Physics -061&062

Torque

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Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label 'O'. We will call the force 'F'. The distance from the pivot point to the point where the force acts is called the moment arm, and is denoted by 'r'. Note that this distance, 'r', is also a vector, and points from the axis of rotation to the point where the force acts.

Torque = r x F = r F sin(theta)

There may be more than one force acting on an object, and each of these forces may act on different point on the object. Then, each force will cause a torque. The net torque is the sum of the individual torques.

Rotational Equilibrium is analogous to translational equilibrium, where the sum of the forces are equal to zero. In rotational equilibrium, the sum of the torques is equal to zero. In other words, there is no net torque on the object.

The SI units of torque is a Newton-metre.

Rolling Motion (Without Slipping)

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images

Angular Acceleration

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Angular Velocity

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image

Angular velocity is generally specified in radians per second (rad/s).

All points in a rigid object rotate with the same angular velocity.

An object such as a wheel rotates about a fixed axis either clockwise or counterclockwise.

The direction can be specified with a + or - sign.

Angular Quantites

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http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html

In purely rotational motion, all points on the object move in circles around the axis of rotation (“O”). The radius of the circle is r. All points on a straight line drawn through the axis move through the same angle in the same time. The angle θ in radians is defined:

s = r θ

where s is the arc length.

Rotational Motion (Chapter 8)

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Objectives

1. Convert angular quantities from revolutions or degrees to radians and vice versa.

2. Write the Greek symbols used to represent angular displacement, angular velocity, and angular acceleration.

3. State the meaning of the symbols used in the kinematics equations for uniformly accelerated angular motion.

4. Write from memory the equations used to describe uniformly accelerated angular motion.

5. Complete a data table using information both given and implied in word problems. Use the completed data table to solve word problems related to angular kinematics.

6. Distinguish between inertia and moment of inertia.

7. Distinguish between linear momentum and angular momentum. State and apply the law of conservation of angular momentum to solve word problems.

8. Calculate the lever arm distance and determine the magnitude and direction of the torque vector if the magnitude and direction of the net force are given.

9. Draw a free body diagram for each object in a system. Locate the forces acting on each object. Use F = ma and τ = Iα to solve for the linear or angular acceleration of each object.

10. Apply the law of conservation of angular momentum to a system where no net external torque acts. Determine the change in angular velocity of a system where the moment of inertia of the objects that make up the system changes.

11. Distinguish between translational kinetic energy and rotational kinetic energy. Apply the Law of Conservation of Energy to solve problems that involve rotational as well as translational kinetic energy.

Practice Questions

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1. An Olympic diver dives off the high-diving platform. The magnitude of his momentum will be a maximum at Point

(A)A
(B)B
(C)C
(D) D

2. Which of the following can best be described as an elastic collision?
(A) A falling tree hitting the earth.
(B) A linebacker stopping a fullback.
(C) A golf club striking a golf ball.
(D) A pie in the face.

3. The Center of Mass of which of the following objects would not lie within the body itself?
(A) baseball
(B) brick
(C) frisbee
(D) paperback book

4. What is the magnitude of the impulse as shown in the following figure?

(A)100 Ns
(B)900 Ns
(C)1000 Ns
(D)1100 Ns

5. If the impulse is applied to a particle at rest, what will be its final momentum?

(A)100 kg m/s
(B)900 kg m/s
(C)1000 kg m/s
(D)1100 kg m/s

6. What is the impulse shown in the following figure?

(A)225 Ns
(B)450 Ns
(C)900 Ns
(D) 1000 Ns

7. If the impulse is applied to a 200 gm model rocket at rest, what will be its final speed?

(A) 2.25 m/s
(B) 22.5 m/s
(C) 2250 m/s
(D) 4500 m/s

8. What constant force acting over the same time as the force in Question 6 would produce the same impulse?

(A)45 N
(B) 50 N
(C) 90 N
(D)100 N

9. If each mass in the figure is 1 kg, what is the x location of the center of mass?

(A) .5
(B) .67
(C)1.0
(D) 1.33

10. If each mass in the figure is 1 kg, what is the y location of the center of mass?

(A) .33
(B) .5
(C) .67
(D) .85

11. If the mass of m1 and m3 in the figure is 1 kg and m2 is 2 kg, what is the x location of the center of mass?

(A) 1.33
(B) 1.0
(C) .67
(D) .33

12. If the mass of m1 and m3 in the figure is 1 kg and m2 is 2 kg, what is the y location of the center of mass?

(A) .75
(B) .5
(C) .33
(D) .25

13. If the mass of m1 is 1 kg and m3 is 2 kg in the figure and m2 is 3 kg, what is the x location of the center of mass?

(A) .67
(B) 1.17
(C) 1.33
(D) 1.67

14. If the mass of m1 is 1 kg and m3 is 2 kg in the figure and m2 is 3 kg, what is the y location of the center of mass?

(A).25
(B).5
(C) .67
(D) .75

15. If all the masses in the figure have a mass of 2 kg, and m2 has speed 2 m/s and m3 has a speed of 4 m/s, what is the x-component of the momentum of the center of mass?

(A)2 kg m/s
(B)4 kg m/s
(C)6 kg m/s
(D)10 kg m/s

16. If all the masses in the figure have a mass of 2 kg, and m2 has speed 2 m/s and m3 has a speed of 4 m/s, what is the y-component of the momentum of the center of mass?

(A)2 kg m/s
(B) 4 kg m/s
(C) 8 kg m/s
(D) 10 kg m/s

17. What is the magnitude of the momentum of the center of mass of the system of particles in Question 15?

(A) 4kg m/s
(B) 8 kg m/s
(C) 8.9 kg m/s
(D)10 kg m/s

18. What is the direction of the momentum of the center of mass of the particles in Question 15, as measured in degrees counter clockwise from East?

(A) 15
(B) 34
(C) 45
(D) 63

19. A 12000 kg railroad car travelling at 10 m/s strikes and couples with a 6000 kg caboose at rest. What is the speed of the final combination?

(A) 3.3 m/s
(B)5.0 m/s
(C)6.7 m/s
(D)10 m/s

20. A 12000 kg railroad car travelling at 10 m/s strikes and couples with a 6000 kg caboose at rest. What is the loss of mechanical energy in the collision?

(A) 50000 J

(B) 100000 J

(D) 250000 J

21. A 12000 kg railroad car travelling at 10 m/s strikes and couples with a 6000 kg caboose at rest. What would be the speed of the combination of railroad cars if the caboose was initially rolling towards the other car with a speed of 2 m/s?

(A) 0 m/s

(B) 6.0 m/s

22. A billiard ball travelling to the right at 6m/s overtakes and strikes an identical ball travelling at 3 m/s. What is the speed of the first ball after the collision?

(A) -6 m/s

(B) -3 m/s

(D) 6 m/s

23. A billiard ball travelling to the right at 6m/s overtakes and strikes an identical ball travelling at 3 m/s. What is the speed of the second ball after the collision?

(A)-6 m/s

(B) -3 m/s

(D) 6 m/s

24. A billiard ball travelling to the right at 6m/s overtakes and strikes an identical ball travelling at 3 m/s. What is the speed of the first ball after collision if the second ball was travelling at it (instead of away from it) with a speed of 3 m/s before collision?

(A) -6 m/s

(B) -3 m/s

(D) 6 m/s

25. What is the speed of the second ball in Question 24 after collision?

(A) -6 m/s

(B) -3 m/s

(D) 6 m/s

Centre of Mass

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Inelastic Collison

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Kinetic Energy lost

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Elastic Collision II

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Elastic Collision

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Conservation of Momentum

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from www.physicsclassroom.com

Momentum is conserved.

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force and momentum

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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

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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 ?

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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

Linear Momentum (Chapter 7)

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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

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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.

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

(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.

(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.

(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

(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

(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

(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.

(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

(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

(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.

(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.

(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.

(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

(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

(D) Greater than R

Power

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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

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Mechanical Energy and its Conservation

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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

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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

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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

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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

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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

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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

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W(net) = KE2 - KE1

Work and Kinetic Energy

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•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

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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

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Fx =const → W = Fx (x2 – x1)

examples

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force, displacement and the angle between them

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A Teacher applies a Force to a Wall.

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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.

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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

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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

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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

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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)

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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.

Practice Questions (Chapter 5)

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1. A stone is whirled in a vertical circle on a cord. Halfway up

(A) the tension is towards the center of the circle and the net force is down.
(B) the weight is down and the net force is towards the center of the circle.
(C) the tension force is towards the center of the circle and the weight is down.
(D) the weight and tension are in the same direction.

2. A roller coaster car is on a track that forms a circular loop in the vertical plane. If the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its centripetal acceleration at this point?

(A) g downward
(B) g upward
(C) 0.5 g downward
(D) 2 g upward

3. Two objects are travelling in circular orbits. Object A is travelling at twice the velocity of object B in a circle with a diameter of twice that of B. The centripetal acceleration

(A) of A and B are the same.
(B) of A is twice that B.
(C) of A is four times that of B.
(D) of A is half that of B.

4. A cannon ball is fired horizontally off a high cliff over a great distance. If air resistance can be ignored the path it follows is

(A) part of a parabola.
(B) part of a circle.
(C) a hyperbola.
(D) part of an ellipse.

5. There are four forces in nature. Which one allows you to close a door by pushing on it?

(A) Gravity
(B) Electric
(C) Nuclear
(D) Weak

6. A ball of mass m is moving in a circle with uniform speed on a horizontal surface with friction at the end of a radial metal rod. The net force is

(A) opposite the friction force.
(B) in the direction of the friction force.
(C) perpendicular to the surface.
(D) is along the rod.

7. Two bodies of equal mass are separated by a distance R. If you double the distance between them the new gravitational force will be

(A) twice the old force.
(B) half the old force.
(C) four times the old force.
(D) one fourth the old force.
(E) The same as the old force.

8. Two bodies of equal mass are separated by a distance R. If you double each mass then the new force will be

(A) twice the old force.
(B) half the old force.
(C) four times the old force.
(D) one fourth the old force.
(E) the same as the old force.

9. Two bodies of equal mass are separated by a distance R. If you double one mass then the new force will be

(A) twice the old force.
(B) half the old force.
(C) four times the old force.
(D) one fourth the old force.
(E) the same as the old force.

10. Two bodies of equal mass are separated by a distance R. If you double each mass and double the distance between them, the new force will be

(A) twice the old force.
(B) half the old force.
(C) four times the old force.
(D) one fourth the old force.
(E) the same as the old force.

11. The Earth moves faster in the winter when its closest to the Sun. This is an example of

(A) Newton's Law of Gravity.
(B) Kepler's First Law.
(C) Kepler's Second Law.
(D) Kepler's Third Law.

12. Which of the following does not belong with the others?

(A) A car speeds up from 30 mph to 40 mph.
(B) A car breaks from 20 mph to 0.
(C) A race car rounds a curve at 120 mph.
(D) A car coasts down a road at constant speed.

13. Kepler's Third Law can be used to relate the motions of the

(A) Moon and the Earth.
(B) Moon and the Space Shuttle.
(C) Earth and the Sun.
(D) Moon and the Sun.

14. If the mass of the Earth was quadrupled and nothing else was changed

(A) g would increase by a factor of 4.
(B) g would increase by a factor of 2.
(C) g would decrease by a factor of 1/2.
(D) g would decrease by a factor of 1/4.

15. A planet with twice the mass and twice its radius has a value of g

(A) four times that of Earth.
(B) two times that of Earth.
(C) 1/2 that of Earth.
(D) 1/4 that of Earth.

16. Four equal masses are arranged in a perfect square. The direction of the force on the one in the lower left hand corner

(A) depends on the value of the mass.
(B) is at 45 degrees.
(C) is at 225 degrees.
(D) zero.

17. The speed of a satellite in orbit does not depend on

(A) its mass.
(B) the mass of the Earth.
(C) its distance above the Earth's surface.

Kepler's Laws

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Satellites and "Weightlessness"

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A satellite is put into orbit by accelerating it to a sufficient high tangential speed with the use of rockets.

If the speed is too high, the spacecraft will not be confined by the Earth's gravity and will escape, never to return.

If the speed is too low, it will return to Earth.

Satellite are usually put into circular (or nearly circular) orbits, because such orbits require the least takeoff speed.

this image is from www.howstuffworks.com

If a satellite stopped moving, it would fall directly to Earth.

But at the very high speed a satellite has, it would quickly fly out into space, if it weren't for the gravitational force of the Earth pulling it into orbit.

In fact, a satellite is falling (accelerating toward Earth), but its high tangential speed keeps it from hitting Earth.

Gravity Near the Earth's Surface: Geophysical Application

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When we apply the gravitaional force to the gravitational force between the Earth and an object at its surface,

m(1) becomes the mass of the Earth m(E) .

m(2) becomes tge mass of the object m.

r becomes the distance of the object from the Earth's centre = radius of the earth

The force of gravity due to the Earth is the weight of the object mg.

Newton's Law of Universal Gravitation

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G = 6.67 x 10-11 N m^2/kg^2

http://www.physicsclassroom.com/Class/circles/U6L3c.html

Centrifugation

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image from http://www.steve.gb.com

Nonuniform Circular Motion

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Any object moving in a circle has a centripetal acceleration,

directed toward the center of that circle.

It has a centripetal force acting on it, also directed toward the center of that circle.

"Directed toward the center"="radially".

Any object moving in a circle has a radial acceleration.

That means it has a radial force acting on it.

In addition, the object may be accelerating tangentially , along the direction tangential to the circle or perpendicular to the radius.

Then there is also a tangential force or the net force has a tangential component as well as the radial component.

The total or net acceleration is the vector sum of the radial and tangential components.

The total or net force is the vector sum of the radial and tangential components.

banked curve

2007-07-12T07:03:00.001-07:00

highway curves

2007-07-12T06:33:00.000-07:00

horizontal circular motion & vertical circular motion

2007-07-12T06:31:00.000-07:00

Force on revolving ball

2007-07-12T06:17:00.000-07:00

T= mv^2/r

object revolving in a circle

2007-07-12T06:14:00.000-07:00

circular motion

2007-07-12T05:52:00.000-07:00

Uniform Circular Motion

2007-07-10T22:00:00.000-07:00

An object moving in a circle is accelerating.

Accelerating objects are objects which are changing their velocity -

either the speed (i.e., magnitude of the velocity vector) or the direction.

An object undergoing uniform circular motion is moving with constant speed; nonetheless, it is accelerating due to its change in direction.

The direction of the acceleration is inwards.

Circular Motion ; Gravitation (Chapter 5)

2007-07-10T21:46:00.000-07:00

Objectives

1. the centripetal acceleration of a point mass in uniform circular motion given the radius of the circle and either the linear speed or the period of the motion.

2. the force that is the cause of the centripetal acceleration and determine the direction of the acceleration vector.

3. Highway Curves, Banked and UnBanked Curves.

4. Nonuniform Circular Motion

5. Centrifugation

6. Newton's universal law of gravitation and explain the meaning of each symbol in the equation.

7.Gravity Near the Earth's Surface; Geophusical Applications

8. Satellites and "Weightlessness".

9. Kepler's Laws and Newton's Synthesis

10.Types Forces in Nature

Practice Questions (Chapter 4)

2007-07-03T07:13:00.000-07:00

1 .A packing crate slides down an inclined ramp at constant velocity. Thus we can deduce that
a frictional force is acting on it.

(A) a net force is acting on it.
( B) it may be accelerating.
( C) Gravity is not acting on it.

2. You fall while skiing and one ski (of weight W) loosens and slides down an icy slope (assume no friction), which makes an angle q with the horizontal. The force that pushes it down along the hill is

(A) zero; it moves with constant velocity.
(B) W cos q
(C) W sin q
(D) it is not acted on by a gravitational force.

3. A decoration, of mass M, is suspended by a string from the ceiling inside an elevator. The elevator is traveling upward with a constant speed. The tension in the string is

(A) equal to Mg.
(B) less than Mg.
(C) greater than Mg.
(D) impossible to tell without knowing the speed.

4. An object is placed on an inclined plane. The angle of incline is gradually increased until the object begins to slide. The angle at which this occurs is q. What is the coefficient of static friction between the object and the plane?

(A) sin q
(B) cos q
(C) tan q
(D) Cannot determine without knowing the mass of the object.

5. A mass on a string attached to a ceiling is pulled to the side and let go. When the string makes an angle q with the vertical, which of the following is not true?

(A) The weight is down.
(B) The tension equals mg*cos(q).
(C) The tension equals mg*sin(q).
(D) The tension and the weight are never in the same direction.

6. A roller coaster car does a loop-the-loop. When it is upside down at the very top, which of the following is true?

(A) The normal force and the weight are in opposite directions.
(B) The normal force and the weight are perpendicular to each other.
(C) The weight is zero.
(D) The normal force and the weight are in the same direction.

7. A roller coaster car does a loop-the-loop. When it is right-side up at the very bottom, which of the following is true?

(A) The normal force and the weight are in opposite directions.
(B) The normal force and the weight are perpendicular.
(C) The weight is zero.
(D) The normal force and the weight are in the same direction.

8. A 2 kg box sits on a 3 kg box which sits on a 5 kg box. The 5 kg box rests on a table top. What is the normal force exerted on the 5 kg box by the table top?

(A) 19.6 N
(B) 29.4 N
(C) 49 N
(D) 98 N

9. A 2 kg box sits on a 3 kg box which sits on a 5 kg box. The 5 kg box rests on a table top. What is the normal force exerted by the 5 kg box on the 3 kg box?

(A) 19.6 N
(B) 29.4 N
(C) 49 N
(D) 98 N

10. A 5 kg (4.9 N) book rests on your desk. Which of the following answers is not correct?

(A) The normal force on the book equals its weight because of Newton's First Law.
(B) The normal force on the book equals its weight because of Newton's Second Law.
(C) The normal force on the book equals the weight because of Newton's Third Law.

11. 5 kg (4.9 N) book rests on your desk. What is the reaction force to the 4.9 N weight (force) of the book?

(A) The 4.9 N normal force from the table on the book.
(B) The 4.9 N reaction force from the book on the table.
(C) The weight of the earth.
(D) A 4.9 N force on the Earth.

12. You stand on a bathroom scale in an elevator accelerating upwards and it reads 200 Newtons. Which of the following is true?

(A) You weigh 200 Newtons.
(B) The scale exert a 200 N force on you.
(C) The acceleration of the elevator is 20.4 m/s^2.
(D) None of the above are correct.

13. During the investigation of a traffic accident, police find skid marks 90 m long. They determine the coefficient of friction between the car's tires and the roadway to be 0.5 for the prevailing conditions. To find the speed of the car you must

(A) know the mass of the car.
(B) find the acceleration of the car.
( C) know the mass of the car and find the acceleration of the car.

14. An object moves left to right (right is positive) with speed decreasing at a constant rate and
its acceleration is positive.

(A) the net force on it is decreasing.
(B) the net force on it is increasing.
(C) its acceleration is negative.

15. A car pulls on a rope tied to a tree with a force of 1000 Newtons. Which of the following is not true.

(A) The rope pulls on the tree with a force of 1000 Newtons.
(B) The tree pulls on the rope with a force of 1000 Newtons.
(C) The tension in the rope is 2000 Newtons.
(D) The tension in the rope is 1000 Newtons.

16. A 5 kg block and a 10 kg block slide down a frictionless incline and
they both have the same acceleration.

(A) the 5 kg block has an acceleration of twice that of the 10 kg block.
(B) the 10 kg block has an acceleration of twice that of the 5 kg block.
(C) their acceleration depends on the normal force.

inclines

2007-07-03T07:10:00.000-07:00

Fig. Forces on an object sliding down an incline.

Coefficients of Friction

2007-07-03T07:01:00.000-07:00

Kinetic Friction

2007-07-03T07:00:00.000-07:00

When an object slides along a rough surface, the force of kinetic friction acts opposite to the direction of the object's velocity.

The magnitude of the force kinetic friction depends on the nature of the two sliding surfaces.

The friction force is approximately proportional to the normal force between the two surfaces, which is the force that either object exerts on the other, perpendicular to their common surface of contact.

We write the proportionality between the friction force F (fr) and the normal force F (N) as an equation by inserting a constant proportionality,

Friction

2007-07-03T06:40:00.001-07:00

When an object is pulled by an applied force , F(A) along a surface, the force of friction F (fr) opposes the motion. The magnitude of F(fr) is proportional to the magnitude of the normal force , F(N).

Two boxes connected by a cord

2007-07-03T05:31:00.000-07:00

Two boxes, A and B , are connected by a lightweight cord and are resting on a smooth (frictionless) table.

Tension in a Flexible Cord

2007-07-02T22:17:00.000-07:00

Tension is a force.

When a flexible cord pulls on an object, the cord is said to be under tension, and the force it exerts on the object is the tension F (T).

Two boxes , A and B , are connected by a cord. A person pulls horizontally on the box A with force F (p) = 40.0 N.

T (upper) = T (lower)

By the 2nd law and the fact that the system is at rest,

T (upper) = T(lower) = W

Pulling the Mystery Box

2007-07-02T18:54:00.000-07:00

Suppose a friend asks to examine the 10.0 kg box , hoping to guess what is inside; and you respond, "Sure, pull the box by the attached cord, as shown in figure, along the smooth surface of the table.

The magnitude of the force exerted by the person is

F (p) = 40.0 N.

It is exerted at a 30 degree angle as shown in figure.

Calculate (a) the acceleration of the box, and (b) the magnitude of the upward force F(N) exerted by the table on the box. Assume that friction can be neglected.

Solution :

1. Draw a sketch : The situation is shown in Fig. (a) , it shows the box and the force applied by the person, F (p).

2. free-body diagram : Fig. (b) shows the free-body diagram of the box. To draw it correctly, we show all the forces acting on the box and only the forces acting on the box.

They are : the force of gravity mg ; the normal force exerted by the table F (N) ; and the force exerted by the person F(p).

3. Choose axes and resolve vectors : We expect the motion to be horizontal , so we choose the x axis horizontal and the y axis vertical . The pull of 40.0 N has components

F (px) = (40.0 N) ( cos 30 degree) = (40.0 N) ( 0.866) = 34.6 N

F (py) = (40.0 N) ( sin 30 degree) = (40.0 N) (0.500) = 20.0 N.

In the horizontal (x) direction, F (N) and mg have zero components.

The horizontal component of the net force is F(px) .

4. (a) Apply Newton's second law to determine the x component of the acceleration;

F (px) = ma (x).

5. (a) Solve :

a (x) = F(px) /m = (34.6 N)/ (10.0 kg) = 3.46 m/s^2.

The acceleration of the box is 3.46 m/s^2 to the right.

4. (b) Apply Newton's second law to the vertical (y) direction , with upward as positive:

F (y) = ma(y)

F (N) - mg + F (py) = ma (y).

5. (b) Solve: We have mg = (10.0 kg ) ( 9.80 m/s^2) = 98.0 N .

F (py) = 20.0 N.

F (py ) is less then mg, the box does not move vertically, so

a (y) = 0.

Thus, F (N) - 98.0 N + 20 N = 0.

F (N) = 78.0 N.

The Honky Puck

2007-07-02T18:49:00.000-07:00

A honky puck is sliding at constant velocity across a flat horizontal ice surface that is assumed to be frictionless. Which of the sketches in the figure is the correct free-body diagram for this punk? What would your answer be if the puck slowed down?

The Resultant Force

2007-07-02T18:42:00.000-07:00

By Newton's Second Law,

the acceleration of an object is proportional to the net force acting on the object.

The net force is the vector sum of all forces acting on the object.

The object will start moving at a 45 degree angle and thus the net force acts at a 45 degree angle.

From the theorem of Pythagoras,

F(R) = ( ( 100 N )^2 + (100 N) ^2 ) ^ 1/2 = 141 N.

Accelerating the Box

2007-07-02T18:18:00.000-07:00

Example :

What happens when a person pulls upward on the box with a force equal to, or greater than , the box's weight , say F (p) = 100.0 N rather than the 40.0 N shown in figure.

Solution: The net force on the box is

F (y) = F (N) - mg + F (p)

= F (N) - 98.0 N + 100.0 N

The box accelerates upward, so the net force is not zero.

Normal force is zero.

The net force = F (y) = F (p) - mg = 100.0 N - 98.0 N

= 2.0 N

a (y) = F (y)/ m = 2.0 N/10.0 kg = 0.20 m/s^2.

Weight, Normal Force, and a Box

2007-07-02T17:53:00.000-07:00

A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table.

(a) Determine the weight of the box and the normal force exerted on it by the table.

(b) Now your friend pushes down on the box with a force of 40.0 N , as in fig (b) .
Again determine the normal force exerted on the box by the table.

(c) If your friend pulls upward on the box with a force of 40.0 N (fig. (c)), what now is the normal force exerted on the box by the table.

Approach : The box is at rest on the table, so the net force on the box in each case is zero. (Newton's Second Law).
The weight of the box equals mg in all three cases.

Solution:

(a) The weight of the box is mg = (10.0 kg) (8.80 m/s^2) = 98.0 N

This force acts downward.

The normal force exerted upward on it by the table. Fig.(a)

The upward direction as the positive y direction;

The net force along y direction on the box = F (y) = F (N) - mg .

The box is at rest, so the net force on it = 0.

F(y) = F (N) - mg = 0

F (N) = mg

The normal force on the box, exerted by the table , is 98.0 N upward. It has magnitude equal to the box's weight.

(b) Your friend is pushing down on the box with a force of 40.0 N .

Now there are three forces acting on th ebox. Fig. (b)

The weight of the box = mg = 98.0 N

The net force = F (y) = F (N) - mg - 40.0 N = 0 ( a = 0)

For Normal Force ,

F(N) = mg + 40.0 N = 98.0 N + 40.0 N = 138.0 N

(c) The box's weight is still 98.0 N and acts downward.

The force exerted by your friend and the normal force
both act upward ( positive direction). Fig. (c)

The box does not move, since your friend's upward force is less than the weight.
The net force = 0 (a =0)

F (y) = F (N) - mg + 40.0 N = 0

F (N) = mg - 40.0 N = 98.0 N - 40.0 N = 58.0 N.

The table does not push against the full weight of the box because of the upeard pull exerted by your friend.

Free-Body Diagrams

2007-07-01T05:43:00.000-07:00

1. A girl is suspended motionless from a bar which hangs from the ceiling by two ropes.

2. An egg is free-falling from a nest in a tree. Neglect air resistance.

3. A flying squirrel is gliding (no wing flaps) from a tree to the ground at constant velocity. Consider air resistance.

4. A rightward force is applied to a book in order to move it across a desk with a rightward acceleration. Consider frictional forces. Neglect air resistance.

5. A rightward force is applied to a book in order to move it across a desk at constant velocity. Consider frictional forces. Neglect air resistance.

6. A college student rests a backpack upon his shoulder. The pack is suspended motionless by one strap from one shoulder.

7. A skydiver is descending with a constant velocity. Consider air resistance.

8. A force is applied to the right to drag a sled across loosely-packed snow with a rightward acceleration.

9. A football is moving upwards towards its peak after having been booted by the punter. Neglect air resistance.

10. A car is coasting to the right and slowing down.

Normal Force

2007-07-01T05:26:00.000-07:00

From Newton's Second Law,

the net force on an object that remains at rest is zero.

There must be another force on the object to balance the gravitational force.

For an object is resting on a table,

the table exerts this upward force.

The table is compressed slightly beneath the object, and due to its elasticity, it pushes up on the object .

The force exerted by the table is often called a contact force, since it occurs when two objects are in contact.

When a contact force acts perpendicular to the common surface of contact, it is referred to as the normal force .

A book is at rest on a table top.

F (G) and F( N) must be equal magnitude and in opposite directions.

Weight-the force of Gravity

2007-07-01T04:56:00.000-07:00

All objects accelerate towards the earth.
Thus the earth exerts a force on these objects - this is called the force of gravity.

From Newton's Second Law,
an object of mass m falling due to gravity;
for the acceleration, a, we use the downward acceleration due to gravity, g.
The gravitational force on an object, F(G) = F = mg

The direction of this force is down toward the centre of the Earth.

The magnitude of the force of gravity on an object is called the object's weight.

In SI units, g= 9.8 m/s^2 = 9.80 N/kg.

Newton's Third Law

2007-06-30T06:25:00.000-07:00

Newton's Third Law

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.

"to every action there is an equal and opposite reaction."

"action" force and "reaction" force are acting on different objects.

Newton's Second Law

2007-06-30T06:10:00.000-07:00

Newton's Second Law of Motion

The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object.

F (net) = m * a

In SI units, with the mass in kilograms, the unit of force is called the newton (N).

1N = 1 kg.m/s^2

Mass

2007-06-30T05:42:00.000-07:00

Mass

Mass is a measure of the inertia of an object.

The more mass an object has, the greater the force needed to give it a particular acceleration.
It is harder to start it moving from rest, or to stop it when it is moving, or to change its velocity sideways out of a straight-line path.

In S.I units, the unit of mass is the kilogram (kg) .

Newton's First Law

2007-06-30T05:06:00.000-07:00

Newton's First Law

Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.

The tendency of an object to maintain its state of rest or of uniform motionin a straight line is called inertia. As a Nweton's first law is often called the law of inertia.

Dynamics: Newton's Laws of Motion (Chapter 4)

2007-06-30T03:57:00.000-07:00

Objectives

1. Newton's three laws of motion
2. the term net force
3. Weight-the Force of Gravity; and the Normal Force
4. free body diagram locating each of the forces acting on an object or a system of objects.
5. free body diagrams and Newton's laws of motion to solve word problems.
6. Tension in a Flexible Cord
7. Firction, Inclines

Practiced Questions (Chapter 3)

2007-06-29T08:29:00.000-07:00

1. If you drive west at 20 km/h for one hour, then drive east at 15 km/h for one hour, your net displacement will be

(A) 5 km east.
(B) 35 km west.
(C) 35 km east.
(D) 5 km west.

2. If the acceleration vector of a body is perpendicular to the velocity vector, which of the following must be true?

(A) The speed is changing.
(B) The direction is changing.
(C) Both the speed and the direction are changing.
(D) The direction is not changing.

3. A vector of magnitude 10 has an angle with the positive x-axis (east) of 120 degrees. What are its components?

(A) 5 and 8.7
(B) -5 and 8.7
(C) 5 and -8.7
(D) -5 and -8.7

4. A vector of magnitude 10 has an angle with the positive x-axis (east) of -60 degrees. What are its components?

(A) 5 and 8.7
(B) -5 and 8.7
(C) 5 and -8.7
(D) -5 and -8.7

5. A vector has components x = 6 m and y = 8 m. What is its magnitude and direction?

(A) 10 m and 30 degrees
(B) 14 m and 37 degrees
(C) 10 m and 53 degrees
(D) 14 m and 53 degrees

6. A vector has components x = -2 m and y = -2 m. What is its direction (angle with respect to east)?

(A) 45 degrees
(B) 135 degrees
(C) 225 degrees
(D) -45 degrees

7. Three vectors have components (x,y): A = (2,3), B = (4,-7) and C = (-6,-3). What is the magnitude of the resultant vector (D = A + B + C)?

(A) -7
(B) 17.7
(C) 15
(D) 7

8. Three vectors have components (x,y): A = (2,3), B = (4,-7) and C = (-6,-3). What is the magnitude of the resultant vector (D = A - B - C)?

(A) 13.6
(B) 15
(C) 52
(D) 7

9. Suppose that several projectiles are fired on level ground. Which one will be in the air longest?

(A) The one with the farthest range, R.
(B) The one with the highest maximum elevation, h.
(C) The one with the greatest initial velocity.

10. Which of the following does not affect the range of a projectile?

(A) V0 Initial Speed
(B) θ0 Launch Angle
(C) y0 Initial Height
(D) x0 Initial Horizontal Position

11. Which of the following must be zero for a projectile to achieve its maximum range?

(A) Vxf
(B) Vyf
(C) xf
(D) yf

12. Which of the following must be zero for a projectile to be at its maximum height?

(A) Vxf
(B) Vyf
(C) yf
(D) xf

13. Taking "up" as positive at the highest point, the acceleration of a projectile is

(A) -9.8 m/s2
(B) 0 m/s2
(C) +9.8 m/s2

14. If the sum of two vectors equals zero, the magnitude of their difference is

(A) equal to the magnitude of either one.
(B) equal to twice the magnitude of either one.
(C) less than twice the magnitude of either one.

motion of motor boat with current

2007-06-29T08:16:00.000-07:00

Relative Velocity

2007-06-29T04:29:00.000-07:00

http://www.physicsclassroom.com

Velocity is not absolute, it is dependant on the observer. That is why we say it is "relative" to the observer's state of motion.

formulas describing projectile motion

2007-06-18T05:14:00.000-07:00

from http://www.mansfieldct.org/,

conceptual notions about projectiles

2007-06-18T05:10:00.000-07:00

* a projectile is any object upon which the only force is gravity,

* projectiles travel with a parabolic trajectory due to the influence of gravity,

* there are no horizontal forces acting upon projectiles and thus no horizontal acceleration,

* the horizontal velocity of a projectile is constant (never changing in value),

* there is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,

* the vertical velocity of a projectile changes by 9.8 m/s each second,

* the horizontal motion of a projectile is independent of its vertical motion.

Characteristics of a Projectile's Trajectory

2007-06-18T05:00:00.000-07:00

From Physics Classroom, http://www.physicsclassroom.com/ ,

Concept of Projectile

2007-06-18T04:51:00.000-07:00

From Physics Classroom , www.physicsclassroom.com

Projectile Motion

2007-06-18T04:45:00.000-07:00

A projectile is an object upon which the only force acting is gravity.

A projectile is any object which once projected continues in motion by its own inertia and is influenced only by the downward force of gravity.

By definition, a projectile has only one force acting upon - the force of gravity

components of vector

2007-06-18T04:42:00.000-07:00

Addition of 3 Vectors , 4 Vectors

2007-06-15T07:52:00.001-07:00

Combining Vector Components

2007-06-15T07:49:00.000-07:00

Polar Form Example

2007-06-15T07:48:00.001-07:00

Example of Vector Components

2007-06-15T07:46:00.000-07:00

Graphical Vector Addition

2007-06-15T07:45:00.001-07:00

Adding two vectors A and B graphically can be visualized like two successive walks, with the vector sum being the vector distance from the beginning to the end point. Representing the vectors by arrows drawn to scale, the beginning of vector B is placed at the end of vector A. The vector sum R can be drawn as the vector from the beginning to the end point.

rectangular components

2007-06-15T07:38:00.000-07:00

Components parallel to the axes of a rectangular system of axes are called rectangular components.
the horizontal axis X
the vertical axis Y

The direction of a vector is given as an angle counter-clockwise from the X-axis.

Method of components

2007-06-15T07:32:00.000-07:00

The components of a vector are those vectors which, when added together, give the original vector.
The sum of the components of two vectors is equal to the sum of these two vectors.

A1, the component in an easterly direction, will have a magnitudeA1 = A cosQ.

A2, the component in a northerly direction, will have a magnitudeA2 = AsinQ

parallelogram method

2007-06-15T07:30:00.001-07:00

Polygon Method

2007-06-15T07:26:00.001-07:00

Addition of Vectors

2007-06-15T07:22:00.001-07:00

Vectors and Scalars

2007-06-15T07:15:00.000-07:00

Scalars are quantities which are fully described by a magnitude alone.
( for example: mass, time, temperature)

Vectors are quantities which are fully described by both a magnitude and a direction.
(for example: velocity)

Kinematics in Two Dimensions; Vectors (Chapter 3)

2007-06-15T07:08:00.000-07:00

Kinematics in Two Dimensions; Vectors (Chapter 3)

Objectives:

1. The magnitude and direction of a vector .
2. Multiply or divide a vector quantity by a scalar quantity.
3. The methods of graphical analysis to determine the magnitude
and direction of the vector resultant in problems involving vector addition
or subtraction of two or more vector quantities.
The graphical methods to be used are the parallelogram method and
the tip to tail method.
4. The trigonometric component method to resolve a vector components in the x and y directions.
5. The trigonometric component method to determine the vector resultant in problems involving vector addition or subtraction of two or more vector quantities.
6. Two dimensional motion of projectiles.

Practice Questions (Chapter 2)

2007-06-06T04:43:00.000-07:00

1. At t = 0 s

(A) Rider C is ahead of rider D.
(B) Rider D is ahead of rider C.
(C) Rider C and D are at the same position.

2. At t = 0 s

(A) C is moving, and D is at rest.
(B) D is moving, and C is at rest.
(C) C and D are both moving.
(D) C and D are both at rest.

3 . At t = 0 s

(A) C has a greater velocity than D.
(B) D has a greater velocity than C.
(C) C and D have the same velocity.
(D) C is accelerating.

4. At t = 10 s

(A) C and D are at the same position.
(B) C and D have the same velocity.
(C) The velocity of D is greater than the velocity of C.
(D) C is in front of D.

5. During the first 8 s

(A) C has decreasing velocity and D has increasing velocity.
(B) C and D both have decreasing velocities.
(C) C and D have the same velocity.
(D) C has the same average velocity as D.

6. During the first 8 s

(A) the magnitude of the acceleration of C is greater than the magnitude of D's acceleration.
(B) their accelerations are equal in magnitude but opposite in sign.
(C) their accelerations are equal in magnitude and equal in sign.

7.Based on all the graphical information

(A) they meet at the same position at t = 8 s.
(B) they will meet at the same position, at t = 10 s.
(C) they will never meet at the same position.
(D) not enough information is given to decide if they meet.

8. Which graph represents an object at rest?

(A)

(B)

(C)

(D)

9 . How is motion in the negative x direction represented on an x vs. t plot?

(A) By a curve to the left of the origin.
(B) By a curve below the horizontal axis.
(C) By a downward sloping curve.
(D) Such a motion cannot be shown on a simple x vs. t graph.

10. What is the meaning of a horizontal line on a plot of v vs. t?

(A) The object is at rest.
(B) The object is moving at constant speed.
(C) The object is speeding up at a constant rate.
(D) The object is accelerating at a constant non-zero rate.

11. If a car accelerates from rest in the positive direction

(A) its final velocity is zero.
(B) its initial position is zero.
(C) its initial velocity is zero.
(D) its acceleration is negative.

12. A car accelerating from rest at 2.0 m/s2 for 5 s

(A) covers a distance of 10 m.
(B) has a final speed of 10 m/s.
(C) covers a distance of 50 m.
(D) has an average velocity of 10 m/s.

13. In this problem "up" is taken to be positive.
A ball is dropped off a high cliff

(A) and its initial velocity is zero.
(B) and its final velocity is positive.
(C) and its acceleration is 9.8 m/s2.
(D) and its acceleration is -9.8 m/s.

14. In this problem "up" is taken to be positive.
A ball is thrown straight up in the air; at the very top its

(A) velocity is negative.
(B) acceleration is zero.
(C) its acceleration is 9.8 m/s2.
(D) its acceleration is -9.8 m/s2.

15. In this problem "up" is taken to be positive.
A vertically upward thrown rock has a

(A) positive initial velocity.
(B) positive final velocity.
(C) positive acceleration.
(D) positive final height.

16. In this problem "up" is taken to be positive.
A falling object has

(A) positive acceleration and positive velocity.
(B) negative acceleration and negative velocity.
(C) positive acceleration and negative velocity.
(D) negative acceleration and positive velocity.

17 . In this problem "up" is taken to be positive.
An object moving upward under the influence of gravity has

(A) positive acceleration and positive velocity.
(B) negative acceleration and negative velocity.
(C) positive acceleration and negative velocity.
(D) negative acceleration and positive velocity.

18. In this problem "up" is taken to be positive.
A ball thrown vertically upward with an initial velocity of 19.6 m/s, after 5 s has a final velocity of

(A) 19.6 m/s.
(B) -19.6 m/s.
(C) 29.4 m/s.
(D) -29.4 m/s.

19 . An object is shot upward at 34.3 m/s. How high does it go?

(A) 60.0 meters
(B) 180.1 meters
(C) 102.9

20 . Ball A is thrown vertically upward with a speed of 19.6 m/s from the topedge of a high building. As it passes the edge on the way down, a second ball, B, is thrown downward at 19.6 m/s. Which of the following is correct?

(A) Ball A hits the ground before B.
(B) The two balls hit the ground at the same time.
(C) Ball B hits the ground before A.

21. A car moving right to left applies the brakes. Its

(A) velocity and acceleration are both positive.
(B) velocity and acceleration are both negative.
(C) velocity is negative and its acceleration is positive.
(D) velocity is positive and its acceleration is negative.

22. A ball is dropped off a ledge. How far does it fall in the 2nd second?

(A) 4.9 meters
(B) 9.8 meters
(C) 14.7 meters
(D) 19.6 meters

23. Ball A is thrown upward at the same time as ball B and with half the speed of ball B. Which of the following is true?

(A) They will hit the ground at the same time.
(B) A will hit before B.
(C) B will hit before A.

24.Ball A is thrown upward at the same time as ball B and with half the speed of ball B. How much higher does B go than A?

(A) 8 times.
(B) 4 times.
(C) 2 times.
(D) Not enough information is given to answer this question.

Graphical Analysis of Linear Motion

2007-06-05T05:10:00.000-07:00

(a)+ Velocity Constant Velocity (b) + Velocity Changing Velocity

(a) Slow, Rightward (+) (b) Fast, Rightward (+)

(a)Slow, Leftward (–) (b)Fast, Leftward (–)

(a)Leftward (–) Velocity (b)Leftward (–) Velocity

Experiment to measure the acceleration due to gravity

2007-06-05T01:45:00.000-07:00

Experiment to measure "g"
http://www.picotech.com/experiments/gravity_acceleration/index.html

Acceleration due to Gravity

2007-06-05T00:07:00.000-07:00

Drop two objects. Which object will reach the ground first ? (i.e. which will fall faster?)
In reality, both objects will reach the ground at the same time since all objects accelerate at the same rate.
The value of this acceleration is known as "g".

g = 9.80 m/s^2

Falling Objects

2007-06-05T00:05:00.000-07:00

At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.