet is placed under a piece of paper and the iron filings are sprinkled on top of the piece of paper (to stop them from just sticking to the magnet).

Figure 1.8: Iron filings surrounding a bar magnet

Activity 1.8 Determining the resultant of two vectors

1. A tennis player kicks a ball at an angle to the ground. If the ball travelled 10 m up and 8 m to the right at the same time, calculate

a. the inclined distance it travelled before beginning to fall.

b. the angle at which the ball was kicked.

2. Two forces are pulling a boulder; assume F₁= 20 N is directed between north and east and F₂= 15 N is directed eastward. If the angle between them is 60⁰, determine the

a. magnitude of the resultant force on the boulder and

b. direction (angle) of its motion with respect to the east.

See Annex A for solutions to activities on vectors. See Annex B for further information on vectors.

Imagine you have two boxes of the same size. One box is filled with feathers, and the other is filled with bricks. Even though the boxes are the same size, they feel very different. The box with the bricks is much heavier, right?

That’s because this volume of bricks contains a much greater mass (more particles) than this volume of feathers. Thus, we say the bricks have a higher density.

Density measures how much stuff is packed into a certain space. It’s like how crowded a room is. If there are a lot of people in a small room, it’s very dense. If there are only a few people in a big room, it’s less dense.

Equal volumes of different substances have different masses. This is due to differences in their densities. Density helps us to understand why some objects float while others sink, and why certain materials feel heavier than others, even if they are the same size.

The density of water is 1000 kgm⁻³. This means that a metre cubed of water has a mass of 1000 kg. In a similar way, an object (e.g. gold) of density 19300 kgm⁻³ means that a metre cubed of gold has a mass of 19300 kg.

Density is defined as the mass per unit volume of a substance.

Mathematically, Density = Mass______ Volume The S. I. Unit of density is kgm⁻³.

Activity 1.9 Discussion on why ships float but metal sinks

1. Identify a friend or friends to form a small learner group

2. Engage in group discussions by asking your friends “why do they think a heavy piece of metal sinks in water, but a large metal ship loaded with goods can float?”

Figure 1.11: Diagram showing a Ship floating

Figure 1.12: Diagram showing Stones Sinking

3. Let your friends think individually for two minutes and write down their thoughts.

4. Share your ideas with your friends to identify similarities and differences.

5. Discuss the similarities and the differences and come out with your conclusions on why you think a heavy piece of metal sinks in water, but a large metal ship loaded with goods can float.

6. Engage your colleagues in a brief discussion about what they think could cause a boat, designed to float, to start sinking. Try to include all of the following key terms in your explanation: density, buoyant force, surface area, and displacement.

Figure 1.13: Diagram of a Ship Sinking

Activity 1.10 Calculating density Read the two worked examples below before attempting the questions that follow.

Worked examples

1. A block of aluminium has a mass of 540 grams and a volume of 200 cm³.

What is the density of the aluminium in gcm⁻³?

2. A piece of gold has a density of 19.3 gcm⁻³and a volume of 10 cm³. What is its mass in grams?

Step-by-Step Solution for Question 1

1. Write the formula for density Density = Mass______ Volume

2. Substitute the given values Density = 540 g/200 cm³3. Perform the division:

Density =2.7 gcm⁻³Step-by-Step Solution for Question 2

1. Rearrange the density formula to solve for mass Mass = Density × Volume

2. Substitute the given values M = 19.3 gcm⁻³×10 cm³3. Perform the multiplication M = 193 g Practice Problems Now, using the worked example as a guide, solve the following problems individually or in groups.

1. A metal block has a mass of 1.5 kg and a volume of 0.0003 m³. Calculate the density of the metal.

2. A gold bar has a volume of 0.0005 m³, and the density of gold is 19,320 kg/ m³. Calculate the mass of the gold bar.

3. A metal object has a mass of 2.4 kg, and its density is 8,000 kg/m³. Calculate the volume of the object.

Activity 1.11 Determining Density of Objects Experimentally

Objective

1. To experimentally determine the density of various regular and irregular- shaped solid objects and of liquids.

2. To understand the relationship between mass, volume, and density.

Materials

1. Regular-shaped objects (e.g., spherical marble, cuboid piece of wood, cube or cuboid plastic block)

2. Irregular-shaped object (e.g., stone, small plastic toy animal)

3. Liquid (cooking oil, water)

4. Balance or digital scale

5. Measuring cylinder or graduated beaker

6. Vernier callipers or ruler

7. Displacement tank or overflow can Procedure

1. Measure mass

a. Place each object on the balance or digital scale and record the mass.

b. Create a table to record the mass of each object.

Table 1.1: Table to record density values of objects Object Mass /g Volume/cm³Density/ g cm⁻³

2. Measure volume

a. Regular-shaped objects

i. Use vernier callipers or a ruler to measure the length, width, and height of each solid object.

ii. Calculate the volume of each solid object using the appropriate formula (e.g., for a cube: Volume = length × width × height, for a sphere: volume = 4/3 × pi × radius³).

b. Liquid

i. Pour the liquid into the measuring cylinder or graduated beaker.

ii. Read the volume of the liquid from the markings on the container.

Note: 1mL = 1cm³c. Irregular-shaped objects: Displacement method

i. Fill the displacement tank or overflow can with water to the brim.

ii. Carefully submerge the irregular-shaped object into the water.

iii. Collect the displaced water in a measuring cylinder and record the volume.

Note: 1mL = 1cm³Figure 1.14: Diagram demonstrating the measurement of volume of the irregular objects

3. Calculate density

1. Use the formula: Density = Mass______ Volume

2. Calculate the density of each object using the measured mass and volume.

3. Record the calculated density in the table.

4. Analyse results

1. Compare the densities of the different materials.

2. Discuss any trends or patterns observed in the densities.

3. Explain the differences in density based on the properties of the materials.

Note:

1. Ensure accurate measurements of mass and volume.

2. Use consistent units for mass and volume (e.g., grams and cubic centimetres).

3. Handle materials carefully, especially when dealing with liquids.

When a body is immersed in a fluid (liquid or gas), the fluid exerts an upward force on the body. This force, known as upthrust or buoyant force, opposes the weight of the body.

Archimedes’ Principle states that, when a body is wholly (fully) or partially immersed in a fluid (liquid or gas), there is an upthrust (upward force) which acts on the body and is equal to the weight of the fluid displaced by the body.

As the weight of an object is equal to its mass, m, multiplied by the gravitational field strength, g, the up thrust, u, on a body is:

u = mg (weight of fluid displaced) As the mass of the displaced fluid depends on its density, ρ, and volume, v, according to:

m = ρv We can combine the two formulae to give the relationship:

u = ρvg Since the volume of an object with a uniform area of cross-section can be found by multiplying the cross-sectional area of its base by its height, we can say:

v = Ah

Figure 1.15: The volume of an object of uniform cross-sectional area Therefore, upthrust on such an object, u = ρAhg

Activity 1.12 Calculating upthrust An object of volume 4 m³is totally immersed in a liquid of density 1080kg/ m³. Calculate the upthrust of the liquid on the object. (g = 10 m/s²) Flotation When the upthrust (buoyant force) is equal to the weight of the body in air (Wₐ), the apparent weight of the body in the fluid (W_(f)) becomes zero.

Wₐ- u = W_(f) = 0 In this case, the body experiences no net downward force, making it appear weightless in the fluid. This results in the body floating.

The principle of floatation states that “A floating body displaces its own weight of fluid in which it floats”.

Activity 1.13 Experimental verification of Archimedes Principle Objective: Verify Archimedes’ principle by measuring the buoyant force on a submerged object. Note: in the absence of practical equipment, use an online simulation to observe Archimedes’ Principle:

ophysics.com/fl1.html Materials needed

1. Spring balance

2. Beaker of water

3. Metal block capable of sinking

4. Large container with water that can serve as a eureka can by creating a hole at the side

5. Measuring cylinder What to do

1. Measure the weight of the object in air using a spring balance.

2. Submerge the object completely in water and measure the apparent weight in water.

3. Collect the water displaced with a beaker and determine the weight of the water displaced

Figure 1.16: Diagram demonstrating experiment to verify Archimedes’ principle

4. Calculate the upthrust or Buoyant Force = Weight in Air−Weight in Water

5. Compare the buoyant force to the weight of water displaced by the object and draw your conclusion

6. You can now verify Archimedes’ principle.

Activity 1.14 Experimental verification of the principle of flotation Objective: To verify the principle of flotation Materials needed

1. Spring balance

2. Beaker of water

3. Wooden block capable of floating

4. Measuring cylinder

5. Large container with water that can serve as a eureka can by creating a hole at the side What to do

1. Measure and record the weight of the block of wood.

2. Gently drop the block of wood into the large container containing the water.

3. Collect the water displaced with a beaker and determine the weight of the water displaced.

4. Compare the weight of the block of wood to the weight of the fluid displaced and draw your conclusion.

5. You can now verify the principle of flotation.

Extension Try repeating Activities 1.13 and 1.14 using fluids other than water (e.g. oil, saltwater). Draw a conclusion as to how the density of the fluid affects the buoyant force on the object, and whether the object sinks or floats as a result.

Activity 1.15 Research on real-life application of density and flotation Now that you have studied density and flotation, research a real-life application of these concepts (e.g., hot air balloons, life jackets and submarines). Work in groups of no more than 4 learners.

Present your research to another group or to the whole class in the form of a written or oral presentation.

Review Questions – Dimensional Analysis and Vectors

1. State three uses of dimensional analysis.

2. Check the validity of the equation: pv = mv²- mgh where P represents momentum by a body, V represent the velocity of the body, m represents the mass of the body, h represents the height of the body above the ground and g represents the acceleration due to gravity.

3. The period T of oscillation of a mass m attached to a spring might depend on: The mass m, the spring constant k, and the acceleration due to gravity

g. Using dimensional analysis, determine how the period T of the mass- spring system depends on these quantities.

4. A car climbs a hill with an acceleration of 5 ms⁻². If the road makes an angle of 20° with the horizontal, calculate the horizontal and vertical components of the car’s acceleration.

5. A hunter lies on the ground and aims his gun at a bird on a tree. The angle between the gun and the ground is 50°. When he shoots at the bird, the bullet travels a total displacement of 50 m. Calculate the

a. distance on the ground between the gun from the base of the tree, and

b. the position of the bird from bottom of the tree

Review Questions – Density and Flotation

1. Define density.

2. State Archimedes’ principle.

3. What condition must be met for an object to float on a fluid?

4. A cube has a side length of 5 cm and a mass of 500 g. What is its density?

5. A solid object has a volume of 0.5 m³ and is completely submerged in a fluid. If the density of the fluid is 1000 kgm⁻³, calculate the buoyant force on the object.

6. An object with a mass of 500 g displaces 400 cm³ of water when floating.

Will the object sink or float in water, and why?

7. Why does ice, which is solid water, float on liquid water, even though both are made of the same substance?

8. You have a solid block of unknown material and are tasked with identifying its composition using only a scale and a graduated cylinder filled with water. How would you determine the material’s identity?

Review Questions - Deformation

1. Define Elastic Deformation.

2. Fill in the table below by grouping the following scenarios under elastic and plastic deformations: A rubber band stretching when pulled, A paperclip bending but returning to shape, A metal wire being twisted, A spring compressing and then expanding, A piece of clay being shaped into a sculpture, A car frame bending in an accident.

Elastic deformation Plastic deformation

Review Questions – Hooke’s Law and Young’s Modulus

1. State Hooke’s law.

2. What are the two physical quantities used to calculate Young’s modulus?

3. A car’s suspension system uses a spring that compresses by 0.15 metres when a load of 3000 N is applied.

a. What is the spring constant of the suspension spring?

b. How much energy is stored in the spring when it is compressed by 0.15 metres?

4. A steel wire of length 2 m and cross-sectional area 1 mm²is stretched by 0.5 mm when a force of 1000 N is applied. Calculate the stress and strain in the wire. Then, determine Young’s modulus for the steel wire.

5. Explain how Young’s modulus defines the stiffness of a material. How does the modulus relate to the ability of a material to resist stretching or compressing when a force is applied?

6. A copper wire has a length of 3 m and a cross-sectional area of 2 mm².

The wire is stretched by 0.25 mm when a force is applied. Given that the Young’s modulus of copper is 1.1×10¹¹Pa, calculate the force applied to the wire.

7. Two springs, A and B, have spring constants of 200 N/m and 300 N/m, respectively. A force of 50 N is applied to both springs. Which spring stretches more, and why? Use Hooke’s Law in your explanation.

8. Two wires, A and B, are made of different materials with Young’s moduli of 200 GPa and 150 GPa respectively. Both wires experience the same tensile stress. Which wire will experience a greater strain, and why? Explain using the relationship between stress, strain, and Young’s modulus.

9. You are designing a container to transport a substance by sea. The material of the container is very dense, but you still want it to float. What strategies can you use to ensure it will float, and why will they work?

Review Questions – Density and Flotation

1. Density is the mass per unit volume of a substance.

2. Archimedes’ Principle states that, when a body is wholly (fully) or partially immersed in a fluid (liquid or gas), there is an upthrust (upward force) that acts on the body and is equal to the weight of the fluid displaced by the body.

3. An object floats on a fluid if the weight of the object is equal to the buoyant force exerted by the displaced fluid. This occurs when the density of the object is less than or equal to the density of the fluid.

4 gcm⁻³4900 N

4. Since the density of the object (1.25 gcm⁻³) is greater than the density of water (1 gcm⁻³), the object will sink.

5. Ice floats on water because it has a lower density than liquid water. When water freezes, its molecules arrange in a crystal structure that takes up more space, making ice less dense than liquid water.

6. First, measure the mass of the block using the scale.

Next, place the block in the graduated cylinder and measure the volume of water displaced. This gives the volume of the block.

Use the formula for density: ρ = m/v , where m is the mass and V is the volume.

Once the density is calculated, compare it to known densities of materials to identify the block’s composition.

Review Questions – Deformation

1. Elastic deformation is a temporary change in shape or size of a material when a stress is applied. The material returns to its original shape once the stress is removed.

2.

Elastic deformation Plastic deformation A rubber band stretching when pulled A metal wire being twisted A paperclip bending but returning to shape A piece of clay being shaped into a sculpture A spring compressing and then expanding A car frame bending in an accident

Review Questions – Hooke’s Law and Young’s Modulus

1. Hooke’s law is stated as provided the elastic limit is not exceeded, the extension, e in an elastic material is proportional to the load or applied force.

2. The two physical quantities used to calculate Young’s modulus are stress (force per unit area) and strain (the ratio of change in length to original length).

3. a. 20,000 N/m

b. 225 J

4. Stress = 1 × 10⁹Nm⁻², Strain = 2.5 × 10⁻⁴, Young’s Modulus = 4 × 10¹² Nm⁻²5. Using Hooke’s Law:

For spring A, the spring will stretch by 0.25m and For spring B: the spring will stretch by =0.167m.

Therefore, Spring A stretches more than spring B because it has a lower spring constant, meaning it’s less stiff.

F = 18,333 N (note 2mm²= 0.000002m²)

6. Young’s modulus defines the stiffness of a material by measuring its ability to resist deformation under stress. It is the ratio of stress (force per unit area) to strain (deformation as a fraction of the original length). A material with a high Young’s modulus is stiff, meaning it deforms very little when subjected to force, while a material with a low modulus is more flexible and deforms more easily.

7. Young’s Modulus is the ratio of tensile stress to tensile strain. Since both wires experience the same tensile stress, the strain is inversely proportional to the Young’s modulus. Wire A has a higher Young’s modulus (200 GPa) than wire B (150 GPa). Therefore, wire B will experience a greater strain because a lower Young’s modulus means more deformation for the same stress.