Homework #5 – due Friday, 10/7Reading assignment:

3.1-3.3.5

Questions: 3.2, 4, 12, 16, 18, 19, 27, 36, 39, 82 – the solutions are on the school website.

Homework – due Tuesday, 10/11 – 11:00 pm

Mastering physics wk 5

Chapter 3Position, velocity, acceleration vectors

Projectile motion

Relative motion

Homework #5 – due Friday, 10/7Reading assignment:

• 3.1-3.3, 3.5

Questions: 3.2, 4, 12, 16, 18, 19, 27, 36, 39, 82 – the solutions are on the school website.

Homework – due Tuesday, 10/11 – 11:00 pm

Mastering physics wk 5

Direction of v is tangent to path of particle at P

(2D)

Example 3.1: calculating average and instantaneous velocity

Test your understanding• In which of these situations would the average

velocity vector vav over an interval be equal to the instantaneous velocity v at the end of interval:

1. A body moving along a curved path at constant speed

2. A body moving along a curved path and speeding up

3. A body moving along a straight line at constant speed

4. A body moving along a straight line and speeding up.

Where b and c are positive constants, when does the velocity vector make an angle of 45.0o with the x- and y-axes?

If r = bt2i + ct3j

example

v = dr/dt = 2bt i + 3ct2 j

= 1

2b3c

t =

a = √ax2 + ay

2 + az2

Average acceleration vs. instantaneous acceleration

Example 3.2

1. Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s.

2. Find the instantaneous acceleration at t = 2.0 s, its magnitude and its direction.

Parallel and perpendicular components of acceleration

The effect of acceleration’s directions

• When acceleration is parallel to particle’s velocity: velocity’s magnitude changes only, its direction remains the same, particle moves in a straight line with changing speed.

• When acceleration is perpendicular to particle’s velocity: velocity’s magnitude does not change, only its direction changes, particle moves in a curved path at constant speed.

Increasing speed decreasing speed

Constant speed

Example 3.3• Given:

• Find the parallel and perpendicular components of the instantaneous acceleration at t = 2.0 s

Test your understanding 3.2• A sled travels over the crest of a snow-covered hill. The

sled slows down as it climbs up one side of the hill and gains speed as it descends on the other side. Which of the vectors (1 though 9) in the figure correctly shows the direction of the sled’s acceleration at the crest? Choice 9 is that the acceleration is zero.)

a = 0

Class work

3.3 projectile motion• A projectile is any body that is given an

initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance.

• Example: a batted baseball, a thrown football, a package dropped from an airplane, a bullet shot from a rifle.

• The path of a projectile is called a trajectory.

x (t) = yo + voxt

y (t) = yo + voyt – ½ gt2

z (t) = 0

Motion graphs – world line

v–t graph

a–t graph

Let’s consider again the skier in example 3.4. What is her acceleration at points G, H, and I after she flies off the ramp? Neglect air resistance.

Example 3.5

The acceleration at points G, H, I are the same:

ax = 0; ay = -9.8 m/s2

Example 3.6:a body projected horizontally

• A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. find the motorcycle’s position, distance and velocity from the edge of the cliff, and velocity after 0.50 s.

• Position:

α0 = 0o; vo = 9.0 m/s; voy = 0; t = 0.5 s

y = vosinα0 ∙t - ½ gt2= - ½ (9.8 m/s2)(0.5 s)2 = -1.2 m

x = vocosα0 ∙t = (9.0 m/s)cos0o(0.5 s) = 4.5 m

• distance:

r = √4.52 + (-1.2 m)2

r = 4.7 m

• velocity:

α0 = 0o; vo = 9.0 m/s; t = 0.5 s

vy = - gt= - (9.8 m/s2)(0.5 s) = - 4.9 m/s

vx = vo = 9.0 m/s

• magnitude:

v = √(9.0m/s)2 + (-4.9 m/s)2 v = 10.2 m/s

• direction: α = 29o below horizontal

Example: 3.7 – height and range of a projectile I – a batted baseball

• Position:

α0 = 53.1o; vo = 37.0 m/s; t = 2.00 s

y = (37.0 m/s)(sin53.1o)(2.00 s) - ½ (9.80 m/s2)(2.00 s)2 = 39.6 m

x = (37.0 m/s)(cos53.1o)(2.00 s) = 44.4 m

• velocity:

vy = (37.0 m/s)(sin53.1o) - (9.80 m/s2)(2.00 s) = 9.99 m/s

vx = (37.0 m/s)(cos53.1o) = 22.2 m/s

a.

v = 24.3 m/s α = 24.2o above horizontal

b. At highest point, vy = 0;

tmax = (37.0 m/s)(sin53.1o)/(9.80 m/s2) = 3.02 s

The height reached at this point:

ymax = ½ [(37.0 m/s)(sin53.1o)] 2 / (9.80 m/s2) = 44.7 m

c. Range

xmax = 134 m

• You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.

Given: y = - 8.0 m; α0 = - 20o; v0 = 10.0 m/s;

Unknown: x = ?

To find t, we use

-8.0 m = (10.0 m/s)sin(-20o)t – ½ (9.8 m/s2)t2

t1 = -1.7 s; t2 = 0.98 s

x = (10.0 m/s)cos(-20o)(0.98 s) = 9.2 m

• A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The clever monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Show that the dart always hits the monkey, regardless of the dart’s muzzle velocity (provided that it gets to the monkey before he hits the ground).

Fire at the monkey with fast speed

Fire at the monkey with slow speed

tanα0 = h / d

• If dart hits the monkey, the position of the dart and the position of the monkey must be the same at time t.

For the dart:

xdart = d = (v0cosα0)t

ydart = (v0sinα0)t – ½ gt2

ydart = (v0sinα0)(d /(v0cosα0)) – ½ gt2

Since tanα0 = h / d

ydart = (dtanα0) – ½ gt2

For the monkey:

xmonkey = d

ymonkey = h – ½ gt2

xdart = d = (v0cosα0)t t = d / (v0cosα0)

For the dart:

ydart = h – ½ gt2 = ymonkey

Check your understanding 3.3• In example 3.10, suppose the tranquilizer dart has a

relatively low muzzle velocity so that the dart reaches a maximum height at a point before striking the monkey. When the dart is at point P, will the monkey be

1. At point A (higher than P)

2. At point B (at the same height as P)

3. At point C (lower than P)?

vSail boat moving at v relative to the ground

ship moving at V relative to the ground

Sail boat moving at v’ relative to the ship. v’ = ?

vSail boat moving at v relative to the ground

ship moving at V relative to the ground

Sail boat moving at v’ relative to the ship. v’ = ?

At any time t

r = r’ + R

v∙t = v’∙ t + V∙t

v’ = v - V

Relative velocity of object A to object B

vAB = vAground – vBground

Note:

vAB = vAground + vgroundB

Relative velocity in one dimension

example

VP/E = VP/A + VA/E

VP/E = VP/A + VA/E

150 km/h

R: 150 km/h

Homework #6 – due Friday, 10/14Reading assignment: • 4.1-4.6Questions: 1, 2, 6, 7, 9, 15, 19, 21, 22, 24,

28, 30, 32, 33 – the solutions are on the school website.

Homework – due Tuesday, 10/18 – 11:00 pm

Mastering physics wk 6

Do now (ap exam 2009)

• All the following are units of power except

a. Watts

b. Joules per second

c. Electron volts per second

d. Newton meters per second

e. Kilogram meters per second 72% correct

do now • A 2 kg ball collides with the floor at an angle θ and

rebounds at the same angle and speed as shown above. Which of the following vectors represents the impulse exerted on the ball by the floor?

θ θB.

A.

C.

D.

E.v1

v2

Impulse = change in momentum = m∆v

Impulse has the same direction as v2 – v1 or v2 + (-v1)

-v1

Do now (ap 2009 exam)• A projectile is launched from level ground with

an initial speed v0 at an angle θ with the horizontal. If air resistance is negligible, how long will the projectile remain in air? (represent t in terms of v0, θ, and g)

t = 2v0sin θ / g 50% correct

Do now (ap exam 2009)

• If air resistance is negligible, what is the speed of a 2 kg sphere that falls from rest through a vertical displacement of 0.2 m?

2 m/s; 70% correct

Air resistance isn’t always negligible.