Day 3
4.4 Nuclear Forces
4.5 Weak forces
4.6 Scope of Classical physics
Points to be Noted
Nuclear forces
The alpha particle is a bare nucleus of Helium. It contains two protons and two neutrons. It is a stable object and once created it can remain intact until it is not made to interact with other objects.
The protons in the nucleus will repel each other due to coulomb force and try to break the nucleus. Why does the Coulomb force fail to break the nucleus?
There are forces called nuclear forces and they are exerted only if the interacting particles are protons or neutrons or both. They are largely attractive, but with a short range. They are weaker than the Coulomb force if the separation between particles is more than 10^-14 m. For separation smaller than this the nuclear force is stronger than the Coulomb force and it holds the nucleus stable.
Radioactivity, nuclear energy (fission, fusion) etc. result from nuclear force.
Weak forces
A neutron can change into proton and simultaneously emit an electron and a particle called antinutrino.
A proton can also change into neutron and simultaneously emit a positron (and a neutrino). The forces responsible for these changes are called weak forces. The effect of this force is experienced inside protons and neutrons only.
Scope of classical physics
Physics based on Newton's Laws of motion, Newton's law of gravitation, Maxwell's electromagnetism, laws of thermodynamics and the Lorentz force is called classical physics. The behaviour of all the bodies of linear sizes greater than 10^-6 m are adequately described by classical physics. Grains of sands and rain drops fall into this range as well as heavenly bodies.
But sub atomic particles like atoms, nuclei, and electrons have sizes smaller than 10^-6 m and they are explained by quantum physics.
The mechanics of particles moving at velocity equal to light are explained by relativistic mechanics formulated by Einstein in 1905.
Tuesday, May 26, 2009
Monday, May 25, 2009
IIT JEE Physics Study Diary - Ch.3 Forces - Day 2
Day 2
4.3 Electromagnetic (EM) forces
Ex. 4.1
Points to note
Electromagnetic force
Apart from gravitational force between any two bodies, the particles may exert upon each other electromagnetic forces.
If two particles having charges q1 and q2 are at rest with respect to the observer, the force between them has a magnitude
F = (1/4πε0)(q1q2/r^2)
Where ε0 = permittivity of air or vacuum = 8.8549 x 10^-12 C² /N-m²
The quantity (1/4πε0) = 9.0 x 10^9 N-m² /C²
q1, q2 = charges
r distance between q1 and q2
This is called coulomb force and it acts along the line joining the particles.
Atoms are composed of electrons, protons and neutrons.
Each electron has 1.6*10^-19 coulomb of negative charge. Each proton has an equal amount of positive charge.
In atoms, the electrons are bound by the electromagnetic force exerted on them by charge on protons. Even the combination of atoms in molecules are brought about by electromagnetic forces only. A lot of atomic and molecular phenomena result from electromagnetic forces between subatomic particles (for example, theory is put forward that charged mesons are responsible for the stability of nucleus).
Examples of electromagnetic force:
1. Bodies in contact: The contact force between bodies in contact arises out of electromagnetic forces acting between the atoms and molecules of the surfaces of the two bodies. The contact force may have a component parallel to the contact surface. This component is known as friction.
2. Tension in a string: Tension in the string is due to electromagnetic forces between atoms or electrons and protons (free electrons and nucleus in metals).
3. Force due to spring: If a spring has natural length x0 and if it is extended to x, it will exert a force
F = k|x-x0| = k|∆x|
k, the proportionality constant is called the spring constant. This force comes into picture due to the electromagnetic forces between the atoms of the material.
Formulas in the session
F = (1/4πε0)(q1q2/r^2)
Where ε0 = permittivity of air or vacuum = 8.8549 x 10^-12 C² /N-m²
The quantity (1/4πε0) = 9.0 x 10^9 N-m² /C²
q1, q2 = charges
r distance between q1 and q2
Each electron has 1.6*10^-19 coulomb of negative charge. Each proton has an equal amount of positive charge.
Force due to spring: If a spring has natural length x0 and if it is extended to x, it will exert a force
F = k|x-x0| = k|∆x|
k, the proportionality constant is called the spring constant.
4.3 Electromagnetic (EM) forces
Ex. 4.1
Points to note
Electromagnetic force
Apart from gravitational force between any two bodies, the particles may exert upon each other electromagnetic forces.
If two particles having charges q1 and q2 are at rest with respect to the observer, the force between them has a magnitude
F = (1/4πε0)(q1q2/r^2)
Where ε0 = permittivity of air or vacuum = 8.8549 x 10^-12 C² /N-m²
The quantity (1/4πε0) = 9.0 x 10^9 N-m² /C²
q1, q2 = charges
r distance between q1 and q2
This is called coulomb force and it acts along the line joining the particles.
Atoms are composed of electrons, protons and neutrons.
Each electron has 1.6*10^-19 coulomb of negative charge. Each proton has an equal amount of positive charge.
In atoms, the electrons are bound by the electromagnetic force exerted on them by charge on protons. Even the combination of atoms in molecules are brought about by electromagnetic forces only. A lot of atomic and molecular phenomena result from electromagnetic forces between subatomic particles (for example, theory is put forward that charged mesons are responsible for the stability of nucleus).
Examples of electromagnetic force:
1. Bodies in contact: The contact force between bodies in contact arises out of electromagnetic forces acting between the atoms and molecules of the surfaces of the two bodies. The contact force may have a component parallel to the contact surface. This component is known as friction.
2. Tension in a string: Tension in the string is due to electromagnetic forces between atoms or electrons and protons (free electrons and nucleus in metals).
3. Force due to spring: If a spring has natural length x0 and if it is extended to x, it will exert a force
F = k|x-x0| = k|∆x|
k, the proportionality constant is called the spring constant. This force comes into picture due to the electromagnetic forces between the atoms of the material.
Formulas in the session
F = (1/4πε0)(q1q2/r^2)
Where ε0 = permittivity of air or vacuum = 8.8549 x 10^-12 C² /N-m²
The quantity (1/4πε0) = 9.0 x 10^9 N-m² /C²
q1, q2 = charges
r distance between q1 and q2
Each electron has 1.6*10^-19 coulomb of negative charge. Each proton has an equal amount of positive charge.
Force due to spring: If a spring has natural length x0 and if it is extended to x, it will exert a force
F = k|x-x0| = k|∆x|
k, the proportionality constant is called the spring constant.
Saturday, May 23, 2009
Physics Study Diary - Ch. 4 Forces - Day 1
Day 1 Study Plan
4.1 Introduction
4.2 Gravitational forces
Points to Note
Force
Force is an interaction between two objects.
Force is exerted by an object A on another object B.
Force is a vector quantity. Hence if two or more forces act on a particle, we can find the resultant force using laws of vector addition.
The SI unit for measuring the force is called a newton.
Newton's third law of motion
If a body A exerts a force F on another body B, then B exerts a force -F on A,the two forces acting along the same line.
Gravitational force
Any two bodies attract each other by virtue of their masses.
The force of attraction between two point masses is
F = Gm1m2/r²
where m1 and m2 are the masses of the particles and r is the distance between them.
G is a universal constant having the value 6.67 x 10^-11 N-m²/kg²
The above rule was given for point masses. But it is analytically found that the gravitational force exerted by a spherically symmetric body of mass m1 on another such body of mass m2 kept outside the first body is Gm1m2/r² where r is the distance between the centres of such bodies.
Thus, for the calculation of gravitational force between two spherically symmetric bodies, they can be treated as point masses placed at their centres.
Gravitational force on small bodies by the earth
For earth, the value of radius R and mass M are 6400 km and 6 x 10^24 kg respectively. Hence, the force exerted by earth on a particle of mass m kept at its surface is, F = GMm/R². The direction of this force is towards the centre of the earth.
The quantity GM/R² is a constant and has the dimensions of acceleration.
It is called acceleration due to gravity and is denoted by letter g.
Hence, g and G are different.
Value of g is approximately 9.8 m/s².
In calculations, we often use 10 m/s².
Force exerted on a small body of mass m, kept near the earth's surface is mg in the vertically downward direction.
Gravitational constant is so small that the gravitational force becomes appreciable only when one of the masses has a very large mass.
HC Verma gives the example of Force exerted by a body of 10 kg on another body of 10 kg when they are separted by a distance of 0.5 m. The force comes out to be 2.7*10^-8 N which can hold only 3 microgram. Such forces can be neglected in practice.
Hence we consider only gravitational force exerted by earth.
Formulas
1. F = Gm1m2/r²
where m1 and m2 are the masses of the particles and r is the distance between them.
G is a universal constant having the value 6.67 x 10^-11 N-m²/kg²
2. Force exerted by earth on a particle of mass m kept at its surface is, F = gm/R².
g = approximately 9.8 m/s².
In calculations, we often use 10 m/s²
4.1 Introduction
4.2 Gravitational forces
Points to Note
Force
Force is an interaction between two objects.
Force is exerted by an object A on another object B.
Force is a vector quantity. Hence if two or more forces act on a particle, we can find the resultant force using laws of vector addition.
The SI unit for measuring the force is called a newton.
Newton's third law of motion
If a body A exerts a force F on another body B, then B exerts a force -F on A,the two forces acting along the same line.
Gravitational force
Any two bodies attract each other by virtue of their masses.
The force of attraction between two point masses is
F = Gm1m2/r²
where m1 and m2 are the masses of the particles and r is the distance between them.
G is a universal constant having the value 6.67 x 10^-11 N-m²/kg²
The above rule was given for point masses. But it is analytically found that the gravitational force exerted by a spherically symmetric body of mass m1 on another such body of mass m2 kept outside the first body is Gm1m2/r² where r is the distance between the centres of such bodies.
Thus, for the calculation of gravitational force between two spherically symmetric bodies, they can be treated as point masses placed at their centres.
Gravitational force on small bodies by the earth
For earth, the value of radius R and mass M are 6400 km and 6 x 10^24 kg respectively. Hence, the force exerted by earth on a particle of mass m kept at its surface is, F = GMm/R². The direction of this force is towards the centre of the earth.
The quantity GM/R² is a constant and has the dimensions of acceleration.
It is called acceleration due to gravity and is denoted by letter g.
Hence, g and G are different.
Value of g is approximately 9.8 m/s².
In calculations, we often use 10 m/s².
Force exerted on a small body of mass m, kept near the earth's surface is mg in the vertically downward direction.
Gravitational constant is so small that the gravitational force becomes appreciable only when one of the masses has a very large mass.
HC Verma gives the example of Force exerted by a body of 10 kg on another body of 10 kg when they are separted by a distance of 0.5 m. The force comes out to be 2.7*10^-8 N which can hold only 3 microgram. Such forces can be neglected in practice.
Hence we consider only gravitational force exerted by earth.
Formulas
1. F = Gm1m2/r²
where m1 and m2 are the masses of the particles and r is the distance between them.
G is a universal constant having the value 6.67 x 10^-11 N-m²/kg²
2. Force exerted by earth on a particle of mass m kept at its surface is, F = gm/R².
g = approximately 9.8 m/s².
In calculations, we often use 10 m/s²
Friday, May 22, 2009
IIT JEE 2011 Physics Study Diary - Ch.3 Rest and Motion - Day 5
Rest and Motion
Day 5 study plan
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Points to Note
The main theme of the section is expressing velocity w.r.t. one Frame into velocity w.r.t. to a different frame
If XOY is one frame called S and X'O'Y' is another frame called S' we can express velocity of a body B w.r.t. S as a combination of velocity of body w.r.t. to S' and velocity of S' w.r.t to S.
V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body w.r.t to S
V(B,S') = velocity of body w.r.t to S'
V(S',S) = velocity of S' w.r.t to S
we can rewrite above equation as
V(B,S') = V(B,S)- V(S',S)
We can interpret the above equation in terms of two bodies. Assume S', and B are two bodies. If we know velocities of two bodies with respect to a common frame (in this case S)we can find the velocity of one body with respect to the other body (V(B,S')
The above expressions for velocity were derived from the relation between position vectors of the body w.r.t. to S and S' and position vector of origin of S' with respect to origin of S.
r(B,S) = r(B,S')+r(S',S)
Differentiating the position vectors with respect to gives respective velocity
Formulas covered in the session
26. r(B,S) = r(B,S')+r(S',S)
Where
r(B,S) = Position vector
r(B,S') = Position vector
r(S',S) = Position vector
27. V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body wrt to S)
V(B,S') = velocity of body wrt to S')
V(S',S) = velocity of S' wrt to S)
we can rewrite above equation as
28. V(B,S') = V(B,S)- V(S',S)
Day 5 study plan
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Points to Note
The main theme of the section is expressing velocity w.r.t. one Frame into velocity w.r.t. to a different frame
If XOY is one frame called S and X'O'Y' is another frame called S' we can express velocity of a body B w.r.t. S as a combination of velocity of body w.r.t. to S' and velocity of S' w.r.t to S.
V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body w.r.t to S
V(B,S') = velocity of body w.r.t to S'
V(S',S) = velocity of S' w.r.t to S
we can rewrite above equation as
V(B,S') = V(B,S)- V(S',S)
We can interpret the above equation in terms of two bodies. Assume S', and B are two bodies. If we know velocities of two bodies with respect to a common frame (in this case S)we can find the velocity of one body with respect to the other body (V(B,S')
The above expressions for velocity were derived from the relation between position vectors of the body w.r.t. to S and S' and position vector of origin of S' with respect to origin of S.
r(B,S) = r(B,S')+r(S',S)
Differentiating the position vectors with respect to gives respective velocity
Formulas covered in the session
26. r(B,S) = r(B,S')+r(S',S)
Where
r(B,S) = Position vector
r(B,S') = Position vector
r(S',S) = Position vector
27. V(B,S) = V(B,S')+V(S',S)
Where
V(B,S) = velocity of body wrt to S)
V(B,S') = velocity of body wrt to S')
V(S',S) = velocity of S' wrt to S)
we can rewrite above equation as
28. V(B,S') = V(B,S)- V(S',S)
Thursday, May 21, 2009
IIT JEE 2011 Physics Study Diary - Ch.3 Rest and Motion - Day 4
Day 4 - Study Plan
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Points to Note
Motion in a plane
Motion in plane is described by x coordinate and y coordinate, if we choose X-Y plane. You can imagine time t is on the third axis.
The position of the particle or the body can be described by x and y coordinates.
r = xi = yj
Displacement during time period t to t+Δt can be represented by Δr
Δr = Δxi = Δyj
Then Δr/Δt = (Δx/Δt)i = (Δy/Δt)j
Taking the limits as Δt tends to zero
v = dr/dt = (dx/dt)i+(dy/dt)j ... (3.15)
Hence x component of velocity is dx/dt
The x-coordinate, the x component of velocity, and the x component of acceleration are related by equations of straight line motion along X axis.
Similarly y components.
Projectile
Projectile motion is an important example of motion in a plane.
What is a projectile? When a particle is thrown obliquely near the earth's surface, it is called a projectile. It moves along a curved path. If we assume the particle is close to the earth and negligible air resistance to the motion of the particle, the acceleration of the particle will be constant. We solve projectile problems with the assumption that acceleration is constant.
Vertical motion of the projectile is the motion along Y axis and horizontal motion is motion along X axis.
Terms used in describing projectile motion
Point of projection
Angle of projection
Horizontal range
Time of flight
Maximum height reached
The motion of projectile can be discussed separately for the horizontal and vertical parts.
The origin is taken as the point of projection.
The instant the particle is projected is taken as t = 0.
X-Y plane is the plane of motion.
The horizontal line OX is taken as the X axis.
Vertical line OY is the Y axis.
Vertically upward direction is taken as positive direction of Y
Initial velocity of the particle = u
Angle between the velocity and horizontal axis = θ
ux – x-component of velocity = u cos θ
ax – x component of acceleration = 0
uy – y component of velocity = u sin θ
ay = y component of acceleration = -g
Horizontal motion of the projectile – Equations of motion
ux = u cos θ
ax = 0
vx = ux +axt = ux = u cos θ (as ax = 0)
Hence x component of the velocity remains constant.
Displacement in horizontal direction = x = uxt+1/2ax t²
As ax = 0, x = ux t = ut cos θ
Vertical motion – Equations of motion
uy = u sin θ
ay = -g
vy = uy – gt
Displacement in y direction = y = uyt – ½ gt²
vy² = uy² - 2gy
22. Time of flight of the projectile = (2u sin θ)/g
23. OB = (u²sin 2θ)/g
24. t = (u sin θ)/g
At t vertical component of velocity is zero.
25. Maximum height reached by the projectile = (u² sin²θ)/2g
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Points to Note
Motion in a plane
Motion in plane is described by x coordinate and y coordinate, if we choose X-Y plane. You can imagine time t is on the third axis.
The position of the particle or the body can be described by x and y coordinates.
r = xi = yj
Displacement during time period t to t+Δt can be represented by Δr
Δr = Δxi = Δyj
Then Δr/Δt = (Δx/Δt)i = (Δy/Δt)j
Taking the limits as Δt tends to zero
v = dr/dt = (dx/dt)i+(dy/dt)j ... (3.15)
Hence x component of velocity is dx/dt
The x-coordinate, the x component of velocity, and the x component of acceleration are related by equations of straight line motion along X axis.
Similarly y components.
Projectile
Projectile motion is an important example of motion in a plane.
What is a projectile? When a particle is thrown obliquely near the earth's surface, it is called a projectile. It moves along a curved path. If we assume the particle is close to the earth and negligible air resistance to the motion of the particle, the acceleration of the particle will be constant. We solve projectile problems with the assumption that acceleration is constant.
Vertical motion of the projectile is the motion along Y axis and horizontal motion is motion along X axis.
Terms used in describing projectile motion
Point of projection
Angle of projection
Horizontal range
Time of flight
Maximum height reached
The motion of projectile can be discussed separately for the horizontal and vertical parts.
The origin is taken as the point of projection.
The instant the particle is projected is taken as t = 0.
X-Y plane is the plane of motion.
The horizontal line OX is taken as the X axis.
Vertical line OY is the Y axis.
Vertically upward direction is taken as positive direction of Y
Initial velocity of the particle = u
Angle between the velocity and horizontal axis = θ
ux – x-component of velocity = u cos θ
ax – x component of acceleration = 0
uy – y component of velocity = u sin θ
ay = y component of acceleration = -g
Horizontal motion of the projectile – Equations of motion
ux = u cos θ
ax = 0
vx = ux +axt = ux = u cos θ (as ax = 0)
Hence x component of the velocity remains constant.
Displacement in horizontal direction = x = uxt+1/2ax t²
As ax = 0, x = ux t = ut cos θ
Vertical motion – Equations of motion
uy = u sin θ
ay = -g
vy = uy – gt
Displacement in y direction = y = uyt – ½ gt²
vy² = uy² - 2gy
22. Time of flight of the projectile = (2u sin θ)/g
23. OB = (u²sin 2θ)/g
24. t = (u sin θ)/g
At t vertical component of velocity is zero.
25. Maximum height reached by the projectile = (u² sin²θ)/2g
Wednesday, May 20, 2009
IIT JEE Physics Study Diary - Ch.3 Rest and Motion - Day 3
Day 3 Study Plan
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3,4,5,6,
Points to Note
Motion in a straight line
As the motion is constrained to move on a straight line, choose the straight line in which motion is taking place as X-axis. Hence x represent the position of the particle at any time instant t. If you want you can imagine a graph between t and x but now t in on the vertical axis and x is on the horizontal axis.
Generally origin is taken at the point where the particle is situated at time t = 0.
Position of the particle at time t is given by x and also x measures displacement (not distance).
Velocity is v = dx/dt (3.9)
acceleration is a = dv/dt (3.10)
a = d²x/dt² (3.11)
Decelaration
If acceleration is negative, then it is along the negative X-axis. It is called deceleration
Motion with constant acceleration
Using integration the formulas for v velocity at any instant, x position at any instant and relation between v,u,x and a are derived in this section.
If acceleration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)
Then v = u+at (3.12)
x = distance moved in time t = ut+½at² (3.13)
Also v² = u²+2ax (3.14)
u,v, and a as well as may take negative or positive values. When u, v and a are negative it shows velocity or acceleration is in the negative X direction.
Example 3.5
a) The question asked is distance travelled. The expression for x gives only displacement. But the remark is that as the particle does not turn back it is equal to distance travelled. Be careful when initial velocity is positive and the acceleration is negative.
Example 3.6
There was a past JEE question which is based on the variable defined in the example.
Freely falling bodies
In this case take the Y axis as the straight line on which the particle or body is moving.
You can take height above the ground as +y and work out the problems.
You can take the starting position of the body as the origin and work out the problem.
The choice may be yours or some choice may be more appropriate in case of some problems, be clear of the formula that you have to use depending on the choice you made.
g is approximately equal to 9.8 m/s², but for convenience in many problems it is given as 10m/s².
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3,4,5,6,
Points to Note
Motion in a straight line
As the motion is constrained to move on a straight line, choose the straight line in which motion is taking place as X-axis. Hence x represent the position of the particle at any time instant t. If you want you can imagine a graph between t and x but now t in on the vertical axis and x is on the horizontal axis.
Generally origin is taken at the point where the particle is situated at time t = 0.
Position of the particle at time t is given by x and also x measures displacement (not distance).
Velocity is v = dx/dt (3.9)
acceleration is a = dv/dt (3.10)
a = d²x/dt² (3.11)
Decelaration
If acceleration is negative, then it is along the negative X-axis. It is called deceleration
Motion with constant acceleration
Using integration the formulas for v velocity at any instant, x position at any instant and relation between v,u,x and a are derived in this section.
If acceleration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)
Then v = u+at (3.12)
x = distance moved in time t = ut+½at² (3.13)
Also v² = u²+2ax (3.14)
u,v, and a as well as may take negative or positive values. When u, v and a are negative it shows velocity or acceleration is in the negative X direction.
Example 3.5
a) The question asked is distance travelled. The expression for x gives only displacement. But the remark is that as the particle does not turn back it is equal to distance travelled. Be careful when initial velocity is positive and the acceleration is negative.
Example 3.6
There was a past JEE question which is based on the variable defined in the example.
Freely falling bodies
In this case take the Y axis as the straight line on which the particle or body is moving.
You can take height above the ground as +y and work out the problems.
You can take the starting position of the body as the origin and work out the problem.
The choice may be yours or some choice may be more appropriate in case of some problems, be clear of the formula that you have to use depending on the choice you made.
g is approximately equal to 9.8 m/s², but for convenience in many problems it is given as 10m/s².
Physics Study Diary for IIT JEE - Ch.3 Rest and Motion - Day 2
Plan for Day 2
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
WOE 3 to 4
Exercises: 1 to 5
Points to Note
Average velocity
Average speed and average velocity of a body over a specified time interval may not turnout to be same.
Example See the worked out example 2 of HC Verma's book.
The teacher made 10 rounds back and forth in the room and the total distance moved is 800 feet (10 rounds back and forth of 40 ft room). As the time taken is 50 minutes, average speed is 800/50 = 16ft/min.
But because he went out of the same door that he has entered, displacement is zero and hence average velocity is zero.
Instantaneous velocity
Average acceleration
Instantaneous acceleration
Position Vector: If we join the origin to the position of a particle by a straight line and put an arrow towards the position of the particle, we get the position vector of the particle.
If the particle moves from position A to position B, we can define position vector of A and position vector of B and OB - OA will give displacement ( a vector quantity).
Another point to note: slope of velocity-time diagram gives the instant acceleration at that point.
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
WOE 3 to 4
Exercises: 1 to 5
Points to Note
Average velocity
Average speed and average velocity of a body over a specified time interval may not turnout to be same.
Example See the worked out example 2 of HC Verma's book.
The teacher made 10 rounds back and forth in the room and the total distance moved is 800 feet (10 rounds back and forth of 40 ft room). As the time taken is 50 minutes, average speed is 800/50 = 16ft/min.
But because he went out of the same door that he has entered, displacement is zero and hence average velocity is zero.
Instantaneous velocity
Average acceleration
Instantaneous acceleration
Position Vector: If we join the origin to the position of a particle by a straight line and put an arrow towards the position of the particle, we get the position vector of the particle.
If the particle moves from position A to position B, we can define position vector of A and position vector of B and OB - OA will give displacement ( a vector quantity).
Another point to note: slope of velocity-time diagram gives the instant acceleration at that point.
Monday, May 18, 2009
Physics Study Diary for IIT JEE 2011 - Ch.3 Rest and Motion
I am planning to study the Physics chapters as per the study plan that I have given. This study would be of help to me in preparing JEE Level Revision problem set for each chapter.
Today (19.5.2009) I did the following portion
Day 1
3.1 Rest and Motion
3.2 Distance and displacement
Ex. 3.1
3.3 average speed and instantaneous speed
Ex. 3.2,3.3
Worked out examples 1,2
Points to note.
3.1 Rest and Motion
Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.
To identify the rest or motion, we need to locate the position of a particle with respect to a frame of reference. The frame of reference will have three mutually perpendicular axes (X-Y-Z) and the particle can be represented by coordinates x,y,z.
If all coordinates remain unchanged as time passes, we say that particle is at rest. If there is change in any of the coordinates with time, we say the particle or the body represented by the particle is having motion.
I some problems or situations frame of reference is specifically mentioned. Otherwise the frame of reference is understood more easily from the context.
Figure 1: A man with a pistol threatening and asking people not to move.
3.2 Distance and Displacement
If a particle moves from position A to Position B in time t, displacement is the length of the straight line joining A to B. The direction of a vector representing this displacement is from A to B. Displacement is a vector quantity. It has both magnitude and direction.
In the movement between positions A and B the particle may take the path ACB. The length of the path ACB will be distance travelled by the particle. It is only scalar quantity. It has not direction.
3.3 Average speed and Instantaneous speed
The average speed of a particle in a specific time interval is defined as the distance travelled by the particle divided by the time interval.
We can plot the distance s as a function of time. In this graph, the instantaneous speed at time t equals the slope of the tangent at the time t. The average speed in a time interval t to t+Δt become equal to the slope of the chord Δs/Δt. As Δt becomes approaches zero, this average speed becomes instantaneous speed and ds/dt becomes the instantaneous speed.
If we plot a speed versus time graph ( v versus t), the distance travelled in time t (t1 to t2) will be equal to the area bounded by the curve v = f(t), x axis, and the two ordinates t = t1 and t = t2.
In terms of integration it can be represented as s = ∫vdt from t1 to t2
For study plan for IIT JEE 2011 for the year 2009-10
http://iit-jee-2011.blogspot.com/2009/04/it-jee-2011-annual-study-plan-for.html
Today (19.5.2009) I did the following portion
Day 1
3.1 Rest and Motion
3.2 Distance and displacement
Ex. 3.1
3.3 average speed and instantaneous speed
Ex. 3.2,3.3
Worked out examples 1,2
Points to note.
3.1 Rest and Motion
Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.
To identify the rest or motion, we need to locate the position of a particle with respect to a frame of reference. The frame of reference will have three mutually perpendicular axes (X-Y-Z) and the particle can be represented by coordinates x,y,z.
If all coordinates remain unchanged as time passes, we say that particle is at rest. If there is change in any of the coordinates with time, we say the particle or the body represented by the particle is having motion.
I some problems or situations frame of reference is specifically mentioned. Otherwise the frame of reference is understood more easily from the context.
Figure 1: A man with a pistol threatening and asking people not to move.
3.2 Distance and Displacement
If a particle moves from position A to Position B in time t, displacement is the length of the straight line joining A to B. The direction of a vector representing this displacement is from A to B. Displacement is a vector quantity. It has both magnitude and direction.
In the movement between positions A and B the particle may take the path ACB. The length of the path ACB will be distance travelled by the particle. It is only scalar quantity. It has not direction.
3.3 Average speed and Instantaneous speed
The average speed of a particle in a specific time interval is defined as the distance travelled by the particle divided by the time interval.
We can plot the distance s as a function of time. In this graph, the instantaneous speed at time t equals the slope of the tangent at the time t. The average speed in a time interval t to t+Δt become equal to the slope of the chord Δs/Δt. As Δt becomes approaches zero, this average speed becomes instantaneous speed and ds/dt becomes the instantaneous speed.
If we plot a speed versus time graph ( v versus t), the distance travelled in time t (t1 to t2) will be equal to the area bounded by the curve v = f(t), x axis, and the two ordinates t = t1 and t = t2.
In terms of integration it can be represented as s = ∫vdt from t1 to t2
For study plan for IIT JEE 2011 for the year 2009-10
http://iit-jee-2011.blogspot.com/2009/04/it-jee-2011-annual-study-plan-for.html
Nobel Prize Winners in Physics from 1901-1925 - IIT JEE - Extra-curricular Study
Year Physics
1901 Röntgen, Wilhelm Conrad
For discovery of X-rays
1902 Lorentz, Hendrik A.
Zeeman, Pieter
1903 Becquerel, Henri;
Curie, Pierre;
Curie, Marie
1904 Rayleigh, Lord
1905 Lenard, Philipp
1906 Thomson, J. J.
1907 Michelson, Albert A.
1908 Lippmann, Gabriel
1909 Braun, Ferdinand
Marconi, Guglielmo
1910 van der Waals, Johannes Diderik
1911 Wien, Wilhelm
1912 Dalén, Gustaf
1913 Onnes, Heike Kamerlingh
1914 von Laue, Max
1915 Bragg, William Henry:
Bragg, William Lawrence
1916 None
1917 Barkla, Charles Glover
1918 Planck, Max
1919 Stark, Johannes
1920 Guillaume, Charles Edouard
1921 Einstein, Albert
1922 Bohr, Niels
1923 Millikan, Robert A.
1924 Siegbahn, Manne
1925 Franck, James
Hertz, Gustav
Sources
http://nobelprize.org/nobel_prizes/physics/laureates/http://history1900s.about.com/library/misc/blnobelphysics.htm
1901 Röntgen, Wilhelm Conrad
For discovery of X-rays
1902 Lorentz, Hendrik A.
Zeeman, Pieter
1903 Becquerel, Henri;
Curie, Pierre;
Curie, Marie
1904 Rayleigh, Lord
1905 Lenard, Philipp
1906 Thomson, J. J.
1907 Michelson, Albert A.
1908 Lippmann, Gabriel
1909 Braun, Ferdinand
Marconi, Guglielmo
1910 van der Waals, Johannes Diderik
1911 Wien, Wilhelm
1912 Dalén, Gustaf
1913 Onnes, Heike Kamerlingh
1914 von Laue, Max
1915 Bragg, William Henry:
Bragg, William Lawrence
1916 None
1917 Barkla, Charles Glover
1918 Planck, Max
1919 Stark, Johannes
1920 Guillaume, Charles Edouard
1921 Einstein, Albert
1922 Bohr, Niels
1923 Millikan, Robert A.
1924 Siegbahn, Manne
1925 Franck, James
Hertz, Gustav
Sources
http://nobelprize.org/nobel_prizes/physics/laureates/http://history1900s.about.com/library/misc/blnobelphysics.htm
Friday, May 15, 2009
Noble Prize Winners in Physics from 2001 - IIT JEE - Extra-curricular Study
Year - Scientist/Physicist
2001 - Eric A. Cornell, Carl E. Wieman, (US), Wolfgang Ketterle (Germany)
2002 - Riccardo Giacconi, Rayond Davis Jr. (US), Masatoshi Koshiba (Jap)
2003 - Alexei A. Abrikov,(US-Rus), Vitaly I. Ginzburg (Rus). Anthony J. Leggett, (UK-US)
2004 - David J. Gross, H. David Politzer, Frank Wilczek (US)
2005 - Roy glauber, John Hall (both US), and Theodor Haensch (Germany)
2006
2007
2008
2001 - Eric A. Cornell, Carl E. Wieman, (US), Wolfgang Ketterle (Germany)
2002 - Riccardo Giacconi, Rayond Davis Jr. (US), Masatoshi Koshiba (Jap)
2003 - Alexei A. Abrikov,(US-Rus), Vitaly I. Ginzburg (Rus). Anthony J. Leggett, (UK-US)
2004 - David J. Gross, H. David Politzer, Frank Wilczek (US)
2005 - Roy glauber, John Hall (both US), and Theodor Haensch (Germany)
2006
2007
2008
Subscribe to:
Posts (Atom)