Wednesday 5 December 2012
Saturday 1 December 2012
CHAPTER 4:HEAT
4.1
Understanding the thermal equilibrium
Define:
| a)Temperature | The measure of the degree of hotness of an object. SI Unit Kelvin K A hot object is at higher temperature than a cold object. |
| b)Heat | Form of energy, measured in Joules J. Heat is transferred from hotter object(higher temperature)to colder object(lower temperature) When an object is heated, it will absorb heat energy and the temperature will increase. When an object is cooled, it will release heat energy and the temperature will decrease |
| c)Thermal contact | Two object are in thermal contact when heat energy can be transferred between them |
| d)Heat transfer | When two object with different degrees of hotness come into thermal contact, heat energy is transferred between the two object |
| e)Mechanism of thermal equilibrium | Energy is transferred at a faster rate from the hotter object to the colder object. Energy also transferred from the colder object to the hotter but at a slower rate There is a net flow from the hotter to the colder object The hotter object cools down while the colder object warm up. After some time, energy is transferred at the same rate between the two object. There is no net heat transfer between the objects. The two objects are said to be in thermal equilibrium. |
Thermal equilibrium
When the two object are in thermal equilibrium, there is no net flow of heat between them.
Two object in thermal equilibrium have the same temperature.
Example of thermal equilibrium
A hot drink can be cooled by adding a few ice cubes to the drinks. Heat from the hot drinks is transferred to the colder ice until thermal equilibrium between the ice and water is reached. The final temperature of the drink equals the final temperature of ice.
UNDERSTANDING SPECIFIC HEAT CAPACITY
Shown here is the thermal motion of a segment of protein alpha helix. Molecules have various internal vibration and rotational degrees of freedom. This is because molecules are complex objects; they are a population of atoms that can move about within a molecule in different ways. This makes molecules distinct from the noble gases such as helium and argon, which are mono-atomic (consisting of individual atoms). Heat energy is stored in molecules’ internal motions which gives them an internal temperature. Even though these motions are called “internal,” the external portions of molecules still move—rather like the jiggling of a water balloon.
Application of Specific Heat capacity
Specific Latent Heat
The specific latent heat of a substance is the energy which is required to change 1 Kg of a substance from a certain physical condition to another physical condition without any change in temperature.
The heat energy absorbed by the ice does not cause the increase in temperature. The energy absorbed is not transferred to the molecules of ice as kinetic energy.
The heat absorbed is known as latent heat.
Heat energy needs to be supplied to change a substance from solid to liquid phase and from liquid to gaseous phase.When a solid melts, heat is absorbed but the temperature remains constant.
When a a liquid is boiling, heat is also absorbed but the temperature remains constant.
From the principle of conservation of energy, latent heat must be given out when a gas condenses to become a liquid and when the liquid solidifies to the solid phase. These two processes also occur at constant temperature.
The four main changes of phase are melting, boiling, condensation and solidification.
Two object in thermal equilibrium have the same temperature.
A hot drink can be cooled by adding a few ice cubes to the drinks. Heat from the hot drinks is transferred to the colder ice until thermal equilibrium between the ice and water is reached. The final temperature of the drink equals the final temperature of ice.
UNDERSTANDING SPECIFIC HEAT CAPACITY
Heat capacity,C | The unit for heat capacity is J°C |
Specific heat capacity,c | The unit of specific heat capacity is Jkg-1°C-1. Therefore, when a body of a mass m and specific heat capacity, c, absorbs a quantity of Heat, Q, then its heat will increase by θ. Therefore Q = mc θ |
The physical meaning of specific heat capacity,c. | When two objects of equal mass are left to cool down,the temperature of the object with smaller heat capacity will drop faster. |
A substances with a small value of specific heat capacity | For example,metal like iron,steel,copper and aluminium is used as pots and pans because they can be quickly heated up when there is only small heat absorption. Sensitive to temperature changes.A thermometer has low specific heat capacity so it enables heat to be easily absorbed and released even when small quantities of heat are involved. |
A substances with a high value of specific heat capacity | Require more heat to raise its temperature by a a specific amount. Can absorb a great amount of heat without a high in temperature. |
Shown here is the thermal motion of a segment of protein alpha helix. Molecules have various internal vibration and rotational degrees of freedom. This is because molecules are complex objects; they are a population of atoms that can move about within a molecule in different ways. This makes molecules distinct from the noble gases such as helium and argon, which are mono-atomic (consisting of individual atoms). Heat energy is stored in molecules’ internal motions which gives them an internal temperature. Even though these motions are called “internal,” the external portions of molecules still move—rather like the jiggling of a water balloon.
Cooking pot a)Copper base b)Wooden handle | Low specific heat capacity.The pot become hot very quickly.This enables quick cooking of the food in the pot.High density.The heavier base ensures that the pot is stable and will not topple over easily Large specific heat capacity.The handle will not become too hot when heat is absorbed.Poor conductor of heat. |
Sea breeze | Land has a smaller specific heat capacity than sea.Faster increase in temperature,land is warmer than the sea. Air above the land is heated up and rises Cooler air from the sea moves from toward the land as the sea breeze. |
Land breeze | At night,heat is lost from the land and the sea. Sea has a larger specific heat capacity so sea is warmer than the land. Cooler air from the land moves towards the sea as land breeze |
The image of land and sea breeze
Specific Latent Heat
The specific latent heat of a substance is the energy which is required to change 1 Kg of a substance from a certain physical condition to another physical condition without any change in temperature.
The heat energy absorbed by the ice does not cause the increase in temperature. The energy absorbed is not transferred to the molecules of ice as kinetic energy.
The heat absorbed is known as latent heat.
Heat energy needs to be supplied to change a substance from solid to liquid phase and from liquid to gaseous phase.When a solid melts, heat is absorbed but the temperature remains constant.
When a a liquid is boiling, heat is also absorbed but the temperature remains constant.
From the principle of conservation of energy, latent heat must be given out when a gas condenses to become a liquid and when the liquid solidifies to the solid phase. These two processes also occur at constant temperature.
The four main changes of phase are melting, boiling, condensation and solidification.
Friday 30 November 2012
CHAPTER 3 FORCES AND PRESSURE
3.1 Pressure
pressure is the force which acts normal per unit area of contact.
P = F / A where P = pressure [Pa]
F = force [N]
A = area [m3]
3.2 Pressure in Liquids
Pressure in liquids are not dependent on the size or shape of the container, only on its depth.
P = hpg where P = pressure [Pa]
h = height of liquid [m]
p = density of liquid [kg m-3]
g = gravitational acceleration [m s-2]
3.3 Atmospheric and Gas Pressure
3.3.1 Atmospheric Pressure
Atmospheric pressure is measured by barometers.
pressure is the force which acts normal per unit area of contact.
P = F / A where P = pressure [Pa]
F = force [N]
A = area [m3]
3.2 Pressure in Liquids
Pressure in liquids are not dependent on the size or shape of the container, only on its depth.
P = hpg where P = pressure [Pa]
h = height of liquid [m]
p = density of liquid [kg m-3]
g = gravitational acceleration [m s-2]
3.3 Atmospheric and Gas Pressure
3.3.1 Atmospheric Pressure
Atmospheric pressure is measured by barometers.
fortin barometer
(simple mercury barometer)
aneroid barometer
Atmospheric pressure measured by a simple mercury barometer can be calculated using the formula P = hpg.
Standard atmospheric pressure at sea level is usually 76 cm Hg or 10 m water. it can also be expressed as 1 bar.
(Note : cm Hg is an accepted unit of atmospheric pressure. Other similar units involving different liquids are also accepted)
For example :
Given that h = 76 and the density of mercury is 13 600 kg m-3.
Answer :
Atmospheric pressure = 76 cm Hg
State the answer in Pascal :
P =hpg = 0.76(13 600)(10) = 103 360 Pa.
3.3.2 Gas Pressure
Gas pressure can be measured using manometer and Bourdon gauges.
1) A hydraulic machine uses a hydraulic pump to transfer hydraulic liquid from one
vessel into a large cylinder to create a movement.
2) The above diagram shows a hydraulic pump used to move the arm of a crane.
Archimedes’ principle dictates that when an object is partially or fully immersed in a
fluid, the object undergoes a buoyant force which is equal to the weight of the fluid
displaced.
where V = volume of fluid displaced [m3]
ρ = density of fluid [kg m-3]
g = gravitational acceleration [m s-2]
If the object is floating stationary:
Weight of object = Weight of fluid displaced.
3.5.1 Submarine
The buoyant force acting on the submarine is always the same because the weight of fluid
displaced is always the same.
When water is pumped into the ballast tank, the weight of the submarine increases.
When the weight of the submarine is greater than the buoyant force, the submarine sinks.
When water is pumped out of the ballast tank and replaced with air, the weight of the submarine decreases. When the weight of the submarine is less than the buoyant force, the submarine floats.
3.5.2 Hydrometer
A hydrometer is used to measure the relative density of a liquid.
When the hydrometer floats stationary on the surface of the liquid,
where m = mass of the hydrometer [kg]
*The higher the hydrometer floats, the denser the liquid.
*The lower the hydrometer floats, the less dense the liquid.
3.5.3 Hot air balloon
1) A hot air balloon displaces a large volume of air.
2) When the weight of the balloon is less than the upward thrust, the hot air balloon will
move upwards.
3.5.4 Ships
Ships are marked with Plimsoll lines at the sides as a guide on the maximum load the
ship can bear in different conditions.
3.6 Bernoulli's Principle.
Bernoulli’s principle dictates that when the velocity of a flowing fluid increases, the
pressure at that point decreases.
Uniform tube
a) when a uniform tube with one end stopped is filled with water, the levels in all three vertical tubes are the same.
water pressure at A = water pressure at B = water pressure at C
b) when the stopper is removed, it is found that the water levels are decreasing.
water pressure at A > water pressure at B > water pressure at C
This is because water flows from high pressure to low pressure.
Venturi tube
- As the aerofoil shape cuts through the air, the air is
forced to travel at a higher speed on top of the
aerofoil compared to below the aerofoil.
- The high velocity area causes the air to have low
pressure.
- Therefore, the higher pressure area pressures
against the aerofoil towards the lower pressure area
causing a lifting force.
- Carburetor mixes petrol and air for burning
in the engine cylinder.
- When air flows through the narrower part of
the tube, the velocity increases thus the
pressure decreases. An area with low
pressure is formed at X.
- The atmospheric pressure, which is greater,
pushes the petrol out from the jet.
- Petrol and air are mixed to create a spray.
- Gas flows out through the narrow jet at a very high velocity.
- A low-pressured area is formed at X.
- The air outside at atmospheric pressure flows in and mixes
with gas to light it.
- Oxygen supply which is enough from the air will ensure a
complete burning to produce a hotter flame and less
luminous.
For example :
Given that h = 76 and the density of mercury is 13 600 kg m-3.
Answer :
Atmospheric pressure = 76 cm Hg
State the answer in Pascal :
P =hpg = 0.76(13 600)(10) = 103 360 Pa.
3.3.2 Gas Pressure
Gas pressure can be measured using manometer and Bourdon gauges.
Manometer
Bourdon gauge
In manometers, pressure can be calculated based on pressure equilibrium.
For example :
Gas pressure = Atmosphere pressure + hpg
3.4 Pascal's Principle
Pascal's principle dictates that pressure which exerted onto the surface of a fluid in a closed container will be transferred uniformly throughout the entire fluid.
F1 / A1 - F2 / A2
All hydraulic system are built based on Pascal's Principle.
3.4.1 Hydraulic jack.
1) when the lever is pushed downwords, valve A closes while valve B opens.
2) Pressure is transferred to the larger cylinder and creates a large force to lift the car.
3) When the lever is pulled upwards, valve B closes while valve A opens. The hydraulic oil flows out from the tank to fill the space in the small cylinder.
4) As every movement downwards by the lever creates a small movement upwards for
the large piston, the lever needs to be pushed and pulled repeatedly before the car can
be raised to a certain height required.
5) The car can be lowered down by opening the release valve so that the hydraulic oil
can flow back into the tank.
3.4.2 Hydraulic brake
1) When the driver steps on the brake pedal, the piston in the main cylinder transfers
pressure through the hydraulic oil in the cylinders.
2) Pressure is transferred uniformly through the hydraulic oil.
3) The pistons in the slave cylinders are larger, therefore a large amount of force is
transferred to the brake pads.
4) The brake pads transfer the large amount of force onto the disc so that the rotation of
the tire is slowed down.
3.4.3 Hydraulic pump
vessel into a large cylinder to create a movement.
2) The above diagram shows a hydraulic pump used to move the arm of a crane.
3.5 Archimedes' Principle
Archimedes’ principle dictates that when an object is partially or fully immersed in a
fluid, the object undergoes a buoyant force which is equal to the weight of the fluid
displaced.
Buoyant force = Weight of fluid displaced.
= Vpg
where V = volume of fluid displaced [m3]
ρ = density of fluid [kg m-3]
g = gravitational acceleration [m s-2]
If the object is floating stationary:
Weight of object = Weight of fluid displaced.
3.5.1 Submarine
The buoyant force acting on the submarine is always the same because the weight of fluid
displaced is always the same.
When water is pumped into the ballast tank, the weight of the submarine increases.
When the weight of the submarine is greater than the buoyant force, the submarine sinks.
When water is pumped out of the ballast tank and replaced with air, the weight of the submarine decreases. When the weight of the submarine is less than the buoyant force, the submarine floats.
3.5.2 Hydrometer
A hydrometer is used to measure the relative density of a liquid.
When the hydrometer floats stationary on the surface of the liquid,
Weight of the hydrometer = Weight of the liquid displaced.
mg = Vpg
V = volume of liquid displaced [m3]
ρ = density of liquid [kg m-3]
g = gravitational acceleration [m s-2]
*The higher the hydrometer floats, the denser the liquid.
*The lower the hydrometer floats, the less dense the liquid.
3.5.3 Hot air balloon
1) A hot air balloon displaces a large volume of air.
2) When the weight of the balloon is less than the upward thrust, the hot air balloon will
move upwards.
3.5.4 Ships
Ships are marked with Plimsoll lines at the sides as a guide on the maximum load the
ship can bear in different conditions.
3.6 Bernoulli's Principle.
Bernoulli’s principle dictates that when the velocity of a flowing fluid increases, the
pressure at that point decreases.
Uniform tube
a) when a uniform tube with one end stopped is filled with water, the levels in all three vertical tubes are the same.
water pressure at A = water pressure at B = water pressure at C
b) when the stopper is removed, it is found that the water levels are decreasing.
water pressure at A > water pressure at B > water pressure at C
This is because water flows from high pressure to low pressure.
Venturi tube
3.6.1 Aerofoil
forced to travel at a higher speed on top of the
aerofoil compared to below the aerofoil.
- The high velocity area causes the air to have low
pressure.
- Therefore, the higher pressure area pressures
against the aerofoil towards the lower pressure area
causing a lifting force.
3.6.2 Carburetor
in the engine cylinder.
- When air flows through the narrower part of
the tube, the velocity increases thus the
pressure decreases. An area with low
pressure is formed at X.
- The atmospheric pressure, which is greater,
pushes the petrol out from the jet.
- Petrol and air are mixed to create a spray.
3.6.7 Bunsen Burner
- A low-pressured area is formed at X.
- The air outside at atmospheric pressure flows in and mixes
with gas to light it.
- Oxygen supply which is enough from the air will ensure a
complete burning to produce a hotter flame and less
luminous.
Saturday 24 November 2012
CHAPTER 2:FORCE AND MOTION
2.1 Linear Motion
Linear motion is the motion in 1 dimension (1-D) or the motion in a straight line.
LINEAR MOTION VIDEO
Distance
- The distance traveled by an object is the total length that is traveled by that object.
- Distance is a scalar quantity.
- The SI unit of distance is m (metre).
Displacement
- Displacement of an object from a point of reference, O is the shortest distance of the object from point O in a specific direction.
- Displacement is a vector quantity.
- The SI unit of displacement is m (metre).
Distance travelled = 200m
Displacement = 120 m, in the direction of Northeast
2.1.2 Speed and Velocity
Speed
- Speed is the rate of change in distance. It is a measure of how fast the distance change in a movement.
- Speed is a scalar quantity.
- The SI unit of speed is m/s (metre per second).
Equation of Speed
Velocity
- Velocity is define as the rate of displacement change. It is the measure of how fast the displacement change of a moving object.
- Velocity is a vector quantity.
- The unit of displacemnet is m/s (metre per second).
Equation of velocity
Positive or Negative Sign of Velocity
- In velocity, the positive/negative sign indicates direction.
- You can take any direction as positive and the opposite as negative.
- For a linear motion, normally we take the motion to the right as positive and hence the motion to the left as negative.
2.1.3 Acceleration
Acceleration
- Acceleration is the rate of velocity change.Acceleration is a vector quantity. It is a measure of how fast the velocity change.
- Acceleration is a vector quantity.
- The unit of acceleration is ms-2.
Equation
Additional NotesAn object moves with a constant velocity if the magnitude and direction of the motion is always constant.
An object experiences changes in velocity if
- the magnitude of velocity changes
- the direction of the motion changes.
An object traveling with a constant acceleration, a, if the velocity changes at a constant rate.
Equation of Uniform Acceleration
Most of the motion problems can be solved by the following equations. Therefore, make sure that you memorise all of them.
How we know when to use the equation?
There are 3 types of motion:- motion with uniform velocity
- motion with uniform acceleration
- motion with changing acceleration
2.2.1 Analyzing Ticker Tape
Ticker Timer
A ticker-timer consists of an electrical vibrator which vibrates 50 times per second. This enables it to make 50 dots per second on a ticker-tape being pulled through it. The time interval between two adjacent dots on the ticker-tape is called one tick. One tick is equal to 1/50 s or 0.02 s.
Uniform Velocity
The distance of the dots is equally distributed.
- All lengths of tape in the chart are of equal length.
- The object is moving at a uniform velocity.
Uniform Acceleration
The distance between the dots increases uniformly.
- The length of the strips of tape in the chart increase uniformly.
- The velocity of the object is increasing uniformly, i.e. the object is moving at a constant acceleration.
The distance between the dots decreases uniformly.
- The length of the strips of tape in the chart decreases uniformly.
- The velocity of the object is decreasing uniformly, i.e. the object is decelerating uniformly.
2.2.2 Finding Velocity from Ticker Tape
Finding Velocity
Velocity of a motion can be determined by using ticker tape through the following equation:
Caution!:
t is time taken from the first dot to the last dot of the distance measured.
Example 1
Diagram 2.4 shows a strip of ticker tape that was pulled through a ticker tape timer that vibrated at 50 times a second. What is the
a. time taken from the first dot to the last dot?
b. average velocity of the object that is represented by the ticker tape?
Answer:
a. There are 15 ticks from the first dot to the last dot, hence
Time taken = 15 × 0.02s = 0.3s
b. Distance travelled = 15cm
\[v = \frac{s}{t} \hfill \\ \]
\[v = \frac{{15cm}}{{0.3s}} = 50cms^{ - 1} \hfill \\ \]
VIDEO FROM YOUTUBE
2.2.3 Finding Acceleration from Ticker Tape
Finding Acceleration
Acceleration of a motion can be determined by using ticker tape through the following equation:
Caution!:
t is time taken from the initial velocity to the final velocity.
Example 1
The ticker-tape in figure above was produced by a toy car moving down a tilted runway. If the ticker-tape timer produced 50 dots per second, find the acceleration of the toy car.
Answer:
In order to find the acceleration, we need to determine the initial velocity, the final velocity and the time taken for the velocity change.
Initial velocity,
\[ u = \frac{s}{t} = \frac{{3cm}}{{0.02s}} = 150cms^{ - 1}\]
\[ v = \frac{s}{t} = \frac{{0.5cm}}{{0.02s}} = 25cms^{ - 1}\]
Time taken for the velocity change,
t = (0.5 + 4 + 0.5) ticks = 5 ticks
t = 5 × 0.02s = 0.1s
Acceleration,
\[a = \frac{{v - u}}{t} = \frac{{25 - 150}}{{0.1}} = - 1250cms^{ - 1}\]
Example 2
A trolley is pushed up a slope. Diagram above shows ticker tape chart that show the movement of the trolley. Every section of the tape contains 5 ticks. If the ticker-tape timer produced 50 dots per second, determine the acceleration of the trolley.
Answer:
In order to find the acceleration, we need to determine the initial velocity, the final velocity and the time taken for the velocity change.
Initial velocity,
\[ u = \frac{s}{t} = \frac{{5cm}}{{5 × 0.02s}} = 50cms^{ - 1}\]
\[ v = \frac{s}{t} = \frac{{1cm}}{{5 × 0.02s}} = 10cms^{ - 1}\]
Time taken for the velocity change,
t = (2.5 + 5 + 5 + 5 + 2.5) ticks = 40 ticks
t = 40 × 0.02s = 0.8s
Acceleration,
\[a = \frac{{v - u}}{t} = \frac{{10 - 50}}{{0.8}} = - 50cms^{ - 1}\]
2.3.1 Displacement-Time Graph
Finding Velocity
Velocity of a motion can be determined by using ticker tape through the following equation:
Caution!:
t is time taken from the first dot to the last dot of the distance measured.
Example 1
Diagram 2.4 shows a strip of ticker tape that was pulled through a ticker tape timer that vibrated at 50 times a second. What is the
a. time taken from the first dot to the last dot?
b. average velocity of the object that is represented by the ticker tape?
Answer:
a. There are 15 ticks from the first dot to the last dot, hence
Time taken = 15 × 0.02s = 0.3s
b. Distance travelled = 15cm
\[v = \frac{s}{t} \hfill \\ \]
\[v = \frac{{15cm}}{{0.3s}} = 50cms^{ - 1} \hfill \\ \]
VIDEO FROM YOUTUBE
2.2.3 Finding Acceleration from Ticker Tape
Finding Acceleration
Acceleration of a motion can be determined by using ticker tape through the following equation:
Caution!:
t is time taken from the initial velocity to the final velocity.
Example 1
The ticker-tape in figure above was produced by a toy car moving down a tilted runway. If the ticker-tape timer produced 50 dots per second, find the acceleration of the toy car.
Answer:
In order to find the acceleration, we need to determine the initial velocity, the final velocity and the time taken for the velocity change.
Initial velocity,
\[ u = \frac{s}{t} = \frac{{3cm}}{{0.02s}} = 150cms^{ - 1}\]
\[ v = \frac{s}{t} = \frac{{0.5cm}}{{0.02s}} = 25cms^{ - 1}\]
Time taken for the velocity change,
t = (0.5 + 4 + 0.5) ticks = 5 ticks
t = 5 × 0.02s = 0.1s
Acceleration,
\[a = \frac{{v - u}}{t} = \frac{{25 - 150}}{{0.1}} = - 1250cms^{ - 1}\]
Example 2
A trolley is pushed up a slope. Diagram above shows ticker tape chart that show the movement of the trolley. Every section of the tape contains 5 ticks. If the ticker-tape timer produced 50 dots per second, determine the acceleration of the trolley.
Answer:
In order to find the acceleration, we need to determine the initial velocity, the final velocity and the time taken for the velocity change.
Initial velocity,
\[ u = \frac{s}{t} = \frac{{5cm}}{{5 × 0.02s}} = 50cms^{ - 1}\]
\[ v = \frac{s}{t} = \frac{{1cm}}{{5 × 0.02s}} = 10cms^{ - 1}\]
Time taken for the velocity change,
t = (2.5 + 5 + 5 + 5 + 2.5) ticks = 40 ticks
t = 40 × 0.02s = 0.8s
Acceleration,
\[a = \frac{{v - u}}{t} = \frac{{10 - 50}}{{0.8}} = - 50cms^{ - 1}\]
2.3.1 Displacement-Time Graph
Displacement - Time Graph
In a Displacement-Time Graph, the gradient of the graph is equal to the velocity of motion.
Analysing Displacement - Time Graph
In a Displacement-Time Graph, the gradient of the graph is equal to the velocity of motion.
Analysing Displacement - Time Graph
 | Gradient = 0 Hence, velocity = 0 |
| Gradient is constant, hence, velocity is Uniform |
 | Gradient is negative and constant, hence velocity is uniform and in opposite direction |
 | Gradient is increasing, hence velocity is increasing. |
 | Gradient is decreasing, hence velocity is decreasing. |
2.3.2 Velocity - Time Graph
Velocity - Time Graph
- The gradient of the velocity-time gradient gives a value of the changing rate in velocity, which is the acceleration of the object.
- The area below the velocity-time graph gives a value of the object's displacement.
Analysing Velocity-Time Graph
 | Uniform velocity |
| Uniform acceleration |
 | Increasing acceleration |
 | Uniform deceleration |
 | Decreasing acceleration |
Converting a Velocity-Time graph to Acceleration-Time graph
In order to convert a velocity-time graph to acceleration time graph, we need to find the gradient of the velocity time graph and plot it in the acceleration-time graph.
Drawing Velocity-Time Graph from Displacement-Time Graph
Drawing Velocity - Time Graph
Launching Object Upward
Dropping Object from High Place
Object Falling and Bounce Back
2.4.1 Mass and Inertia
Pulling a Book
When the book is pulled out, the books on top will fall downwards.
Explanation:
Pulling a Thread
1. Pull slowly - Thread A will snap.
Explanation:
2. Yank quickly - Thread B will snap.
Explanation:
Larger Mass - Greater Inertia
Bucket filled with sand is more difficult to be moved. It's also more difficult to be stopped from swinging.
Explanation:
An empty cart is easier to be moved compare with a cart full with load. This is because a cart with larger mass has larger inertia to resist the attempt to change the state of motion.
2.3.3 Graph of Free Fall
Free falling is a motion under gravitational force as the only force acting on the moving object. In SPM, you need to know the graphs of free falling of the following movement
- Launching object upward.
- Dropping Object from High Place
- Object Falling and Bounce Back
Launching Object Upward
| Velocity-Time Graph | Acceleration-Time Graph |
 |  |
Dropping Object from High Place
| Velocity-Time Graph | Acceleration-Time Graph |
 |  |
Object Falling and Bounce Back
| Velocity-Time Graph | Acceleration-Time Graph |
 |  |
2.4.1 Mass and Inertia
Mass
Mass is defined as the amount of matter. The SI unit of mass is kilogram (kg)
Mass is a scalar quantity.
Inertia
Inertia is the property of a body that tends to maintain its state of motion.
Newton's First Law
In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line).
Jerking a Card
When the cardboard is jerked quickly, the coin will fall into the glass.
Explanation:
Mass is defined as the amount of matter. The SI unit of mass is kilogram (kg)
Mass is a scalar quantity.
Inertia
Inertia is the property of a body that tends to maintain its state of motion.
Newton's First Law
In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line).
Jerking a Card
When the cardboard is jerked quickly, the coin will fall into the glass.
Explanation:
- The inertia of the coin resists the change of its initial state, which is stationary.
- As a result, the coin does not move with the cardboard and falls into the glass because of gravity.
Pulling a Book
When the book is pulled out, the books on top will fall downwards.
Explanation:
- Inertia tries to oppose the change to the stationary situation, that is, when the book is pulled out, the books on top do not follow suit.
Pulling a Thread
1. Pull slowly - Thread A will snap.
Explanation:
- Tension of thread A is higher than string B.
- Tension at A = Weight of the load + Pulling Force
2. Yank quickly - Thread B will snap.
Explanation:
- The inertia of the load prevents the force from being transmitted to thread A, hence causing thread B to snap.
Larger Mass - Greater Inertia
Bucket filled with sand is more difficult to be moved. It's also more difficult to be stopped from swinging.
Explanation:
- Object with more mass offers a greater resistance to change from its state of motion.
- Object with larger mass has larger inertia to resist the attempt to change the state of motion.
An empty cart is easier to be moved compare with a cart full with load. This is because a cart with larger mass has larger inertia to resist the attempt to change the state of motion.
2.5.1 Momentum
Momentum
Momentum is defined as the product of mass and velocity.
Momentum is a vector quantity. It has both magnitude and direction.
The SI unit of momentum is kgms-1
Formula:
Example 1A student releases a ball with mass of 2 kg from a height of 5 m from the ground. What would be the momentum of the ball just before it hits the ground?
Answer:
In order to find the momentum, we need to know the mass and the velocity of the ball right before it hits the ground.
It's given that the mass, m = 2kg.
The velocity is not given directly. However, we can determine the velocity, v, by using the linear equation of uniform acceleration.
This is a free falling motion,
The initial velocity, u = 0
The acceleration, a = gravirational acceleration, g = 10ms-2
The displacement, s = high = 50m.
The final velocity = ?
From the equation
v2 = u2 + 2as
v2 = (0)2 + 2(10)(5)
v = 10ms-1
The momentum,
p = mv =(2)(10) = 20 kgms-1
video about this chapter : Youtube Video
2.5.2 Principle of Conservation of Momentum
Principle of Conservation of Momentum
The principle of conservation of momentum states that in a system make out of objects that react (collide or explode), the total momentum is constant if no external force is acted upon the system.
Sum of Momentum Before Reaction = Sum of Momentum After Reaction
Formula
Example 1 - Both Object are in the Same Direction before Collision
A Car A of mass 600 kg moving at 40 ms-1 collides with a car B of mass 800 kg moving at 20 ms-1 in the same direction. If car B moves forwards at 30 ms-1 by the impact, what is the velocity, v, of the car A immediately after the crash?
Answer:
m1 = 600kg
m2 = 800kg
u1 = 40 ms-1
u2 = 20 ms-1
v1 = ?
v2 = 30 ms-1
According to the principle of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
(600)(40) + (800)(20) = (600)v1 + (800)(30)
40000 = 600v1 + 24000
600v1 = 16000
v1 = 26.67 ms-1
Example 2 - Both Object are in opposite direction Before Collision
A 0.50kg ball traveling at 6.0 ms-1 collides head-on with a 1.0 kg ball moving in the opposite direction at a speed of 12.0 ms-1. The 0.50kg ball moves backward at 14.0 ms-1 after the collision. Find the velocity of the second ball after collision.
Answer:
m1 = 0.5 kg
m2 = 1.0 kg
u1 = 6.0 ms-1
u2 = -12.0 ms-1
v1 = -14.0 ms-1
v2 = ?
(IMPORTANT: velocity is negative when the object move in opposite direction)
According to the principle of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
(0.5)(6) + (1.0)(-12) = (0.5)(-14) + (1.0)v2
-9 = - 7 + 1v2
v2 = -2 ms-1
Explosion
The principle of conservation of momentum states that in a system make out of objects that react (collide or explode), the total momentum is constant if no external force is acted upon the system.
Sum of Momentum Before Reaction = Sum of Momentum After Reaction
Formula
Example 1 - Both Object are in the Same Direction before Collision
A Car A of mass 600 kg moving at 40 ms-1 collides with a car B of mass 800 kg moving at 20 ms-1 in the same direction. If car B moves forwards at 30 ms-1 by the impact, what is the velocity, v, of the car A immediately after the crash?
Answer:
m1 = 600kg
m2 = 800kg
u1 = 40 ms-1
u2 = 20 ms-1
v1 = ?
v2 = 30 ms-1
According to the principle of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
(600)(40) + (800)(20) = (600)v1 + (800)(30)
40000 = 600v1 + 24000
600v1 = 16000
v1 = 26.67 ms-1
Example 2 - Both Object are in opposite direction Before Collision
A 0.50kg ball traveling at 6.0 ms-1 collides head-on with a 1.0 kg ball moving in the opposite direction at a speed of 12.0 ms-1. The 0.50kg ball moves backward at 14.0 ms-1 after the collision. Find the velocity of the second ball after collision.
Answer:
m1 = 0.5 kg
m2 = 1.0 kg
u1 = 6.0 ms-1
u2 = -12.0 ms-1
v1 = -14.0 ms-1
v2 = ?
(IMPORTANT: velocity is negative when the object move in opposite direction)
According to the principle of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
(0.5)(6) + (1.0)(-12) = (0.5)(-14) + (1.0)v2
-9 = - 7 + 1v2
v2 = -2 ms-1
Explosion
| Before explosion both object stick together and at rest. | After collision, both object move at opposite direction. |
| Total Momentum before collision Is zero | Total Momentum after collision : m1v1 + m2v2 |
| From the law of conservation of momentum:
Total Momentum Before collision = Total Momentum after collision
0 = m1v1 + m2v2 m1v1 = - m2v2 (-ve sign means opposite direction) | |
Examples or Application of Conservation of Momentum in Explosion
- Fire a pistol or rifle
- Launching a rocket
- Application in jet engine
- Fan boat
Example 3
A man fires a rifle which has mass of 2.5 kg. If the mass of the bullet is 10 g and it reaches a velocity of 250 m/s after shooting, what is the recoil velocity of the pistol?
Answer
This is a typical question of explosion.
m1 = 2.5 kg
m2 = 0.01 kg
u1 = 0 ms-1
u2 = 0 ms-1
v1 = ?
v2 = 250 ms-1
By using the equation of conservation of momentum principle
0 = m1v1 + m2v2
0 = (2.5)v1 + (0.01)(250)
(2.5)v1 = -2.5v1 = -1 ms-1
video about this chapter : Youtube Video
2.5.3 Elastic and Inelastic Collision
Elastic Collision
Elastic collision is the collision where the kinetic energy is conserved after the collision.
Total Kinetic Energy before Collision = Total Kinetic Energy after Collision
Additional notes:
Elastic collision is the collision where the kinetic energy is conserved after the collision.
Total Kinetic Energy before Collision = Total Kinetic Energy after Collision
Additional notes:
- In an elastic collision, the 2 objects separated right after the collision, and
- the momentum is conserved after the collision.
- Total energy is conserved after the collision.
Rocket
- 1Mixture of hydrogen and oxygen fuels burn in the combustion chamber.
- Hot gases are expelled through the exhausts at very high speed .
- The high-speed hot gas produce a high momentum backwards.
- By conservation of momentum, an equal and opposite momentum is produced and acted on the rocket, pushing the rocket upwards.
Force
- A force is push or pull exerted on an object.
- Force is a vector quantity that has magnitude and direction.
- The unit of force is Newton ( or kgms-2).
When the forces acting on an object are not balanced, there must be a net force acting on it. The net force is known as the unbalanced force or the resultant force.
When a force acts on an object, the effect can change the
- size,
- shape,
- stationary state,
- speed and
- direction of the object.
Newton's Second Law
The rate of change of momentum of a body is directly proportional to the resultant force acting on the body and is in the same direction.
Implication:
When there is resultant force acting on an object, the object will accelerate (moving faster, moving slower or change direction).
Formula of Force
From Newton's Second Law, we can derived the equation
(IMPORTANT: F Must be the net force)
Summary of Newton's 1st Law and 2nd Law
Newton's First Law:
When there is no net force acting on an object, the object is either stationary or move with constant speed in a straight line.
Newton's Second Law:
When there is a net force acting on an object, the object will accelerate.
Example 1
A box of mass 150kg is placed on a horizontal floor with a smooth surface; find the acceleration of the box when a 300N force is acting on the box horizontally.
Answer:
F = ma
(300) = (150)a
a = 2 ms-2
Example 2
A object of mass 50kg is placed on a horizontal floor with a smooth surface. If the velocity of the object changes from stationary to 25.0 m/s in 5 seconds when is acted by a force, find the magnitude of the force that is acting?
Answer:
We know that we can find the magnitude of a force by using the formula F = ma. The mass m is already given in the question, but the acceleration is not give directly.
We can determine the acceleration from the formula
\[a = \frac{{v - u}}{t} \hfill \\ \]
\[a = \frac{{25 - 0}}{5} = 5ms^{ - 2} \hfill \\}\]
From the formula
F = ma = (50)(5) = 250N
The force acting on the box is 250N.
2.7.1 Impulse
Impulse
Impulse is defined as the product of the force (F) acting on an object and the time of action (t). Impulse exerted on an object is equal to the momentum change of the object. Impulse is a vector quantity.
Formula of impulse
Impulse is the product of force and time.
Impulse = F × t
Impulse = momentum change
Impulse = mv - mu
2.7.2 Impulsive Force
Impulsive Force
Impulsive force is defined as the rate of change of momentum in a reaction.
It is a force which acts on an object for a very short interval during a collision or explosion.
Example 1
A car of mass 1000kg is traveling with a velocity of 25 m/s. The car hits a street lamp and is stopped in0.05 seconds. What is the impulsive force acting on the car during the crash?
Answer:
\[\begin{gathered} {\text{Impulsive Force}} \hfill \\ F = \frac{{m(v - u)}}{t} \hfill \\ F = \frac{{(1000)(0 - 25)}}{{0.05}} \hfill \\ F = - 500N \hfill \\ \end{gathered} \]
Effects of impulse vs Force
Impulsive force is defined as the rate of change of momentum in a reaction.
It is a force which acts on an object for a very short interval during a collision or explosion.
Example 1
A car of mass 1000kg is traveling with a velocity of 25 m/s. The car hits a street lamp and is stopped in0.05 seconds. What is the impulsive force acting on the car during the crash?
Answer:
\[\begin{gathered} {\text{Impulsive Force}} \hfill \\ F = \frac{{m(v - u)}}{t} \hfill \\ F = \frac{{(1000)(0 - 25)}}{{0.05}} \hfill \\ F = - 500N \hfill \\ \end{gathered} \]
Effects of impulse vs Force
- A force determines the acceleration (rate of velocity change) of an object. A greater force produces a higher acceleration.
- An impulse determines the velocity change of an object. A greater impulse yield a higher velocity change.
- Playing football
- Playing badminton
- Playing tennis
- Playing golf
- Playing baseball
- The long jump pit is filled with sand to increase the reaction time when atlete land on it.
- This is to reduce the impulsive force acts on the leg of the atlete because impulsive force is inversely proportional to the reaction time.
(This image is licenced under the GNU Free Document Licence. The original file is from the Wikipedia.org.)
- During a high jump, a high jumper will land on a thick, soft mattress after the jump.
- This is to increase the reaction time and hence reduces the impulsive force acting on the high jumper.
Jumping
A jumper bends his/her leg during landing. This is to increase the reaction time and hence reduce the impact of impulsive force acting on the leg of the jumper.
2.8.1 Safety Features in Vehicles
Crumble Zone
The crumple zone increases the reaction time of collision during an accident.
This causes the impulsive force to be reduced and hence reduces the risk of injuries.
Seat Belt
Prevent the driver and passengers from being flung forward or thrown out of the car during an emergency break.
Airbag
The inflated airbag during an accident acts as a cushion to lessen the impact when the driver flings forward hitting the steering wheel or dashboard.
Head Rest
Reduce neck injury when driver and passengers are thrown backwards when the car is banged from backward.
Windscreen
Shatter-proof glass is used so that it will not break into small pieces when broken. This may reduce injuries caused by scattered glass.
Padded Dashboard
Cover with soft material. This may increases the reaction time and hence reduce the impulsive force when passenger knocking on it in accident.
Collapsible Steering Columns
The steering will swing away from driver’s chest during collision. This may reduce the impulsive force acting on the driver.
Anti-lock Braking System (ABS)
Prevent the wheels from locking when brake applied suddenly by adjusting the pressure of the brake fluid. This can prevents the car from skidding.
Bumper
Made of elastic material so that it can increases the reaction time and hence reduces the impulsive force caused by collision.
Passenger Safety Cell
The body of the car is made from strong, rigid stell cage.
This may prevent the car from collapsing on the passengers during a car crash.
2.9.1 Gravity
Gravitational Field
A gravitational field as a region in which an object experiences a force due to gravitational attraction.
Gravitational Field Strength
The gravitational field strength at a point in the gravitational field is the gravitational force acting on a mass of 1 kg placed at that point.
The unit of gravitational field strength is N/kg.
The gravitational field strength is denoted by the symbol "g".
Gravitational Field Strength Formula
Gravitational Acceleration
The gravitational acceleration is the acceleration of an object due to the pull of the gravitational force.
The unit of gravitational acceleration is ms-2
Gravitational acceleratio is also denoted by the symbol "g".
Symbol: g
Important notes:
A gravitational field as a region in which an object experiences a force due to gravitational attraction.
Gravitational Field Strength
The gravitational field strength at a point in the gravitational field is the gravitational force acting on a mass of 1 kg placed at that point.
The unit of gravitational field strength is N/kg.
The gravitational field strength is denoted by the symbol "g".
Gravitational Field Strength Formula
Gravitational Acceleration
The gravitational acceleration is the acceleration of an object due to the pull of the gravitational force.
The unit of gravitational acceleration is ms-2
Gravitational acceleratio is also denoted by the symbol "g".
Symbol: g
Important notes:
- Gravitational acceleration does not depend on the mass of the moving object.
- The magnitude of gravitational acceleration is taken to be 10ms-2.
Gravitational Field Strength vs. Gravitational Acceleration
- Both the gravitational field strength and gravitational acceleration have the symbol, g and the same value (10ms-2) on the surface of the earth.
- When considering a body falling freely, the g is the gravitational acceleration.
- When considering objects at rest, g is the Earth’s gravitational field strength acting on it.
Weight
The weight of an object is defined as the gravitational force acting on the object.
The SI unit of weight is Newton (N)
Differences between Weight and Mass
| Weight | Mass |
| Depends on the gravitational field strength | Independent from the gravitational field strength |
| Vector quantity | Scalar Quantity |
| Unit Newton (N) | Unit: Kilogram (kg) |
2.9.2 Free Falling
Free Falling
- Free falling is a motion under force of gravity as the only force acting on the moving object.
- Practically, free falling can only take place in vacuum.
- The gravitational acceleration is the acceleration of an object due to the pull of the gravitational force. It has the unit of ms-2
- The symbol of gravitational acceleration is " g ".
- Gravitational acceleration does not depend on the mass of the moving object.
- The magnitude of gravitational acceleration is taken to be 10ms-2.
Gravitational Field Strength vs. Gravitational Acceleration
- Both the gravitational field strength and gravitational acceleration have the symbol, g and the same value (10ms-2) on the surface of the earth.
- When considering a body falling freely, the g is the gravitational acceleration.
- When considering objects at rest, g is the Earth’s gravitational field strength acting on it.
When an object is released from a high place,- its initial velocity, u = 0.
- its acceleration is equal to the gravitational acceleration, g, which taken to be 10ms-2 in SPM.
- the displacement is the of the object when it reaches the ground is equal to the initial height of the object, h.
If an object is launched up vertically,- the acceleration = -g (-10ms-2)
- the velocity become zero when the object reaches the highest point.
- the displacement of the object at highest point is equal to the vertical height of object, h
- the time taken for the object to move to the maximum height = the time taken for the object to fall from the maximum point to its initial position.
Vector and Scalar Quantity
A scalar quantity is a quantity which can be fully described by magnitude only.
A vector quantity is a quantity which is fully described by both magnitude and direction.
Vector Diagram
The arrow shows the direction of the vector.
The length representing the magnitude of the vector.
Equal Vector
Two vectors A and B may be defined to be equal if they have the same magnitude and point in the same direction.
2.10.2 Vector Addition
Vector Addition - Triangle Method
Join the tail of the 2nd vector to the head of the 1st vector. Normally the resultant vector is marked with double arrow.
Vector Addition - Parallelogram Method
Join the tail of the 2nd vector to the tail of the 1st vector. Normally the resultant vector is marked with double arrow.
Addition of 2 Perpendicular Vectors
If 2 vectors (a and b) are perpendicular to each others, the magnitude and direction of the resultant vector can be determined by the following equation.
Example 1
Two forces, P and Q of magnitude 10N and 12N are perpendicular to each others. What is the magnitude of the resultant force if P and Q are acting on an object?
Answer:
Magnitude of the resultant force
\[ |F| = \sqrt {10^2 + 12^2 } = \sqrt {244} = 15.62N \hfill \\ \]
Example 2
Diagram above shows that four forces of magnitude 2N, 4N, 5N and 8N are acting on point O. All the forces are perpendicular to each others. What is the magnitude of the resulatant force that acts on point O?
Answer:
The resultant force of the horizntal component = 5 - 2 = 3N to the right
The resultant force of the vertical component = 8 - 4 = 4N acting downward.
Therefore, the magtitude of these 2 force components,
\[|F| = \sqrt {3^2 + 4^2 } = \sqrt {25} = 5 \]
2.10.3 Vector Resolution
Vector Resolution
A vector can be resolve into 2 component which is perpendicular to each others.
Example 1
Diagram above shows a lorry pulling a log with an iron cable. If the tension of the cable is 3000N and the friction between the log and the ground is 500N, find the horizontal force that acting on the log.
Answer:
Horizontal component of the tension = 3000 cos30o =2598N
Friction = 500N
Resultant horizontal force = 2598N - 500N =2098N
Example 2
Diagram above shows two forces of magnitude 25N are acting on an object of mass 2kg. Find the acceleration of object P, in ms-2.
Answer:
Horizontal component of the forces = 25cos45o + 25cos45o = 35.36N
Vertical component of the forces = 25sin45o - 25sin45o = 0N
The acceleration of the object can be determined by the equation
F = ma
(35.36) = (2)a
a = 17.68 ms-2
Inclined Plane
Weight component along the plane = Wsinθ.
Weight component perpendicular to the plane = Wcosθ.
Example 3
A block of mass 2 kg is pulling along a plane by a 20N force as shown in diagram above. Given that the fiction between block and the plane is 2N, find the magnitude of the resultant force parallel to the plane.
Answer:
First of all, let's examine all the forces or component of forces acting along the plane.
The force pulling the block, F = 20N
The frictional force Ffric = 2N
The weight component along the plane = 20sin30o = 10N
The resultant force along the plane = 20 - 2 - 10 = 8N
2.10.4 Forces in Equilibrium
Vectors in Equilibrium
When 3 vectors are in equilibrium, the resultant vector = 0. After joining all the vectors tail to head, the head of the last vector will join to the tail of the first vector.
Forces in equilibrium
Forces are in equilibrium means the resultant force in all directions are zero.
When the forces acting on an object are balanced, they cancel each other out. The net force is zero.
Effect :
When 3 vectors are in equilibrium, the resultant vector = 0. After joining all the vectors tail to head, the head of the last vector will join to the tail of the first vector.
Forces in equilibrium
Forces are in equilibrium means the resultant force in all directions are zero.
When the forces acting on an object are balanced, they cancel each other out. The net force is zero.
Effect :
- an object at rest is continuely at rest [ velocity = 0]
- a moving object will move at constant velocity [ a = 0]
Example 1
Diagram above shows a load of mass 500g is hung on a string C, which is tied to 2 other strings A and B. Find the tension of string A.
Answer:
Tension of string C, TC = weight of the load = 5N
All forces in the system are in equilibrium, hence
Vertical component of tension A (TA) = TC
TAcos60o = TC
TA = TC/cos60o
TA = 5/cos60o = 10N
2.11.1 Work
Work
- Work done by a constant force is given by the product of the force and the distance moved in the direction of the force.
- The unit of Nm(Newton metre) or J(Joule).
- Work is a scalar quantity.
When the direction of force and motion are same, θ = 0o, therefore cosθ = 1
Work done,
W = F × s
Example 1
A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force.
Answer:
Work done,
W = F × s × cos θ
W = 50 × 3.0 × cos30o = 129.9J
W = F × s × cos θ
W = 50 × 3.0 × cos30o = 129.9J
Example 2
Diagram above shows a 10N force is pulling a metal. The friction between the block and the floor is 5N. If the distance travelled by the metal block is 2m, find
- the work done by the pulling force
- the work done by the frictional force
Asnwer:
(a) The force is in the same direction of the motion. Work done by the pulling force,
W = F × s = (10)(2) = 20J
(b) The force is not in the same direction of motion, work done by the frictional force
W = F × s × cos180o= (5)(2)(-1) = -10J
Work Done Against the Force of Gravity
Example 3
Ranjit runs up a staircase of 35 steps. Each steps is 15cm in height. Given that Ranjit's mass is 45kg, find the work done by Ranjit to reach the top of the staircase.
Answer:
In this case, Ranjit does work to overcome the gravity.
Ranjit's mass = 45kg
Vertical height of the motion, h = 35 × 0.15
Gravitational field strength, g = 10 ms-2
Work done, W = ?
Ranjit's mass = 45kg
Vertical height of the motion, h = 35 × 0.15
Gravitational field strength, g = 10 ms-2
Work done, W = ?
W = mgh = (45)(10)(35 × 0.15) = 2362.5J
Finding Work from Force-Displacement Graph
In a Force-Displacement graph, work done is equal to the area in between the graph and the horizontal axis.Example 4
The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m. Find the work done by the force to move the trolley.
Answer:
In a Force-Displacement graph, work done is equal to the area below the graph. Therefore, work done
2.11.2 Potential Energy
Energy
Energy is defined as the capacity to do work. Work is done when energy is converted from one form to another.
Nm or Joule(J)
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e., height).
Formula:
Example 1
A ball of 1kg mass is droppped from a height of 4m. What is the maximum kinetic energy possessed by the ball before it reached the ground?
Answer
According to the principle of conservation of energy, the amount of potential energy losses is equal to the amount of kinetic energy gain.
Maximum kinetic energy
= Maximum potentila energy losses
= mgh = (1)(10)(4) = 40J
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing.
Formula:
Example 2
Diagram above shows a spring with a load of mass 0.5kg. The extention of the spring is 6cm, find the energy stored in the spring.
Answer:
The energy stored in the spring is the elestic potential energy.
\[ E_P = \frac{1}{2}Fx \hfill \\ \]
\[ E_P = \frac{1}{2}(5)(0.06) = 0.15J \hfill \\ \]
2.11.3 Kinetic Energy
Kinetic Energy
Kinetic energy is the energy of motion.
Equation of Kinetic Energy
Example 1
Kinetic energy is the energy of motion.
Equation of Kinetic Energy
Example 1
Determine the kinetic energy of a 2000-kg bus that is moving with a speed of 35.0 m/s.
\[ E_K = \frac{1} {2}mv^2 \hfill \\ \]
\[ E_K = \frac{1} {2}(2000)(35)^2 \hfill \\ \]
\[ E_K = 1225000J \hfill \\ \]
Answer:
Kinetic Energy,\[ E_K = \frac{1} {2}mv^2 \hfill \\ \]
\[ E_K = \frac{1} {2}(2000)(35)^2 \hfill \\ \]
\[ E_K = 1225000J \hfill \\ \]
During a conversing of energy,
Amount of Work Done = Amount of Energy Converted
Example
A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops at B after 4 seconds. What is the work done to overcome the friction?
Answer:
In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence
\[\begin{array}{l}{\rm{Work Done}} \\{\rm{ = Kinetic Energy Loss}} \\= \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}(5)(4)^2 - \frac{1}{2}(5)(0)^2 \\= 40J \\\end{array}\]
2.12.1 Power
Power
Power is the rate at which work is done, which means how fast a work is done.
Formula:
Example 1An electric motor takes 20 s to lift a box of mass 20kg to a height of 1.5 m. Find the amount of work done by the machine and hence find the power of the electric motor.
Answer:
Work done,
W = mgh = (20)(10)(1.5) = 300J
Power,
\[ P = \frac{W}{t} = \frac{{300}}{{20}} = 15J\]
2.12.2 Efficiency
Efficiency
The efficiency of a device is defined as the percentage of the energy input that is transformed into useful energy.
Example
In the example above, the input power is 100J/s, the desire output power (useful energy) is only 75J/s, the remaining power is lost as undisire output. Therefore, the efficiency of this machine is
75/100 x 100% = 75%
Air Conditioner
The efficiency of a device is defined as the percentage of the energy input that is transformed into useful energy.
Example
In the example above, the input power is 100J/s, the desire output power (useful energy) is only 75J/s, the remaining power is lost as undisire output. Therefore, the efficiency of this machine is
75/100 x 100% = 75%
Air Conditioner
- Switch off the air conditioner when not in use.
- Buy the air conditioner with suitable capacity according to the room size.
- Close all the doors and windows of the room to avoid the cool air in the room from flowing out.
- Always remember to close the door of refrigerator.
- Open the refrigerator only when necessarily.
- Always keep the cooling coil clean.
- Defrost the refrigerator regularly.
- Choose the refrigerator with capacity suitable for the family size.
- Refrigerator of large capacity is more efficient compare with refirgerator of small capacity.
- Use fluorecent bulb rather than incandescent bulb. Fluorescent bulbs are much more efficient than incandescent bulbs.
- Use a lamp with reflector so that more light is directed towards thr desirable place.
- Use front-loading washing machine rather than top-loading wahing machine because it uses less water and electricity.
- Use washing machine only when you have sufficient clothes to be washed. Try to avoid washing small amount of clothes.
2.13.1 Elasticity
Elasticity
Elasticity is the ability of a sub-stance to recover its original shape and size after distortion.
Forces Between Atoms
The intermolecular forces consist of an attractive force and a repulsive force.
Elasticity is the ability of a sub-stance to recover its original shape and size after distortion.
Forces Between Atoms
The intermolecular forces consist of an attractive force and a repulsive force.- At the equilibrium distance d, the attractive force equal to the repulsive force.
- If the 2 atoms are brought closer, the repulsive force will dominate, produces a net repulsive force between the atoms.
- If the 2 atoms are brought furhter, the attractive force will dominate, produces a net attractive force between the atoms.
Graph of Forces Between 2 atoms
x0 = Equilibrium Distance
When the particles are compressed, x < x0, the attractive force between the particles increases.
If the distance x exceeds the elastic limit, the attractive force will decreases.
2.13.2 Hooke's Law
Hooke's Law
Hooke's Law states that if a spring is not stretched beyond its elastic limit, the force that acts on it is directly proportional to the extension of the spring.
Elastic Limit
The elastic limit of a spring is defined as the maximum force that can be applied to a spring such that the spring will be able to be restored to its original length when the force is removed.
Equation derived from Hooke's Law
From Hook's Law, we can derived that
Spring Constant
Spring constant is defined as the ratio of the force applied on a spring to the extension of the spring.
It is a measure of the stiffness of a spring or elastic object.
Graph of Streching Force - Extension
Gradient = Spring constant
Area below the graph = Work done
F-x graph and spring constant
The higher the gradient, the greater the spring constant and the harder (stiffer) spring.
For example, the stiffness of spring A is greater than spring B.
2.13.3 Spring
| Arrangement in series: | Arrangement in parallel: |
| Extension = x × number of spring Stiffness decreases Spring constant = k/number of spring | Extension = x ÷ number of spring Stiffness increases Spring constant = k × number of spring |
Factors Affecting the Stiffness of Spring
| Stiffer | Less stiff | |
| Material type of spring (A steel spring is stiffer than a copper spring) |  |  |
| Diameter of wire of spring (The greater the diameter of the wire, the stiffer the spring) |  |  |
| Diameter of the spring (The smaller the diameter of spring, the stiffer the spring) |  |  |
| Length of the string (Shorter spring is stiffer) |  | |
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