Tuesday, 27 August 2013
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Archimedes Principle
Archimedes Principle
- 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.
- Upthrust/Buoyant force is an upward force exerted by a fluid on an object immersed in it.
- Mathematically, we write
F = Upthrust/Buoyant Force
ρ = Density of the liquid
V = Volume of the displaced liquid
g = Gravitational field strength
Bernoulli’s Principle
Bernoulli’s Principle
Bernoulli's Principle states that as the speed of a moving fluid (liquid or gas) increases, the pressure within the fluid decreases.
Venturi Effect
The Venturi effect is the fluid pressure that results when an incompressible fluid flows through a constricted section of a pipe.
Experiment 1
Figure above shows that when water flow from left to right, the water level decreases from left to right. This indicates that, the water pressure decreases from left to right.
Explanation:
Liquids flow from places with higher pressure to places with lower pressure.
However, if the experiment is repeated by using a Venturi tube where the diameter at B is made smaller than A and C as in the diagram above, the water level become lowest at B.
Explanation:
The pressure at B is the lowest because the liquid flow the fastest at B. According to Bernoulli's Principle, the faster the water flow, the lower the water pressure.
Explanation:
Liquids flow from places with higher pressure to places with lower pressure.
Explanation:
The pressure at B is the lowest because the liquid flow the fastest at B. According to Bernoulli's Principle, the faster the water flow, the lower the water pressure.
Experiment 2
Figure above shows some air is blow through a tube from left to right. The water level in the capillary tube increases from left to right.
This indicates that the pressure in the tube decreases from left to right.
Explanation:
Gases flow from places with higher pressure to places with lower pressure.
However, if the tube is replaced by a Venturi tube, the water level become highest at B. This indicates that, the pressure of the air is the lowest at B.
Explanation:
The pressure at B is the lowest because the gas flow the fastest at B. According to Bernoulli's Principle, the faster the gas flow, the lower the gas pressure.
Figure above shows some air is blow through a tube from left to right. The water level in the capillary tube increases from left to right.
This indicates that the pressure in the tube decreases from left to right.
Explanation:
Gases flow from places with higher pressure to places with lower pressure.
Explanation:
The pressure at B is the lowest because the gas flow the fastest at B. According to Bernoulli's Principle, the faster the gas flow, the lower the gas pressure.
Pascal's Principle
Pascal’s Principle
- Pascal's principle states that in a confined fluid, an externally applied pressure is transmitted uniformly in all directions.
- Pascal's principle is also known as the principle of transmission of pressure in a liquid.
Q & A
Q: Suggest an experiment to prove Pascal’s Principle.A:
- When the plunger is pushed in, the water squirts equally from all the holes.
- This shows that the pressure applied to the plunger has been transmitted uniformly throughout the water.
Hydraulic System
- A hydraulic system applies Pascal's principle in its working mechanism. It can be used as a force multiplier.
- In this hydraulic system, a small force, Fl is applied to the small piston X results in a large force, F2 at the large piston Y. The pressure, due to the force, F1, is transmitted by the liquid to the large piston.
- According to Pascal’s principle,
Change of Oil Level in a Hydraulic System
In the diagram to the left, when piston-X is pressed down, piston-Y will be push up. The change of the piston levels of the 2 pistons is given by the following equation:
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.
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.
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
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
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