Sunday, September 30, 2007

Inelastic Collison




Kinetic Energy lost


Elastic Collision II


Elastic Collision


Conservation of Momentum

from www.physicsclassroom.com





























Momentum is conserved.







force and momentum















The rate of change of momentum of an object is equal to the net force applied to it.

In order to change the momentum of an object, a force must be applied .
The time rate of change of momentum of an object is equal to the net force acting on it.

Impulse and Momentum






If momentum changes, it’s because mass or velocity change.

Most often mass doesn’t change so velocity changes and that is acceleration.

And mass x acceleration = force

Applying a force over a time interval to an object changes the momentum

Force x time interval = Impulse

Impulse = F x t or Ft = mv

What’s Momentum ?

Momentum = mass x velocity

momentum = mv

if direction is not an important factor : . . momentum = mass x speed

So, A really slow moving truck and an extremely fast roller skate can have the same momentum.

Unit: kgm/s

Thursday, September 20, 2007

Linear Momentum (Chapter 7)

Objectives

1. Define linear momentum and write the mathematical formula for linear momentum from memory.

2. Distinguish between the unit of force and momentum.

3. Write Newton's Second Law of Motion in terms of momentum.

4. Define impulse and write the equation that connects impulse and momentum.

5. State the Law of Conservation of Momentum and write, in vector form, the law for a system involving two or more point masses.

6. Distinguish between a perfectly elastic collision and a completely inelastic collision.

7. Apply the laws of conservation of momentum and energy to problems involving collisions between two point masses.

8. Define center of mass and center of gravity and distinguish between the two concepts.

Practice Questions VI

1. Two men, Joel and Jerry, push against a wall. Jerry stops after 10 min, while Joel is able to push for 5 min longer. Compare the work against the wall they each do.
(A) Joel does 50% more work than Jerry.
(B) Jerry does 50% more work than Joel.
(C) Joel does 75% more work than Jerry.
(D) Neither of them do any work.



2. A simple pendulum, consisting of a mass m and a string, swings upward, making an angle θ with the vertical. The work done by the tension force is

(A) zero.

(B) mg.

(C) mg cos theta

(C) mg sin theta






3. A simple pendulum, consisting of a mass m, is attached to the end of a 1.5 m length of string. If the mass is held out horizontally, and then released from rest, its speed at the bottom is




(A) 4.4 m/s


(B) 5.4 m/s


(C) 9.8 m/s


(D) 17 m/s




4. A 4-kg mass moving with speed 2 m/s, and a 2-kg mass moving with a speed of 4 m/s, are gliding over a horizontal frictionless surface. Both objects encounter the same horizontal force, which directly opposes their motion, and are brought to rest by it. Which statement best describes their respective stopping distances?
(A) The 4-kg mass travels twice as far as the 2-kg mass before stopping.


(B) The 2-kg mass travels twice as far as the 4-kg mass before stopping.


(C) Both masses travel the same distance before stopping.


(D) The 2 kg mass travels greater than twice as far.




5. A 4-kg mass moving with speed 2 m/s and, an otherwise identical, 2-kg mass moving with a speed of 4 m/s, are gliding over a horizontal surface with friction and are brought to rest by it. Which statement best describes their respective stopping distances?




(A) The 4-kg mass travels twice as far as the 2-kg mass before stopping.


(B) The 2-kg mass travels twice as far as the 4-kg mass before stopping.


(C) Both masses travel the same distance before stopping.


(D) The 2-kg mass travels greater than twice as far.






6. A force that Object A exerts on Object B is observed over a 10-second interval, as shown on the graph. How much work did Object A do during that 10 s?
(A) Zero

(B) 12.5 J

(C) 25 J

(D) 50 J










7. A force that Object A exerts on Object B is observed over a 10-second interval, as shown on the graph. What is the average power output of A into B?

(A) O W

(B) 1.3 W

(C) 2.5 W
(D) 5 W










8. The resultant force you exert on a shopping cart, for a 10 s period, is plotted on the graph shown. How much work did you do during this 10 s interval?
(A) Zero

(B) 12.5 J

(C) 25 J

(D) -25 J






9. The resultant force you exert on a shopping cart, for a 10 s period, is plotted on the graph, shown. Which of the following statements are true?

(A) The average power input into B is greater than zero.

(B) The average power input into A is the same in the first half as the power input in the second half.

(C) The average power equals the instantaneous power.

(D) The average power is zero.
















10. How much work was required to bring the 1000-kg roller coaster from Point P to rest at Point Q at the top of the 50 m peak?

(A) 32,000 J

(B) 50,000 J

(C) 245,000 J

(D) 490,000 J










11. If the roller coaster leaves Point Q from rest, how fast is it traveling at Point R?

(A) 22.1 m/s

(B) 31.3 m/s
(C) 490 m/s

(D) 980 m/s


12. What was the total work done on you by all the forces in the universe between the time just before you awoke this morning and right now?

(A) Don't have a clue

(B) Greater than zero.

(C) Zero.
(D) Can not be calculated.

13. Two cars, starting from rest at the same place, travel by different routes to the same destination. One of the cars passes the other as they drive through it. Which of the following statements will be true?

(A) The work done by friction during the trip was the same for both
(B) The total work done on both is the same.

(C) The work done by gravity is the same on both.

(D) The work done by gravity on both is positive.


14. The work done by friction, f,

(A) equals -fd, where d is the total distance moved.

(B) equals fd, where d is the total distance.

(C) can't easily be calculated because it depends on the angle between f and d.

(D) can't easily be calculated.
15. A 102 kg man climbs a 5.0 meter high stair case at constant speed. How much work does he do?

(A) 510 J

(B) 49 J

(C) 5000 J

(D) 2500 J









15. A ball is released, from rest, at the left side of the loop-the-loop, at the height shown (h = 2R). If the radius of the loop is R and there is no friction, what vertical height does the ball rise to on the other side?

(A) Less than R

(B) R

(C) 2R

(D) Greater than R

Power

P = W / t (work per unit time)

Since W = F d, P = F d / t = F v


Unit: 1 Watt (W) = 1 J/s

problem solving




Mechanical Energy and its Conservation

The change in total mechanical energy is the work done by non-conservative forces.
In case there aren’t any (no friction, etc.):

W(nc) = change in KE + change in PE


The total mechanical energy doesn’t change. It is
conserved in the absence of non-conservative forces.

W(nc) = change in KE + change in PE

non-conservative forces



For non-conservative force work done does depend on
the chosen path.

We learned Wfr= -Ffr ∙ d → longer path requires more work.
Coefficient of friction may be different.

Conservative forces

A force F is acting on an object.

If the work done by F to get from A to B is independent of the path, then F is a said to be a conservative force.

The force of gravity is a conservative force.

Gravitational force: mechanical work against it depends
just on difference in elevation not how an object is
lifted.

Elastic force: work against spring only depends on
length change.

Hooke's Law


F(p) = kx
F(s) = -kx

k = Spring constant [N/m]
F is not constant,
the average force = (0 + 1/2 kx)
W = 1/2 kx^2
elastic PE= 1/2 kx^2

Conservation of Energy



In a closed system energy is conserved.


The position of the blue ball ,there is the Potential Energy (PE) while the Kinetic Energy (KE) = 0.


As the blue ball is approching the purple ball position the PE is decreasing while the KE is increasing. At exactly halfway between the blue and purple ball position the PE = KE.


The position of the purple ball is where the Kinetic Energy is at its maximum while the Potential Energy (PE) = 0.

At this point, theoretically, all the PE has transformed into KE> Therefore now the KE = 19.6J while the PE = 0.


The position of the pink ball is where the Potential Energy (PE) is once again at its maximum and the Kinetic Energy (KE) = 0.


PE + KE = 0

PE = -KE


The sum of PE and KE is the total mechanical energy:

Total Mechanical Energy = PE + KE

Potential Energy



The gravitational potential energy depends on the height
of an object over some reference level.


The potential energy depends only on the position of the object.

PE = mgh


where
PE = Energy (in Joules)
m = mass (in kilograms)
g = gravitational acceleration of the earth (9.8 m/sec2)
h = height above earth's surface (in meters)

the net work done on the hammer

W(net) = KE2 - KE1

Work and Kinetic Energy



•An object’s kinetic energy can be likened to the work that could be done if object were brought to rest (so, the K.E. is like potential work content)

–The moving hammer has kinetic energy and can do work on the nail .

Kinetic Energy



Kinetic Energy KE is the energy associated with the motion of an object .
Kinetic Energy exists whenever an object which has mass is in motion with some velocity. Everything you see moving about has kinetic energy.
KE = 1/2 mv(square)


where
KE = Energy (in Joules)
m = mass (in kilograms)
v = velocity (in meters/sec)

work done by a varying force




Fx =const → W = Fx (x2 – x1)

Wednesday, September 19, 2007

examples











force, displacement and the angle between them


A Teacher applies a Force to a Wall.




W = F d


d = 0

The wall is not displaced.

Work Done is Zero when the Displacement is Zero.

A teacher applies a force to a wall and becomes exhausted. But No Work is Done.

The force exerts the brakes to stop the car.




The direction of the force is in the opposite direction that the object moves.

W = F cos 180˚ d


cos 180 ˚ = - 1

W= - F d

The Work Done on the car is negative because of the frictional force.

Frictional Force is always opposing the relative motion of the body.

A Waiter Carrying A Tray




W = F cos 90 ˚d


W = 0


No Work is Done.


•When a particular force is perpendicular to the motion, no work is done by that force.

Unit of Work

SI system: 1 Joule (J) = 1 N.m


cgs system: 1 erg = 1 dyne.cm = 10-7 J


1 calorie (cal) = 4.19 J


British units 1 foot-pound (ft.lb) = 1.36 J

Work done by a constant force




The work W done by a constant force on an object is
the product of the magnitude of the displacement
and the component of the force parallel to the displacement.


W = F// d


W = Fd cos (theta)


vector F = the constant force

d = the magnitude of the displacement of the object

In S.I unit, 1 Nm = 1 J

Work Done is a scalar quantity.

Work and Energy (Chapter 6)

Objectives

1. Distinguish between work in the scientific sense as compared to the colloquial sense.

2. Write the definition of work in terms of force and displacement and calculate the work done by a constant force when the force and displacement vectors are at an angle.

3. Use graphical analysis to calculate the work done by a force that varies in magnitude.

4. Define each type of mechanical energy and give examples of types of energy that are not mechanical.

5. State the work energy theorem and apply the theorem to solve problems.

6. Distinguish between a conservative and a nonconservative force and give examples of each type of force.

7. State the law of conservation of energy and apply the law to problems involving mechanical energy.

8. Define power in the scientific sense and solve problems involving work and power.