Chapter 4 – Kinematics (cont.)

Recap:

Dropping a PackageYou drop a package

from a plane flying at

constant speed in a

straight line. Without

air resistance, the

package will:

a) quickly lag behind the plane while falling

b) remain vertically under the plane while falling

c) move ahead of the plane while falling

d) not fall at all

You drop a package

from a plane flying at

constant speed in a

straight line. Without

air resistance, the

package will::

a) quickly lag behind the plane while falling

b) remain vertically under the plane while falling

c) move ahead of the plane while falling

d) not fall at all

Both the plane and the package have

the same horizontal velocity at the

moment of release. They will

maintain this velocity in the x-

direction, so they stay aligned.

Follow-up: : what would happen if air resistance is present?

Dropping a Package

a b

c) both at the same time

A battleship simultaneously fires two shells at two

enemy submarines. The shells are launched with the

same magnitude of initial velocity. If the shells follow

the trajectories shown, which submarine gets hit first ?

a b

c) both at the same time

A battleship simultaneously fires two shells at two

enemy submarines. The shells are launched with the

same magnitude of initial velocity (i.e., speed). If the

shells follow the trajectories shown, which submarine

gets hit first ?

The flight time is fixed by the motion in the y-direction. The higher an object goes, the longer it stays in flight. The shell hitting submarine #2 goes less high, therefore it stays in flight for less time than the other shell. Thus,

submarine #2 is hit first.

Follow-up: Did you need to know that they had the same initial speed?

Range: the horizontal distance a projectile travels

As before, use

and

Eliminate t and solve for x when y=0

(y = 0 at landing)

θ

Sinθ

Range is maximum at 45o

sin(2 x 45o) = sin(90o) = 1

Range Gun

If the range gun launches a ball 6 meters with a launch angle of 45 degrees, at which of these angles should a ball be launched to land in a bucket at 3 meters?

a) 10 degrees

b) 22.5 degrees

c) 60 degrees

d) 75 degrees

e) one would also need the

launch velocity of the range

gun to know

Range Gun

If the range gun launches a ball 6 meters with a launch angle of 45 degrees, at which of these angles should a ball be launched to land in a bucket at 3 meters?

a) 10 degrees

b) 22.5 degrees

c) 60 degrees

d) 75 degrees

e) one would also need the launch velocity of the range gun to know

The range is proportional to sin2θ, so to travel half the distance, the ball would need to be launched with sin2θ = 0.5.

sin(75o) = 0.5, so θ=75o

Note: there are two angles that would work [sin(30o) = 0.5 also]. How are the two solutions different?

Symmetry in projectile motion

Dropping the Ball III

A projectile is launched from the ground at an angle of 30°. At what point in its trajectory does this projectile have the least speed?

a) just after it is launched

b) at the highest point in its flight

c) just before it hits the ground

d) halfway between the ground and the highest point

e) speed is always constant

A projectile is launched from the ground at an angle of 30º. At what point in its trajectory does this projectile have the least speed?

a) just after it is launched

b) at the highest point in its flight

c) just before it hits the ground

d) halfway between the ground and the highest point

e) speed is always constant

The speed is smallest at

the highest point of its

flight path because the

y-component of the

velocity is zero.

Dropping the Ball III

Which of the

three punts

has the

longest hang

time??

Punts I

d) all have the same hang time

a b c

h

Which of the

three punts

has the

longest hang

time?d) all have the same hang time

a b c

h

The time in the air is determined by the vertical

motion! Because all of the punts reach the same

height, they all stay in the air for the same time.

Follow-up: Which one had the greater initial speed?

Punts I

•• On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 0.616 s, how high above the water was she when she let go of the rope?

y = (v0 sinθ) t - 1/2 g t2

y = 2.25 m/s * sin(35o) * (0.616 s) - 1/2 (9.8 m/s2) (0.616 s)2 = - 1.07 m

Time to hit the water: t=0.616Initial velocity: 2.25 m/s at 35o above the horizontal

At time t= 0.616 s, the girl is 1.07 m below her starting position, so her initial position was 1.07m above the water.

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 34° above the horizontal. (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall?

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 34° above the horizontal. (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall?

d = 3.8 mv0 = 14 m/sθ = 34o

v0x = (14 m/s) cos(34o) = 11.6 m/s

(a) d = v0x t t = d/v0x = (3.8m) / (11.6 m/s) = 0.33 s

v0y = (14 m/s) sin(34o) = 7.83 m/s

(b) h = v0y t - 1/2 g t2

= (7.83 m/s) (0.33s) -1/2 (9.8 m/s2) (0.33 s)2 = 2.0 m

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall (3.8 m away) with a speed of 14 m/s at an angle of 34° above the horizontal. (c) what are the magnitude and direction of the ball's velocity when it

strikes the wall? (d) Has the ball reached the highest point of its trajectory at this time?

Explain.

d = 3.8 m v0 = 14 m/s θ = 34o

v0x = (14 m/s) cos(34o) = 11.6 m/s v0y = (14 m/s) sin(34o) = 7.83 m/s t = d/v0x = (3.8m) / (11.6 m/s) = 0.33 s

vy = v0y - g t = 7.83 m/s - (9.8m/s2) (0.33s) = 4.6 m/s

vx = v0x

The vertical component of velocity is still positive, that is, the ball is still going up. So the ball has not yet reached its highest point.

Kinematics: Assumptions, Definitions and Logical

Conclusions• Defined displacement, velocity, acceleration (also

position, distance, speed)...

• Defined scalers (like speed) and vectors (like velocity)

• Laid out assumptions about free-fall

• noticed that 2-dimensional motion is really just two, simultaneous, 1-dimensional motions.

Used this to shoot a monkey, range out a small cannon, etc.

This wasn’t physics. This was preparing the language needed to talk about physics.

What have we done so far?

Chapter 5

Newton’s Laws of Motion(sections 5.1-5.4)

Newton’s Laws

How can we consistently and generally describe the way objects move and

interact?

Isaac Newton1643-1727

Newton in a 1702 portrait by Godfrey KnellerNewton in a 1689 portrait by Godfrey Kneller

Nature and nature's laws lay hid in night;

God said "Let Newton be" and all was light.

Newton’s epitaph Alexander Pope (English poet, 1688-1744)

I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the sea-shore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me.

from a memoir by Newton

Force

Force: push or pull

Force is a vector – it has magnitude and direction

Mass

Mass is the measure of how hard it is to change an object’s velocity.

Mass can also be thought of as a measure of the quantity of matter in an object.

Newton’s First Law of Motion

If you stop pushing an object, does it stop moving?

Only if there is friction!

In the absence of any net external force, an object at rest will remain at rest.

In the absence of any net external force a moving object will keep moving at a constant speed in a straight line.

This is also known as the Law of Inertia.

Inertia

Newton’s First Law

a) more than its weight

b) equal to its weight

c) less than its weight but more than zero

d) depends on the speed of the puck

e) zero

A hockey puck slides on ice at constant velocity. What is the net force acting on the puck?

The puck is moving at a constant velocity, and

therefore it is not accelerating. Thus, there must

be no net force acting on the puck.

Newton’s First Law

a) more than its weight

b) equal to its weight

c) less than its weight but more than zero

d) depends on the speed of the puck

e) zero

A hockey puck slides on ice at constant velocity. What is the net force acting on the puck?

Follow-up: Are there any forces acting on the puck? What are they?

a) a net force acted on it

b) no, or insufficient, net force acted on it

c) it remained at rest

d) it did not move, but only seemed to

e) gravity briefly stopped acting on it

Newton’s First Law

You put your book on

the bus seat next to

you. When the bus

stops suddenly, the

book slides forward off

the seat. Why?

a) a net force acted on it

b) no, or insufficient, net force acted on it

c) it remained at rest

d) it did not move, but only seemed to

e) gravity briefly stopped acting on it

The book was initially moving forward (because it was

on a moving bus). When the bus stopped, the book

continued moving forward, which was its initial state of

motion, and therefore it slid forward off the seat.

Newton’s First Law

You put your book on

the bus seat next to

you. When the bus

stops suddenly, the

book slides forward off

the seat. Why?

Calibrating force

Two equal weights exert twice the force of one; this can be used for calibration of a spring:

Experiment: Acceleration vs Force

Now that we have a calibrated spring, we can do more experiments.

Acceleration is proportional to force:

Experiment: Acceleration vs Mass

Acceleration is inversely proportional to mass:

Newton’s Second Law of Motion

Combining these two observations gives

Or, more familiarly,

Acceleration is proportional to force:

Acceleration is inversely proportional to mass:

Newton’s Second Law of Motion

SI unit for force Newton is defined using this equation as:

1 N is the force required to give a mass of 1 kg an acceleration of 1 m/s2

An object may have several forces acting on it; the acceleration is due to the net force:

Units of force: Newtons

The weight of an object is the force acting on it due to gravity

Weight: W = Fg = ma = mg vertically downwards

Since , the weight of an object in Newtons is approximately 10 x its mass in kg

Force of Gravity

adult human

700 N ~ 160 lbs. 70 kg

There is no “conversion” from kg to pounds!(Unless you specify what planet you are assuming)

WHERE?

Newton’s First and Second LawsIn order to change the velocity of an object – magnitude or direction – a net force is required.

(I)

(II)

What about the bus... From the perspective of someone who didn’t know they were on the bus?

Inertial Reference Frames

Newton’s First and Second Laws do not work in an accelerating frame of reference

In order to change the velocity of an object – magnitude or direction – a net force is required.

(I)

(II)

An inertial reference frame is one in which the first and second laws are true.

Accelerating reference frames are not inertial.Was the bus an inertial reference

frame?Is the earth an inertial reference frame?No, but acceleration due to earth’s rotation around

Its axis (0.034 m/s2), and due to earth’s rotation around sun (smaller) are negligible compared to g; so approximately yes.

Analyzing the forces in a system

Free-body diagrams:

A free-body diagram shows every force acting on an object.

• Sketch the forces

• Isolate the object of interest

• Choose a convenient coordinate system

• Resolve the forces into components

• Apply Newton’s second law to each coordinate direction

Free-body Diagram

Example of a free-body diagram:

Newton’s First Law a) there is a net force but the book has too

much inertia

b) there are no forces acting on it at all

c) it does move, but too slowly to be seen

d) there is no net force on the book

e) there is a net force, but the book is too heavy to move

A book is lying at

rest on a table.

The book will

remain there at

rest because:

There are forces acting on the book, but the

only forces acting are in the y-direction.

Gravity acts downward, but the table exerts an

upward force that is equally strong, so the two

forces cancel, leaving no net force.

Newton’s First Law a) there is a net force but the book has too

much inertia

b) there are no forces acting on it at all

c) it does move, but too slowly to be seen

d) there is no net force on the book

e) there is a net force, but the book is too heavy to move

A book is lying at

rest on a table.

The book will

remain there at

rest because:

Newton’s Third Law of Motion

Forces always come in pairs, acting on different objects:

If object 1 exerts a force on object 2, then object 2 exerts a force – on object 1.

These forces are called action-reaction pairs.

Some action-reaction pairs

Action-reaction pair?

a) Yes

b) No

Newton’s 3rd: F12 = - F21

action-reaction pairs are equal and opposite, but they act on different bodies

Newton’s Third Law of Motion

Although the forces are the same, the accelerations will not be unless the objects have the same mass.

Q: When skydiving, do you exert a force on the earth? Does the earth accelerate towards you?

Is the magnitude of the acceleration of the earth the same as the magnitude of your acceleration?

Newton’s Third Law of Motion

Contact forces:

The force exerted by one box on the other is different depending on which one you push.

Assume the mass of the two objects scales with size, and the forces pictured are the same. In which case is the magnitude of the force of box 1 on box 2 larger?

Truck on Frozen Lake

A very large truck sits on a frozen lake. Assume there is no friction between the tires and the ice. A fly suddenly smashes against the front window. What will happen to the truck?

a) it is too heavy, so it just sits there

b) it moves backward at constant speed

c) it accelerates backward

d) it moves forward at constant speed

e) it accelerates forward

When the fly hit the truck, it exerted a force on the truck

(only for a fraction of a second). So, in this time period,

the truck accelerated (backward) up to some speed.

After the fly was squashed, it no longer exerted a force,

and the truck simply continued moving at constant

speed.

Truck on Frozen Lake

A very large truck sits on a frozen lake. Assume there is no friction between the tires and the ice. A fly suddenly smashes against the front window. What will happen to the truck?

a) it is too heavy, so it just sits there

b) it moves backward at constant speed

c) it accelerates backward

d) it moves forward at constant speed

e) it accelerates forward

Follow-up: What if the fly takes off, with the same speed in the direction from whence it came?

Contact Force

2m m

F

Two blocks of masses 2m and m

are in contact on a horizontal

frictionless surface. If a force F0

is applied to mass 2m, what is

the force on mass m ?

a) 2F

b) F

c) ½F

d) 1/3F

e) ¼F

The force F0 leads to a specific

acceleration of the entire system. In

order for mass m to accelerate at the

same rate, the force on it must be

smaller! For the two blocks together,

F0 = (3m)a. Since a is the same for

both blocks, Fm = ma

Contact Force

Two blocks of masses 2m and m

are in contact on a horizontal

frictionless surface. If a force F0

is applied to mass 2m, what is

the force on mass m ?

a) 2F

b) F

c) ½ F

d) 1/3 F

e) ¼ F

2m m

F

A 71-kg parent and a 19-kg child meet at the center of an ice rink. They place their hands together and push.

(a) Is the force experienced by the child more than, less than, or the same as the force experienced by the parent?

(b) Is the acceleration of the child more than, less than, or the same as the acceleration of the parent? Explain.

(c) If the acceleration of the child is 2.6 m/s2 in magnitude, what is the magnitude of the parent’s acceleration?

On vacation, your 1300-kg car pulls a 540-kg trailer away from a stoplight with an acceleration of 1.9 m/s2

(a) What is the net force exerted by the car on the trailer? (b) What force does the trailer exert on the car? (c) What is the net force acting on the car?

An archer shoots a 0.022-kg arrow at a target with a speed of 57 m/s. When it hits the target, it penetrates to a depth of 0.085 m.

(a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the

target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain.