Using Newton’s Laws

Section 3 page 86
Essential Questions

 How does Newton’s 1st law explains what happens in a car crash?

 How does Newton’s 2nd law explain the effect of air resistance?

 When is momentum conserved?

Main Idea

 Newton’s laws can be used to explain everyday events,

such as falling and collision.
Understanding Car Crashes:
It’s Basic Physics!

 Watch the video

https://www.youtube.com/watch?v=yUpiV2I_IRI
 What happens in a crash?
 The law of inertia can explain what

happens in a car crash.

 When a car traveling about 50 km/h

collides head-on with something solid,

the car crumples, slows down, and

stops within approximately 0.1 s.

 What happens in a crash?
Any passenger not wearing a safety belt
continues to move forward at the same
speed the car was traveling.
Within about 0.02 s (1/50 of a second) after
the car stops, unbelted passengers slam into
the dashboard, steering wheel, windshield,
or the backs of the front seats.
The force needed to slow a person from 50
km/h to zero in 0.1 s is equal to 14 times the
force that gravity exerts on the person.
What happens in a crash?

 The belt loosens a little as it restrains the person, increasing the time
it takes to slow the person down.
 This reduces the force exerted on the person.
 The safety belt also prevents the person from being thrown out of
the car.
 Safety Belts
 Air bags also reduce injuries in car crashes by providing a
cushion that reduces the force on the car's occupants.
 When impact occurs, a chemical reaction occurs in the air
bag that produces nitrogen gas.
 The air bag expands rapidly and then deflates just as quickly
as the nitrogen gas escapes out of tiny holes in the bag.
Concept Test
You are a passenger in a car and not wearing your seat belt.
Without increasing or decreasing its speed, the car makes a sharp left
turn, and you find yourself colliding with the right-hand door.
Which is the correct analysis of the situation? ...
Concept Test
1. Before and after the collision, there is a rightward force pushing
you into the door.
2. Starting at the time of collision, the door exerts a leftward force
on you.

2. Starting at the time of collision, the

3. both of the above
4. neither of the above
door exerts a leftward force on
you.
Newton’s 2nd law and Gravitational
acceleration
 Air resistance
 Free fall
 Newton’s 2nd Law proves that different masses
accelerate to the earth at the same rate, but with
We know that objects different forces.
with different masses
accelerate to the
ground at the sameIn vacuum, there is
rate. no air resistance, as
a result the 2
However, becausedifferent
of objects will
fall with the same
the 2 Law we know
nd acceleration
that they don’t hit the
ground with the same
force.

F = ma F = ma
98 N = 10 kg x 9.8 9.8 N = 1 kg x 9.8
m/s/s m/s/s
Elephant and Feather - Air Resistance
Terminal Velocity: maximum speed an object
will reach when falling through air
 Velocity-Time Graph
Parachute opens – diver
Speed
slows down
increases…

Terminal
Velocity

velocity
reached…

New, lower terminal velocity Time

reached Diver hits the ground
 Questions
1. The diagram shows a skydiver – two forces act on the skydiver (X and Y)

a) What is the equation which links weight, gravitational field strength and mass?
b) What causes force X?
c) As the skydiver falls the size of force X increases. What happens to the size of
force Y?
d) Describe the motion of the skydiver when force X is smaller than force Y; and
when force X is equal to force Y
 Answers

a) What is the equation which links weight, gravitational field

strength and mass? Weight = mass x gravity

a) What causes force X? Drag (air resistance / friction)

a) As the skydiver falls the size of force X increases. What happens to

the size of force Y? Stays the same

a) Describe the motion of the skydiver when force X is smaller than

force Y; and when force X is equal to force Y. When force X is
smaller than force Y the skydiver accelerates downwards. When
the forces are equal the skydiver moves at a constant speed
Weightlessness

 People aboard the space station appear to be weightless (They float

around like there is no gravity).
 However, they are not out of Earth’s gravitational field (gravity is still pulling
on them).
Freefall

 The space station is actually in freefall around the planet.

 That is what orbit is (constant free fall around the planet).
 It is moving so quickly forward, it clears the planet.
Frame of Reference

 The “weightlessness” has to do with the frame of reference.

 If everything is falling at the same rate, it appears as though it is floating
with no gravity.
 Similar to being on an amusement park ride that drops you straight down.
 In the case of the space station everything is falling with you (air included),
so you can’t tell you are falling.
Centripetal
forces keep
these children
moving in a
circular path.
 Uniform Circular Motion

Uniform circular motion is motion along a circular path in which there

is no change in speed, only a change in direction.

v Constant velocity tangent

Fc to path.

Constant force toward

center.

Question: Is there an outward force on the ball?

 Uniform Circular Motion (Cont.)

The question of an outward force can be resolved by asking what

happens when the string breaks!

Ball moves tangent to path, NOT

v
outward as might be expected.

When central force is removed, ball continues

in straight line.

Centripetal force is needed to change direction.

 Examples of Centripetal Force

You are sitting on the seat next

to the outside door. What is the
direction of the resultant force
on you as you turn? Is it away
from center or toward center of
the turn?
 Car going around a curve.
Fc

Force ON you is toward the center.

 Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each
other without colliding, such as gravity) momentum of the system
(both objects together) is conserved. This mean the total momentum
of the objects is the same before and after the collision.
(Choosing right as the +
before: p = m1 v1 - m2 v2
direction, m2 has - momentum.)
v1 v2
m1 m2

m1 v1 - m2 v2 = - m1 va + m2 vb
after: p = - m1 va + m2 vb
va vb
m1 m2
 Directions after a collision
On the last slide the boxes were drawn going in the opposite direction
after colliding. This isn’t always the case. For example, when a bat hits
a ball, the ball changes direction, but the bat doesn’t. It doesn’t really
matter, though, which way we draw the velocity vectors in “after”
picture. If we solved the conservation of momentum equation (red box)
for vb and got a negative answer, it would mean that m2 was still moving
to the left after the collision. As long as we interpret our answers
correctly, it matters not how the velocity vectors are drawn.
v1 v2
m1 m2

m1 v1 - m2 v2 = - m1 va + m2 vb
va vb
m1 m2
Sample Problem 1
35 g
7 kg
700 m/s
v=0
A rifle fires a bullet into a giant slab of butter on a frictionless surface.
The bullet penetrates the butter, but while passing through it, the bullet
pushes the butter to the left, and the butter pushes the bullet just as hard
to the right, slowing the bullet down. If the butter skids off at 4 cm/s
after the bullet passes through it, what is the final speed of the bullet?
(The mass of the rifle matters not.)

35 g
7 kg
v=? 4 cm/s
continued on next slide
 Sample Problem 1 (cont.)
Let’s choose left to be the + direction & use conservation of
momentum, converting all units to meters and kilograms.
35 g
p before = 7 (0) + (0.035) (700) 7 kg
700 m/s
= 24.5 kg · m /s v=0

35 g p after = 7 (0.04) + 0.035 v

7 kg
v=? 4 cm/s = 0.28 + 0.035 v

p before = p after 24.5 = 0.28 + 0.035 v v = 692 m/s

v came out positive. This means we chose the correct

direction of the bullet in the “after” picture.
Sample Problem 2
35 g
7 kg 700 m/s
v=0
Same as the last problem except this time it’s a block of wood rather than
butter, and the bullet does not pass all the way through it. How fast do
they move together after impact?

v
7. 035 kg

(0.035) (700) = 7.035 v v = 3.48 m/s

Note: Once again we’re assuming a frictionless surface, otherwise there
would be a frictional force on the wood in addition to that of the bullet,
and the “system” would have to include the table as well.