Garam

 

  1. Garam
  2. Garam Terlarut dan Garam Tak Terlarut
  3. Penyediaan Garam Terlarut
  4. Penyediaan Garam Tak Terlarut
  5. Ciri-ciri Fizik Hablur
  6. Penyediaan Persamaan Ion Melalui Kaedah Perubahan Berterusan
  7. Penghitungan Melibatkan Persamaan Stoikiometri
  8. Analisis Kualitatif Garam
  9. Ujian Gas
  10. Tindakan Haba Terhadap Garam
  11. Ujian Anion
  12. Ujian Kation

Archimedes’ Principle

In this lesson, we are going to discuss Archimedes’ Principle, where we will discuss what is upthrust? Why objects float in liquid? The relationship between upthrust and density, and what is stated in Archimedes’ Principle. After that we will discuss the equation of Archimedes Principle. Under the equation, we will learn how to find the volume of the displaced liquid and how to find weight of an object in a liquid and the upthrust exerted on it by using the same equation.

After that, we will discuss forces exerted on an object immerses in water in 3 difference cases. Last but not least, we will discuss the applications of Archimedes’ Principle such as hydrometer, hot air balloon, submarine and airship.

Click on the branch “uptrust” or the “next slide” button to start.

In coming slides, we will discuss why objects float in liquid, upthrust and density and what is stated in Archimedes’ Principle and what does it mean.

Click on any of the question clouds to see the explanation.

If a block is immersed in water, the pressure exerted on the upper surface of the block will be the atmospheric pressure, while the pressure exerted on the bottom of the block will be equal to the atmospheric pressure + the liquid pressure. As we can see, the pressure exerted on lower surface is higher than the pressure exerted on the upper surface. The difference of the pressure produces a force heading upward, called the upthrust or buoyant force. This is the force that cause objects float on water.

If an object is partially or fully immerse in water, 2 forces are acting on it: its own weight and the upthrust.

If the density of the object is greater than the density of the liquid, its weight will be greater than the upthrust, and the object will sink to the bottom of the liquid.

If the density of the object is lower than the water, the object will partially immerse in the liquid and the upthrust is equal to the weight of the object.

If the density of the object is equal to the density of the liquid, the object will fully immerse in the water and the upthrust is equal to the weight of the object. The object can stays in any position inside the water.

Archimedes Principle states that when a body is wholly or partially immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced.

For example, in the figure, there is an object of weight 10N hanging on a spring balance. If the object is immerse in water, we will find that the reading on the spring balance will decrease. This is because part of the weight is supported by the upthrust produced by the water. Let’s say, the new reading is 8N. So, what is the magnitude of the upthrust? The weight reduced by 2N owing to the upthrust. Therefore, the upthrust must also be 2N.

At the same time, some water will be displaced by the object and flows into the beaker. If we measure the weight of the water displaced, we will find that it is also 2N.

According to Archimedes’ Principle, the magnitude of the upthrust must be equal to the weight of the displaced fluid. If the weight of the displaced fluid is 2N, then the upthrust must also be 2N. This is what stated in Archimedes Principle. It said that the upthrust is equal to the weight of the displaced fluid. A fluid can be liquid or gas.

According to the Archimedes’ Principle, the upthrust exerted on an object immersed in a fluid is give by the equation, F = rVg, where F is the upthrust; r is the density of the fluid; V is the volume of the displaced fluid and g is the gravitational field strength. Take notes that the rV is actually equal to the mass of the fluid. A mass of an object can be calculated by the equation m = rV. You should have learned this in your PMR science.

When you want to use this equation to solve problems related to object immersed in fluid, you need to know 2 important notes. One is related to the volume of the fluid and another is related to the density of the fluid. Let’s begin with the volume of the displaced fluid.

According to Archimedes Principle, the upthust is equal to the weight of the displaced fluid, which is given by a formula, rVg. Take notes that, the V is the volume of the displaced fluid. However, in practice, sometime, students may confuse with the volume of the displaced fluid and the volume of the object.

Let’s look at these 2 cases. If an object is fully immersed in a fluid, the volume of the displaced fluid is equal to the volume of the object. Normally, students will have no problem with this.

However, if the object is partially immerse in the fluid, , like the one in the diagram, the volume of the displaced fluid is not equal to the volume of the object, but equal to the volume of the object immerse in the fluid. For example, if the volume of the block is 100cm3, 80 cm3 is in the water, 20cm3 is above the water level, then the volume of the displaced fluid is 80 cm3, but not 100 cm3. If an object is partially or fully immerse in water, 2 forces are acting on it: its own weight and the upthrust.

The upthrust is given by the formula F = rVg. The weight of the object can be determined by the formula, W = mg. We have just learned that, m = rV. Therefore, the weight of the object is also equal to rVg. Students always confuse with these 2 formulae because it look almost the same. Is there any difference between this 2 formulae?

Well, the upthrust is equal to the weight of the displaced fluid. Therefore, for upthrust, the density must be the density of the fluid but not the density of the object and the volume V must be the volume of the displaced fluid.

For the weight of the object, the 8density must be the density of the object and the 8volume V must be the volume of the object.

In short, upthrust is equal to the weight of the displaced fluid, therefore the r must be the density of the fluid and the V must be the volume of the displaced fluid. For the weight of the object, the r must be the density of the object and the V must be the volume of the object.

Click on the “example” button” to see an example of question related to this. You may skip the example and proceed to “Forces in 3 cases” by clicking on the “continue” button. The weight of a copper block is 50N. Find the upthrust that act on the block when it is fully immerse in water. [ Density of copper =   9000 kg m-3; Density of water =  1000kg m-3 ]

Well, in this question, we are asked to find the upthrust. The upthrust is equal to rVg. g is the gravitational field strength, equal to 10N/kg. Now, we have 2 densities given in the question: the density of the copper, equal to 9000 kg/m3 and the density of the liquid, equal to 1000kg/m3.  Which density should we use for our upthrust? Well, since upthrust is equal to the weight of the displaced water, therefore, we should use the density of the water to find the upthrust, but not the density of the copper.

V is not given in the question. Instead, we are given the density and weight of the block. Since the block is fully immersed in water, hence the volume of the displaced water will be equal to the volume of the object. In previous slide, we have learned that the weight of the object is equal to rVg as well. The weight is 300N; Density is 9000kg/m3. Since this is the weight of the copper block, hence we must use the density of copper, but not the density of water. g is equal to 10N/kg. Solve the equation, we get V = 0.005m3.

Let’s come back to the upthrust. The density is 1000; the volume is 0.005; the g is 10. Therefore, the upthrust is equal to 50N. I suppose we have learned almost everything that you need to know about Archimedes principle. Here we will discuss a few cases involve a block in water. In these cases, we should pay attention on the forces involve. This is important for us to solve the calculation questions.

  1. Let’s start with the first case. If an object which is less dense than water is place into water, it will partially immerse in the water. Since the object is not moving, hence the force must be in equilibrium. Therefore, the upthrust is equal to the weight of the object.

However, if the object is tied to an iron bar by a string, it may sink to the bottom. Now, in this case, there are 2 forces pulling the object down: the weight and the tension of the string. Since the object is not moving, hence all the forces must also be in equilibrium. Therefore, the upthrust is equal to the weight + the tension. In exam, normally you will be asked to find the tension of the string.

Let’s look at another case. An iron block is hung in water by a string. There are 3 forces acting on the block: the weight, the tension of the string and the upthrust. Both the tension and the upthrust are pointing upward. The block is not moving, which means all the forces are in equilibrium. Therefore we conclude that the tension + the upthrust = the weight.

Knowing the forces acting on the object may help you to solve some of the problems related to Archimedes’ principle.

 

Gas Pressure and Atmospheric Pressure

In previous lesson, we have discussed liquid pressure, where we discussed the formula of liquid pressure, characteristics of liquid pressure and some applications of liquid pressure.

In gas pressure, we are going to explain how gas pressure is produced by using the kinetic theory of gas. After that, we will shift to atmospheric pressure where we will discuss the characteristics of atmospheric pressure, proofs or phenomena involving atmospheric pressure, how to measure atmospheric pressure and some applications of atmospheric pressure.

Let’s begin with kinetic theory of gas. We use kinetic theory of gas to explain how gas pressure is produced.

There are many assumptions in kinetic theory of gas. However, in SPM, we only focus on 2 assumptions, namely:

  1. The gas consists of very small particles, all with non-zero mass.
  2. These particles are at constant and random movements; hence constantly collide with the walls of the container.

Now, let’s use these 2 assumptions to explain how gas pressure is produced in a container

Gases consists of very small particles which are at constant and random movements. The particles constantly collide with the walls of the container. When the particles collide with the wall of the container and bounce back, they experience a change in momentum. The momentum change exerts a force on the wall and hence produces a pressure on the wall of the container. This is how the gas pressure is produced.

In SPM, you need to memorise this explanation because in exam you may be asked to explain how the gas pressure is produced by using the kinetic theory of gas. We have just discussed kinetic theory of gas. Let’s continue with atmospheric pressure.

Atmospheric pressure is the pressure exerted on the surface of any objects by the atmospheric gas. It is caused by the weight of the atmospheric gases above the earth surface. In SPM, you need to know some experiments that prove the existence of atmospheric pressure, including the crushing can experiment, the Magdeburg hemisphere and experiment where water covered with a cardboard in an upside down glass does not flow out.

We have just discussed the proofs of existence of atmospheric pressure. In this slide, we are going to discuss how to measure atmospheric pressure. Atmospheric pressure can be measure by Aneroid Barometer, Fortin Barometer and mercury barometer. In SPM, most of the questions related to measuring atmospheric pressure are questions regarding the mercury barometer. Therefore, the focus of our discussion is on the mercury barometer.

The diagram shows the illustration of a mercury barometer. A mercury barometer is also called a simple barometer. The green colour substance represents the mercury and there is vacuum at the upper end of the tube.

From my teaching experience, I know many students facing problem in finding atmospheric pressure by using a mercury barometer. Finding atmospheric pressure by using mercury barometer is actually very simple. If you follow this explanation step by step, you will find that it is just piece of cake.

Now, in order to learn how to use a mercury barometer to find atmospheric pressure, firstly, we must know how to find the pressure “in the mercury” in unit of cmHg if the column above the mercury is vacuum.

Second, we need to know how to find the pressure “in the mercury” in unit of cmHg if the column above the mercury is not vacuum.

Third, we need to know how to convert the unit of pressure from cmHg to Pascal.

Forth, we need to know some important notes regarding the gas pressure and liquid pressure. Last but not least, we learn how to find the atmospheric pressure by using the mercury barometer.

Next, we will learn how to find the pressure in the mercury in the unit of cmHg.

First of all, we need to know that the pressure of the vacuum is 0. At point A, the surface of the mercury is exposed to the vacuum, hence its pressure is equal to the pressure of the vacuum, which is 0.

The pressure of any point in the mercury is equal to the depth of the point from the vacuum. For example, point B is 24cm below the surface. Therefore, the pressure at B is 24 cmHg. How about point C? Point C is 75cm below the surface. The pressure is 75 cmHg. Lastly, point D. Point D is 83cm below the surface, therefore the pressure is 83cmHg.

If some gas is trapped in the space inside the tube, the pressure inside the tube is not 0 anymore. This will affect the pressure inside the mercury. The pressure inside the mercury will be the gas pressure + the mercury pressure.

For example, if the pressure of the gas is 20 cmHg, then the pressure at point B will become 20 + 24 = 44 cmHg. How about point C? Pressure at point C will become 20 + 75 = 95 cmHg. Lastly, point D. Pressure at point D is  103 cmHg.

So far, we have learned that the pressure (in the unit of cmHg) in the mercury of a mercury barometer is equal to the depth of the mercury from the vacuum. Therefore the pressure at B, C and D are 24, 75 and 83 cmHg respectively. However, sometime, we are asked to find the pressure in the unit of Pascal (Pa), but not cmHg. How do we convert the unit in cmHg to Pa?

We can convert the unit in cmHg to Pa by using the equation of pressure, P = hrg. For example, at point B, the depth, h = 24cm or 0.24m. The density of mercury, r = 13,600 kgm-3 (this will be given in the question) and the gravitational field strength, g = 10 Nkg-1. Substitute these values into the equation. Pressure at B equals to 32,640 Pa. At C, the depth is 0.75m, density and gravitational field strength is still the same. Substitute these values into the equation; the pressure at C is 102,000 Pa. The depth at D is 0.83m. Substitute this into the equation. Pressure at D is 112,880 Pa. This is how we convert the unit of pressure from cmHg to Pascal.

Sometime, the density may be given in g/cm3. You need to convert it to kg/m3 before finding the value of the pressure.

In SPM, usually, the density of an object is given in g/cm3 or kg/m3. Sometime, you will need to convert these units from one to another. You should have learned how to convert the unit of density in Chapter 1, introduction to physics.

However, we can convert these 2 units of density faster and easier by using a factor of conversion.

In chapter 1, you should have learned that 1g/cm3 is equal to 1000kg/m3. Therefore, we can convert g/cm3 to kg/m3 by multiplying a factor of 1000. For example, 1 g/cm3 = 1000kg/m3; 2 g/cm3 = 2000kg/m3; 3 g/cm3 = 3000kg/m3; 2.5 g/cm3 = 2500kg/m3…….

In opposite, we can convert kg/m3 to g/cm3 by dividing 1000. For example, 3000kg/m3 = 3g/cm3; 1500kg/m3 = 1.5g/cm3; 800kg/m3 = 0.8g/cm3; 200kg/m3 = 0.2g/cm3……

You are advised to memorise this factor of conversion because conversion of the unit of density is very common in Physics and Chemistry.

Before we learned how to find the atmospheric pressure by using a mercury barometer, there are a few things that we need to know.

  1. The difference of gas pressure at different height can be ignored if the difference of the height is not too big. For example, we can assume that the gas pressure at point D is equal to the gas pressure at point E even though point D is higher than point E. This is also true for the gas inside the container. The pressure at point A will be equal to the pressure at point F.
  2. The pressure on the surface of a liquid is equal to the pressure of the gas it exposed. For example, in the diagram, the pressure of the gas inside the container is 200,000Pa. Surface A is exposed to the gas inside the container. Therefore, the pressure at point A is also 200,000 Pa. The atmospheric pressure is 100,000Pa. Surface B exposes to the atmosphere, hence the pressure at point B is equal to the atmospheric pressure, which is 100,000 Pa.
  3. Pressure of a given liquid is the same at the same level. For example, A and C are at the same level. Therefore pressure at C is equal to pressure at A, which is 200,000Pa

These are the 3 important points that you need to know when using a mercury barometer to find atmospheric pressure. So, how do we determine the atmospheric pressure from a mercury barometer?

Well, we have just learned that the pressure on the surface of a liquid is equal to the gas pressure of the gas it exposed. Point A exposes to the atmosphere, hence the pressure at A is equal to the atmospheric presssure. Point B is at the same level as A, therefore, pressure at B is equal to pressure at A, which is also equal to the atmospheric pressure. We write, PA = atmospheric pressure. We demote atmospheric pressure by using the symbol Patm.

Now, let’s look at point B. Point B are at the same level as point A. Therefore, their pressure should be the same. Write this as PA = PB = the atmospheric pressure. Now, our problem is, how much is the pressure at point B? Point B is 76cm below the vacuum. Therefore, the pressure at point B is 76 cmHg.

Since the pressure at point B is also equal to the atmospheric pressure, hence we can deduce that the atmospheric pressure is equal to 76 cmHg. Yes! This is how we find the atmospheric pressure by using a mercury barometer. The atmospheric pressure is equal to the height of the mercury column in the tube.

We can convert the unit from cmHg to Pascal by using the formula P=hρg. In this case, h = 0.76m. The unit of h must be in meter. The density of mercury, ρ = 13,600 kg/m3. This will be given in the question. g is 10 N/kg. Plug in all these numbers into the equation. The atmospheric pressure is 103360 Pa. We have just finished our discussion on mercury barometer. Now we are going to continue with Aneroid barometer and Fortin barometer.

 

Pressure and Liquid Pressure

Force and Pressure is the 3rd chapter in SPM form 4 Physics. There are 6 main sub-topics in this chapter, namely pressure, liquid pressure, gas pressure, Pascal’s Principle, Archimedes Principle and Bernoulli’s Principle.

In pressure, we will discuss what pressure is, the formula of pressure and applications of high and low pressure.

In liquid pressure, we will discuss the formula of liquid pressure, characteristics of liquid pressure and applications of liquid pressure.

In gas pressure, we will discuss how to measure gas pressure. We will also discuss atmospheric pressure where we will focus our discussion on the characteristic of atmospheric pressure, proofs of existence of atmospheric pressure, how to measure atmospheric pressure and applications of atmospheric pressure.

In Pascal’s Principle, we will discuss the physics formula related to Pascal’s Principle and some applications of Pascal’s Principle. In Archimedes’ Principle, again we will discuss the formula and then its applications. In Bernoulli’s Principle, we will discuss what is stated in Bernoulli’s principle and its applications.

In SPM, there are a few things that we need to know about pressure. First, we need to know what pressure is? Pressure is defined as perpendicular force acting on a surface per unit area. The unit of pressure is Newton per meter square or Pascal, denoted by the symbol Pa.

Second, we need to know the formula of pressure. Pressure is given by the formula P = F/A, where P is pressure, F is the perpendicular force and A is the area. From the formula we know that pressure is directly proportional to the perpendicular force but inversely proportional to the area the force exerted on. In other words, the greater the perpendicular force, the greater the pressure; the greater the area, the lower the pressure.

Third, we also need to know some applications of high and low pressure. We are going to discuss this in the very next slide.

In SPM, you need to know some applications of high and low pressure. Applications of high pressure include knifes, spike shows or soccer shoes and ice skates. Applications of low pressure include ski, seatbelt, foundation of building, tyre of tractors and strap of handbag.

After this, we are going to discuss liquid pressure. There are a few points that we need to know about liquid pressure.

First, we need to know how liquid pressure is produced. Second, we need to know what is the equation related to liquid pressure and third we need to know what is the difference between liquid pressure and pressure in liquid? Click on the “play” or “next slide” button to start.

Liquid pressure is the pressure caused by a liquid. Liquid pressure exists due to the weight of the liquid. The pressure is caused by the weight of the liquid pressing on a surface.

In SPM, one of the most important things that you need to know about liquid pressure is the formula of liquid pressure. There are a lot of physics problems in this chapter are related to this formula. Liquid pressure is given by the equation P = hρg where P is the pressure, h is the depth from the surface, rho(ρ) is the density and g is the gravitational field strength, which is taken to be 10Nkg-1 in SPM.

One thing that you need to be careful about the formula of liquid pressure is the difference between liquid pressure and pressure in liquid. Liquid pressure is the pressure at a point in the liquid caused by the liquid wheareas liquid in pressure is the total pressure at a point in the liquid.

Let’s see this diagram. There is a fish h metre below the pond. The weight of the water is pressing on the body of the fish, causing a pressure called liquid pressure. There is also another pressure caused by the atmospheric gases above the pond called the atmospheric pressure. As a result, the total pressure pressing on the fish is the liquid pressure + the atmospheric pressure. Therefore, the liquid pressure exerts on the fish is P = hρg  and the pressure in liquid is P = Patm + hρg  where Patm is the atmospheric pressure and g  is the liquid pressure.

Click on the example button to see the example question. If you think you have already well master this topic, you may skip the example and continue with “characteristic of liquid pressure”.

Figure above shows a fish is at a place 2 metre below the surface of a lake. Find

  1. the water  pressure exerts on the fish.
  2. The pressure experiences by the body of the fish.

(Density of water = 1000kgm-2; Atmospheric pressure = 100,000 Pa)

For a., water pressure P = hρg, h is 2m, density is 1000 kgm-3, g is 10 N/kg. Therefore, P is 20,000 Pa.

In b., the question want us to find the pressure experiences by the body of the fish, it is the total pressure exerts on the body of the fish. In previous slide, we have learned that the total pressure in water is equal to the atmospheric pressure + the liquid pressure. Atmospheric pressure is equal to 100,000Pa.Therefore, the pressure exerts on the fish is 100,000  + 20,000 , which is equal to 120,000 Pa.

This example shows the difference between liquid pressure and pressure in liquid.

 

Understanding Momentum

In previous lesson, we have discussed inertia, including what is inertia, Newton’s first law of motion, situation involving inertia, mass and inertia and reducing negative effects of inertia. In this lesson, we will discuss momentum, including the principle of conservation of momentum and its applications. Under principle of conservation of momentum, we will discuss how to solve problems related to collision and explosion.

Under momentum, we will discuss what is momentum (including its unit and formula), the principle of conservation of momentum and the applications of the principle of conservation of momentum.

Under the principle of conservation of momentum, we will discuss the formula that describes the conservation principle. We will also discuss the 2 cases of reaction: collision and explosion. Under collision, we will discuss the elastic collision and inelastic collision.

In applications, we will include rocket, jet engine, and the moving mechanism of a jelly fish.

In SPM, there are a few things that you need to know about momentum. First, you need to know the definition of momentum. Momentum is defined as the product of mass and velocity.

Second, you need to know the formula of momentum. The formula of momentum is p = mv, where p is momentum, m is mass and v is velocity.

Third, you need to know the unit of momentum. In SPM, you need to state the unit of the physical quantity of your answer. Momentum is equal to mass multiply by velocity. The SI unit of mass is kg, and the SI unit of velocity is ms-1. Therefore, the unit of momentum is kgms-1.

Last but not least, you need to know that momentum is a vector quantity. Therefore, you need to be caution about the direction of the motion. If an object move in opposite direction, its momentum is negative.

Principle of Conservation of Momentum

Let’s see this example. 2 balls of mass 4kg and 2kg, respectively, moving in the same direction in a straight line. The velocity of the ball at the back is 3 m/s, higher than the velocity of the ball in front. If the ball at the back moves faster than the ball in front, and both of the balls are moving in a straight line, collision will happen. After collision, the balls move at 1m/s and 6m/s, respectively.

Let’s calculate the momentum of the balls before and after collision. Before collision, the momentum of the 4kg ball is 4kg x 3m/s, equal to 12 kg m/s. By using the same equation, we can find the momentum of the 2kg ball, which equal to 4kgms-1. The sum of momentum before collision is 12 + 4, equal to 16 kg m/s.

How about after the collision? The momentum of the 4kg ball became 4 kg m/s while the momentum of the 2 kg ball became 12 kg m/s, and the sum of momentum is 16 kg m/s.

Well, we can see that the sums of momentum before and after the collision are equal. In scientific term, we say the momentum is conserve after the collision. The word conserve means remain unchanged.

Sum of momentum before the collision is equal to the sum of momentum after the collision. This is the physics concept that we going to discuss in the principle of conservation of momentum.

The principle of conservation of momentum states that in a system make out of objects that react, the sum of the momentum is constant if no external force is acted on the system. The reaction can be a collision or explosion. The word constant means remain unchanged.

In other words, in a collision or explosion, if no external force acted on the system, the sum of the momentum before the collision or explosion will be equal to the sum of the momentum after the collision or explosion.

In SPM, there are a few things you need to know about this principle of conservation of momentum.

First, you need to remember this statement because in exam, you may be asked to state this principle of conservation of momentum.

Second, you need to know the equation that describer this principle of conservation of momentum and its applications in elastic collision, inelastic collision and explosion.

Third, you need to know the difference between elastic and elastic collision.

Forth, you need to know the application of principle of conservation of momentum in rocket and jet airplane.

Equation of Principle of Conservation of Momentum

If 2 balls of mass m1 and m2 collide with each other, the velocity before the collision is u1 and u2 while the velocity after the collision is v1 and v2 respectively, then the sum of momentum before the collision will be m1u1 + m2u2 while the sum of momentum after the collision is m1v1 + m2v2,

According to the principle of conservation of momentum, the sum of momentum before reaction equal to the sum of momentum after reaction. Therefore, m1u1 + m2u2 is equal to m1v1 + m2v2.

We can use this equation to solve almost all the problems involving collision or explosion in SPM physics.

In SPM, almost all the numerical problems related to collision and explosion can be solved by the principle of conservation of momentum.

Let’s see this example: The diagram shows 2 objects P and Q of mass 1kg and 2kg which are about to collide. Given that object P move at 1 m/s after the collision, find the velocity of Q after the collision.

We know that a question related to collision can be solved by the equation of principle of conservation of momentum: m1u1 + m2u2 = m1v1 + m2v2. When solving numerical problems, the first thing we need to do is to list down all the information that we have. Listing down all given information is an important step.

The mass m1 is 1kg, m2 is 2kg. The initial velocity u1 is 4 m/s and u2 is 1 m/s. Opps, wait a minute! This 1 m/s is not correct. Well, we must be very careful when writing the value of the velocity.  Velocity is a vector quantity. Velocity is negative if the object moves in opposite direction. In this case, Q moves in the opposite direction. Therefore u2 is -1 m/s but not 1 m/s. After the collision, the velocity v1 become 1 m/s and we are asked to find the velocity of Q.

Plug in all these numbers into the equation. Clean up the equation, we get 4 – 2 = 1 + 2v2. Therefore, v2 is equal to 0.5 m/s. The velocity of Q is positive. This positive value of Q shows that it moves to the right after the collision.

By definition, elastic collision is the collision where the kinetic energy is conserved after the collision. The word conserve means remain unchanged. If the kinetic energy is not conserved, then the collision is inelastic.

For example: 2 balls of mass 3kg and 1 kg move in opposite direction. Let’s say the kinetic energy of the 3kg ball is 12J while the kinetic energy of the 1 kg ball is 6J. After the collision, the kinetic energy of the 2 balls becomes 10J and 8J respectively. Do you think this collision is elastic or inelastic? Yes, you are correct. The sum of the kinetic energy before the collision is 18J and the sum of kinetic energy after the collision is also 18J. The sum of the kinetic energy before the collision equal to the sum of the kinetic energy after the collision. The kinetic energy is conserved in the collision. Therefore, this is an elastic collision. There are 2 types of collision: the elastic collision and the inelastic collision. In both collisions, the momentum is conserved. For elastic collision, the kinetic energy is conserved and the colliding objects bounce away after the collision. For inelastic collision, the kinetic energy is not conserved after the collision. In a perfectly inelastic collision, the 2 colliding objects attach and move with same speed after collision.

 

Understanding Inertia

Introduction

Force is a subtitle in Force and Motion, the second chapter in SPM form 4 physics. In force, we have 5 main subtopics: inertia, momentum, effects of force, gravity and force as a vector.

In inertia, we will discuss Newton’s First Law, situation involving inertia, mass and inertia and reducing negative effects of inertia.

In momentum, we will discuss conservation of momentum, where we will discuss 2 cases of reaction: collision and explosion. We will also discuss the applications of conservation of momentum.

In effects of force, we will discuss Newton’s 2nd Law, Newton’s 3rd Law, impulse and impulsive force and safety feature of vehicles.

In gravity, we will discuss gravitational field, free falling, lift and pulley.

In force as a vector, we will discuss addition of vectors, resolution of vectors, inclined plane and equilibrium of forces.

In this lesson, we will begin with inertia.

In inertia, we will discuss what is inertia, some situations involving inertia, mass and inertia and reducing negative effects.

Inertia

Inertia is the property of a body that tends to maintain its state of motion.

Well, let’s see what does this mean. Let’s take the example of a driver in a car. When the car star moving, and the driver start speeding up the car, he will find that his body tends to move backward. According to the concept of inertia, before the car move, his body is in stationary state. When the car start moving forward, his body tends to maintain its state of motion, which is stationary. Well, if a car want to move forward, but your body don’t‘ want to move, the result will be your body being left behind by the car. This make the driver looks like being thrown backward when the car accelerate. This is what inertia mean. The body of the driver tends to maintain its state of motion and resist any attempt to change its state of motion.

Let’s look at another situation. When the car is moving and the driver stop the car suddenly, his body will be thrown forward. This can also be explained by the concept of inertia. When the car is moving, the driver in the car is also moving at the same speed as the car. When the car stop, according to the concept of inertia, the body of the driver will tend to maintain its state of motion, and keep on moving forward at the same speed. In other words, the car want to stop, but the body of the driver don’t want to stop, due to the effect of inertia. As a result, the body is thrown forward.

In conclusion, according to the concept of inertia, a non-moving object will tend to become not moving and resist any attempt to make it move. A moving object will keep on moving in a straight line at a constant speed and tend to resist any attempt to change its speed or direction.

Stopwatch, Thermometer, Voltmeter and Ammeter

In previous lesson, we have discussed micrometer. In this lesson, we will discuss a few more measuring instruments, namely: stopwatch, ammeter, voltmeter and thermometer.

In stopwatch, we will discuss 2 types of stopwatch: the analogue stopwatch and the digital stopwatch. We will also discuss the sensitivity of stopwatches and how to take reading from a stopwatch.

In voltmeter and ammeter, we will discuss how a voltmeter and ammeter is connected to a circuit. Under thermometer, we will discuss the sensitivity of thermometers and the precaution steps need to be taken when using a thermometer.

There are 2 types of mercury thermometer used in school science lab, one with range -10oC – 110oC and another one with range 0oC – 360oC.

The sensitivity of the one with range -10oC – 110oC  is 1oC while the sensitivity of the thermometer with range 0oC – 360oC is 2oC.

Stopwatches

Sensitivity of digital and analogue stopwatches are different. The sensitivity of normal school use digital stopwatches is 0.01s. Sensitivity of analogue stopwatches are much lower than this.

There are 2 types of analogue stopwatches, one with sensitivity of 0.2s and one with sensitivity of 0.1s. This is the stopwatch of sensitivity 0.2s. Take notes that the pointer makes 1 revolution in 60s or 1 minute. Let’s have a closer look at the stopwatch. We can see that there are 5 divisions in every one second, which means each division gives a reading of 0.2s. Therefore the sensitivity is 0.2s.

Let’s see the analogue stopwatch of sensitivity 0.1s. Take notes that the pointer makes 1 revolution in 30s. Let’s have a closer look at the stopwatch. We can see that there are 10 divisions in every one second. Therefore, each division gives a reading of 0.1s. The sensitivity is 0.1s.

There are 2 scales in an analogue stopwatch: the minute scale and the second scale. The minute scale give the reading of minute and the second scale give the reading of second. For example, in the stopwatch above, the minute scale shows a reading of 4 minutes while the second scale shows a reading of 14.4s. Therefore the reading of the stopwatch is 4 minutes and 14.4s.

Let’s see another example. In this stopwatch, the minute scale shows a reading of 1.5 minute while the second scale shows a reading of 17.2s. 1.5 minute is equal to 1 minute 30s. Therefore, the reading of the stopwatch is 1 minute 30 seconds + 17.2s, which is equal to 1 minute 47.2s.

One of the common mistakes when students taking reading from a stopwatch is they overlook the minute scale. They just take the reading from the second scale, but ignore the reading of the minute scale. So, every time when you take a reading, check whether there is any reading at the minute scale.

You should have learned how to convert seconds to minutes or minutes to second in primary school. However, I believe that most of you do not realize that you can also do the conversion by using your scientific calculator.

The conversion of unit of time can be done by using the key “degree, minute and second”. The idea is pretty simple. Let’s compare the unit of angle with the unit of time. This is the symbol used for degree, minute and second. 1 degree = 60 minutes, and 1 minute = 60 seconds.

How about time? For time, 1 hour = 60 minutes, and 1 minute = 60 seconds. Can you see that? Amazingly, they are actually the same. In the operation or conversion of unit of time, what you need to do is just assume that the degree is equivalent to hour. In the very next slide, we are going to discuss how to convert the time in seconds to hour, minute and seconds.

Let’s see this example. 8266s is equal to how many hours, minutes and seconds.

Let’s see how to do this by using your scientific calculator. Before we start, we should keep in mind that we assume 1 degree is equal to 1 hour. In the calculator, we key in 0 hour, 0 minute, 8266 seconds, and then press equal. We get the answer. 8266s is equal to 2 hours, 17 minutes, 46 seconds.

After this, we are going to discuss how to convert hours to minutes or seconds.

Let’s see another example. 2 hours 54 minutes and 26 seconds is equal to how many seconds?

Before we start, we need to know that 1 hour is equal to 60 minutes times 60 seconds, which is equal to 3600 seconds. In order to convert hours to seconds, we multiply the time by 3600s. Let’s see how to do this in a calculator.

Let’s key in 2 hours 54 minutes 26 seconds. In order to convert this time to second, we times 3600, and then press equal. 2 hours 54 minutes 26 seconds is equal to 10466 seconds.

Thermometer

There are 2 types of mercury thermometer used in school science lab, one with range -10oC – 110oC and another one with range 0oC – 360oC.

The sensitivity of the one with range -10oC – 110oC  is 1oC while the sensitivity of the thermometer with range 0oC – 360oC is 2oC.

Ammeter and Voltmeter

Ammeter is a measuring instrument used to measure electric current in a circuit. The circuit symbol of an ammeter is a circle with a letter A inside. An ammeter must be connected in series in an electric circuit.

Voltmeter is a measuring instrument used to measure voltage or potential difference in a circuit. The circuit symbol of an ammeter is a circle with a letter V inside. A voltmeter must be connected in parallel in an electric circuit.

Micrometer Screw Gauge

In previous lesson, we have discussed vernier caliper. In this lesson, we will focus on micrometer, where we will discuss the sensitivity of micrometer, the label of the parts of a micrometer and the function of some of the parts. We will also discuss how to take readings from a micrometer such as how to read the main scale, how to read the thimble scale and how to determine the zero error.

This is a micrometer screw gauge. One of the famous question related to micrometer screw gauge in SPM exams is naming the part of the micrometer. Therefore, make sure that you memorise the name of all the parts of a micrometer.

Let’s see the anvil and spindle. Both of these are to hold the measured object. On the sleeve there is the main scale. Each division on the main scale equivalent to 1mm. The range of the main scale is from 0 – 25mm.

There is a thimble. The scale on it is called the thimble scale. Each division on the thimble scale represents 0.01mm.

The ratchet knob. The function of the ratchet knob is to exert appropriate amount of pressure on the measured object so that the object is not compressed. Asking about the function of the ratchet knob is also a famous question in SPM exam, especially in the structure questions in paper 2.

The lock. Its function is to tighten and hold the spindle stationary so that a reading can be taken easily.

Last but not least, the frame, a C-shaped body that holds the anvil and sleeve in constant relation to each other.

A micrometer scale comprises a main scale and a thimble scale. The reading of a micrometer is equal to the sum of the main scale and the thimble scale.

Let’s see the main scale. Each division in the main scale represent 1 mm. Below the line are the marks for 0.5 mm. Therefore, starting from zero, there is 0.5 mm, there is 1.0 mm, there is 1.5, 2.0, 2.5, 3.0 and so on.

The reading of the main scale is taken by referring to the edge of the thimble. For example, in the micrometer above, the edge of the thimble shows that the reading of the mains scale is 5.5 mm.

Let’s see the thimble scale. Each division on the thimble scale represents 0.01 mm. The reading of the thimble scale is taken by referring to the straight line on the main scale. In this case, the reading of the thimble scale is 0.28mm.

As we mentioned before, the reading of a micrometer is equal to the sum of the main scale and the thimble scale.  Therefore, the reading is 5.5 + 0.28 = 5.78 mm.

As all other measuring instruments, micrometers are also subjected to zero error.

When the anvil and spindle are firmly closed, the reading of the micrometer should be 0.00mm. However, in some micrometers, the reading does not start from zero and we call this zero error.

Let’s see the  image above. The reading is slightly higher than zero. This is a positive zero error. The reading of the main scale is zero and the reading of the thimble scale is 0.02mm. The zero error is +0.02mm.

Image above shows an example of negative zero error. The reading of the thimble scale is 0.47mm. However, this is not the zero error. If you still remember, the negative zero error must be read from the back of the scale. In this case, it must be read from the “0” mark. Therefore, the zero error is -0.03mm.

The correct reading of a micrometer is equal to the reading obtained – zero error. For example, if the reading obtained = 2.34 mm, the zero error is +0.02 mm, then the correct reading is 2.34 – (+0.02), which is equal to 2.32 mm. If the reading obtained is 2.34 mm and the zero error is -0.03 mm, then the correct reading is 2.34 – (-0.03), minus a negative become plus, and the answer is 2.37 mm.