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Showing posts with label Rest and Motion. Show all posts
Showing posts with label Rest and Motion. Show all posts
Sunday, April 8, 2012
Friday, November 7, 2008
3. Rest and Motion: Kinematics - Revision Facilitator
3. Rest and Motion: Kinematics
Sections
3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average acceleration and instantaneous acceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame
Mind Map
Rest - Motion – displacement – Speed – Velocity – Acceleration – Frame of Reference
Displacement – Distance moved
Speed - average speed - instantaneous speed
Velocity - Average velocity - instantaneous velocity – Acceleration
Acceleration - Average acceleration - instantaneous acceleration
Motion - straight line - Motion in a plane - Projectile motion
Frame of Reference – Change in Frame of Reference
Sections
3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average acceleration and instantaneous acceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame
Mind Map
Rest - Motion – displacement – Speed – Velocity – Acceleration – Frame of Reference
Displacement – Distance moved
Speed - average speed - instantaneous speed
Velocity - Average velocity - instantaneous velocity – Acceleration
Acceleration - Average acceleration - instantaneous acceleration
Motion - straight line - Motion in a plane - Projectile motion
Frame of Reference – Change in Frame of Reference
Wednesday, May 7, 2008
Concept Review - Chapter 3 Rest and Motion: Kinematics
Rest, Motion and Reference Frame
Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.
For example, a book placed on the table remains on the table and we say that it is not moving, it is at rest. However, if we station ourselves on the moon, the whole earth is changing its position, and os the room, the table and the book are all continuously changing their positions.
To locate the position of a particle we need a frame of reference. We can fix up three mutually perpendicular axes, name them X,Y and Z and then specify the position of the particle with respect to that frame. Then if the coordinates of the particle change with respect to time, we can say that the body is moving with respect to this frame.
Many times the choice of frame is clear from the context.
Displacement
The magnitude of the displacement is the length of the straight line joining the initial and final position.
Displacement has magnitude as well as direction (initial position and final position).
It is a vector quantity.
Displacements add according to triangle rule of vectors.
For example, if a book kept on table is displaced and the table is also displaced. The net displacement of the book is obtained by the vector addition of the two displacements.
Distance
Average speed
Instantaneous speed
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
Motion in a straight line
Choose the line as the X-axis
Position of the particle at time t is given by x.
Velocity is v = dx/dt
acceleration is a = dv/dt = d²x/dt²
If accelaration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)
The v = u+at
x = distance moved in time t = ut+½at²
Also
v² = u²+2ax
Motion in a plane
Motion in plane is described by x coordinate and y coordinate.
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.
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 – 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
Time of flight of projectile
A projectile is projected from the ground at point O, and after some travel in the space, it reaches the ground at point B. The time taken for this travel is called time of flight of the projectile. Over this time, the displacement in y direction becomes zero.
Hence we can write y = uyt – ½ gt² = u sin θ*t - ½ gt²
solving we get T = (2u sin θ)/g
Time of flight of the projectile = (2u sin θ)/g
Range of the projectile:
The distance OB travelled by the projetile in the horizontal direction is the range.
OB = (u²sin 2θ)/g
Maximum height reached
When the projectile reaches the highest level in vertical direction, the vertical component of the velocity becomes zero.
The time taken for the vertical component of velocity to become zero is
vy = 0 = uy - gt = u sin θ - gt
So t = (u sin θ)/g
So time taken for vertical component of velocity to become zero is (u sin θ)/g.
Note that this time is half of Time of flight.
so maximum height reached in t = (u sin θ)/g = (u² sin²θ)/2g
Expressing velocity w.r.t. one Frame 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 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 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
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')
Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.
For example, a book placed on the table remains on the table and we say that it is not moving, it is at rest. However, if we station ourselves on the moon, the whole earth is changing its position, and os the room, the table and the book are all continuously changing their positions.
To locate the position of a particle we need a frame of reference. We can fix up three mutually perpendicular axes, name them X,Y and Z and then specify the position of the particle with respect to that frame. Then if the coordinates of the particle change with respect to time, we can say that the body is moving with respect to this frame.
Many times the choice of frame is clear from the context.
Displacement
The magnitude of the displacement is the length of the straight line joining the initial and final position.
Displacement has magnitude as well as direction (initial position and final position).
It is a vector quantity.
Displacements add according to triangle rule of vectors.
For example, if a book kept on table is displaced and the table is also displaced. The net displacement of the book is obtained by the vector addition of the two displacements.
Distance
Average speed
Instantaneous speed
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
Motion in a straight line
Choose the line as the X-axis
Position of the particle at time t is given by x.
Velocity is v = dx/dt
acceleration is a = dv/dt = d²x/dt²
If accelaration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)
The v = u+at
x = distance moved in time t = ut+½at²
Also
v² = u²+2ax
Motion in a plane
Motion in plane is described by x coordinate and y coordinate.
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.
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 – 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
Time of flight of projectile
A projectile is projected from the ground at point O, and after some travel in the space, it reaches the ground at point B. The time taken for this travel is called time of flight of the projectile. Over this time, the displacement in y direction becomes zero.
Hence we can write y = uyt – ½ gt² = u sin θ*t - ½ gt²
solving we get T = (2u sin θ)/g
Time of flight of the projectile = (2u sin θ)/g
Range of the projectile:
The distance OB travelled by the projetile in the horizontal direction is the range.
OB = (u²sin 2θ)/g
Maximum height reached
When the projectile reaches the highest level in vertical direction, the vertical component of the velocity becomes zero.
The time taken for the vertical component of velocity to become zero is
vy = 0 = uy - gt = u sin θ - gt
So t = (u sin θ)/g
So time taken for vertical component of velocity to become zero is (u sin θ)/g.
Note that this time is half of Time of flight.
so maximum height reached in t = (u sin θ)/g = (u² sin²θ)/2g
Expressing velocity w.r.t. one Frame 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 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 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
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')
Friday, February 8, 2008
IIT JEE Physics Formula Revision 3. Rest and motion :kinematics
1. vav = s/(t2 - t1)
where vav = average speed
2. v = limΔt→0 Δs/Δt = ds/dt
where v = instantaneous speed
3. s = ∫vdt from t1 to t2
where s = distance travelled during time t1 to t2
4. vav = (r2 - r1)/(t2-t1)
Where
vav = average velocity
r2, r1 = position vectors of a particle
t2, t1 = time instants
5. v = lim Δt→0 Δr/Δt = dr/dt
6. v = |dr|/dt = ds/dt
Where v = magnitude of velocity
s = displacement, for small time intervals magnitude of displacement will be equal to distance.
7. aav = (v2 - v1)/(t2-t1)
8. a = lim Δt→0 Δv/Δt = dv/dt
For Motion in straight line
9. Velocity is v = dx/dt
10. a = dv/dt
11. acceleration is a = dv/dt = d²x/dt²
For constant acceleration
12. v = u+at
13. x = distance moved in time t = ut+½at²
14. v² = u²+2ax
For motion in plane
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 (in t = (u sin θ)/g) = (u² sin²θ)/2g
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)
where vav = average speed
2. v = limΔt→0 Δs/Δt = ds/dt
where v = instantaneous speed
3. s = ∫vdt from t1 to t2
where s = distance travelled during time t1 to t2
4. vav = (r2 - r1)/(t2-t1)
Where
vav = average velocity
r2, r1 = position vectors of a particle
t2, t1 = time instants
5. v = lim Δt→0 Δr/Δt = dr/dt
6. v = |dr|/dt = ds/dt
Where v = magnitude of velocity
s = displacement, for small time intervals magnitude of displacement will be equal to distance.
7. aav = (v2 - v1)/(t2-t1)
8. a = lim Δt→0 Δv/Δt = dv/dt
For Motion in straight line
9. Velocity is v = dx/dt
10. a = dv/dt
11. acceleration is a = dv/dt = d²x/dt²
For constant acceleration
12. v = u+at
13. x = distance moved in time t = ut+½at²
14. v² = u²+2ax
For motion in plane
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 (in t = (u sin θ)/g) = (u² sin²θ)/2g
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)
Tuesday, October 23, 2007
Study guide H C Verma JEE Physics Ch. 3 REST AND MOTION - KINEMATICS
Inertial and uniformly accelerated frames of reference;
-----------
HCV Chapter Sections
3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average accleration and instantaneous aceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame
-------
Study Plan
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 example 1
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-2011.html
Day 2
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
Exercises: 1 to 5
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-ch3.html
Day 3
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3 to 6
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/iit-jee-physics-study-diary-ch3-rest.html
Day 4
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Day 5
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Day 6
Formula Revision
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
WOE 7 to 10, and 13,15, 19,20
Day 7
Exercises 6 to 15
Day 8
Exercises 16 to 25
Day 9
Exercises 26 to 35
Day 10
Exercises 36 to 45
Revision Period
Day 11
Exercises 46 to 50
Day 12
Exercises 51,52
Objective I
Day 13
Exercises
Objective II
Day 14
Exercises
Question for short answer
Day 15
Concept review
http://iit-jee-physics.blogspot.com/2008/05/concept-review-chapter-3-rest-and.html
Day 16
Formula review
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
Day 17 to 20
Test paper problems 10 per day can be solved.
Inertial and accelerated frames of reference.
Section 3.9 covers material related to it. But it is not using the terms inertial frame of reference.
------
Past JEE question
1982
A particle is moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
a) zero
b) [1/Sqrt(2)]m/s^2 towards northwest
c) 1/2 m/s^2 towards northwest
d) 1/2 m/s^2 towards north
Answer b.
You can find acceleration in the direction of west and check that velocity become zero in east direction in 5 seconds and also velocity in north direction becomes 5 m/s.
----------------------
Audio visual lectures
http://www.curriki.org/nroc/Introductory_Physics_1/lesson01/Container.html
Motion in two dimensions (plane)
http://www.curriki.org/nroc/Introductory_Physics_1/lesson02/Container.html
web sites
one dimensional motion
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-15-05.pdf
-----------
HCV Chapter Sections
3.1 Rest and Motion
3.2 Distance and displacement
3.3 average speed and instantaneous speed
3.4 Average velocity and instantaneous velocity
3.5 Average accleration and instantaneous aceleration
3.6 Motion in a straight line
3.7 Motion in a plane
3.8 Projectile motion
3.9 Change of frame
-------
Study Plan
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 example 1
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-2011.html
Day 2
4. Average velocity and instantaneous velocity
Ex. 3.4
Worked out example 2
3.5 Average accleration and instantaneous aceleration
Exercises: 1 to 5
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/physics-study-diary-for-iit-jee-ch3.html
Day 3
3.6 Motion in a straight line
Ex. 3.5.3.6, 3.7
WOE 3 to 6
Points to Note in the session
http://iit-jee-physics.blogspot.com/2009/05/iit-jee-physics-study-diary-ch3-rest.html
Day 4
3.7 Motion in a plane
Ex. 3.8
3.8 Projectile motion
Ex. 3.9
WOE 11,12, 14
Day 5
3.9 Change of frame
Ex. 3.10, 3.11
WOE 16,17, 18
Day 6
Formula Revision
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
WOE 7 to 10, and 13,15, 19,20
Day 7
Exercises 6 to 15
Day 8
Exercises 16 to 25
Day 9
Exercises 26 to 35
Day 10
Exercises 36 to 45
Revision Period
Day 11
Exercises 46 to 50
Day 12
Exercises 51,52
Objective I
Day 13
Exercises
Objective II
Day 14
Exercises
Question for short answer
Day 15
Concept review
http://iit-jee-physics.blogspot.com/2008/05/concept-review-chapter-3-rest-and.html
Day 16
Formula review
http://iit-jee-physics.blogspot.com/2008/02/iit-jee-physics-formula-revision-3-rest.html
Day 17 to 20
Test paper problems 10 per day can be solved.
Inertial and accelerated frames of reference.
Section 3.9 covers material related to it. But it is not using the terms inertial frame of reference.
------
Past JEE question
1982
A particle is moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
a) zero
b) [1/Sqrt(2)]m/s^2 towards northwest
c) 1/2 m/s^2 towards northwest
d) 1/2 m/s^2 towards north
Answer b.
You can find acceleration in the direction of west and check that velocity become zero in east direction in 5 seconds and also velocity in north direction becomes 5 m/s.
----------------------
Audio visual lectures
http://www.curriki.org/nroc/Introductory_Physics_1/lesson01/Container.html
Motion in two dimensions (plane)
http://www.curriki.org/nroc/Introductory_Physics_1/lesson02/Container.html
web sites
one dimensional motion
http://plasma4.sr.unh.edu/ng/phys407/phys407-6-15-05.pdf
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