Download m/s

Motion & Forces
Lesson 3
Action and Reaction
Newton’s Third Law
 Momentum
 Conservation of Momentum

Newton’s Third Law

Newton’s Third Law of Motion
 When one object exerts a force on
a second object, the second object
exerts an equal but opposite force
on the first.
Newton’s Third Law

Problem:
 How
can a horse
pull a cart if the cart
is pulling back on
the horse with an equal but
opposite force?
 Aren’t
these “balanced forces”
resulting in no acceleration?
NO!!!
Newton’s Third Law

Explanation:
 forces
are equal and opposite but
act on different objects
 they are not “balanced forces”
 the movement of the horse
depends on the forces acting on
the horse
Action and Reaction



When a force is applied in nature, a
reaction force occurs at the same time.
When you jump on a trampoline, for
example, you exert a downward force on
the trampoline.
Simultaneously, the trampoline exerts
an equal force upward, sending you high
into the air.
Action and Reaction Forces
Don’t Cancel


According to the third law of motion,
action and reaction forces act on
different objects.
Thus, even though the forces are equal,
they are not balanced because they act
on different objects.
Action and Reaction Forces
Don’t Cancel


For example, a swimmer “acts” on the
water, the “reaction” of the water
pushes the swimmer forward.
Thus, a net force, or unbalanced force,
acts on the swimmer so a change in his
or her motion occurs.
Rocket Propulsion
In a rocket engine, burning fuel
produces hot gases. The rocket engine
exerts a force on these gases and
causes them to escape out the back of
the rocket.
 By Newton’s third law, the
gases exert a force on the
rocket and push it forward.

Newton’s Third Law

Action-Reaction Pairs


The hammer
exerts a force on
the nail to the
right.
The nail exerts
an equal but
opposite force on
the hammer to
the left.
Momentum



A moving object has a property called
momentum that is related to how much
force is needed to change its motion.
The momentum of an object is the
product of its mass and velocity
Momentum is given the symbol p and can
be calculated with the following
equation:
Momentum

Momentum
 quantity of motion
• Use the unit
kg·m/s
p
m v
p:
m:
v:
p = mv
momentum (kg·m/s)
mass (kg)
velocity (m/s)
Force and Changing
Momentum


By combining these two relationships,
Newton’s second law can be written in
this way:
In this equation mvf is the final
momentum and mvi is the initial
momentum
Law of Conservation of
Momentum



The momentum of an object doesn’t
change unless its mass, velocity, or both
change.
Momentum, however, can be transferred
from one object to another.
The law of conservation of momentum
states that if a group of objects exerts
forces only on each other, their total
momentum doesn’t change.
Law of Conservation of
Momentum
• The results of a collision depend on the
momentum of each object.

When the first puck hits the second
puck from behind, it gives the second
puck momentum in the same direction.
When Objects Collide
• If the pucks are speeding toward each
other with the same speed, the total
momentum is zero.
Newton’s Third Law

Action-Reaction Pairs
 Both objects accelerate.

The amount of acceleration depends on
the mass of the object.
F
a 
m

Small mass  more acceleration

Large mass  less acceleration
Conservation of Momentum

Law of Conservation of Momentum
 The total momentum in a group of
objects doesn’t change unless
outside forces act on the objects.
pbefore = pafter
Conservation of Momentum


Elastic Collision
 KE is conserved
Inelastic Collision
 KE is not conserved
Momentum

Find the momentum of a bumper car if it has a
total mass of 280 kg and a velocity of 3.2 m/s.
WORK:
GIVEN:
p=?
m = 280 kg
v = 3.2 m/s
p = mv
p = (280 kg)(3.2 m/s)
p
m v
p = 896 kg·m/s
Momentum

The momentum of a second bumper car is 675
kg·m/s. What is its velocity if its total mass
is 300 kg?
GIVEN:
p = 675 kg·m/s
m = 300 kg
v=?
p
m v
WORK:
v=p÷m
v = (675 kg·m/s)÷(300
kg)
v = 2.25 m/s
Conservation of Momentum

A 5-kg cart traveling at 1.2 m/s strikes a
stationary 2-kg cart and they connect. Find
their speed after the collision.
BEFORE
Cart 1:
p = 21 kg·m/s
m = 5 kg
v = 4.2 m/s
Cart 2 :
m = 2 kg
v = 0 m/s
p=0
pbefore = 21 kg·m/s
AFTER
Cart 1 + 2:
m = 7 kg
v=?
p
m v
v=p÷m
v = (21 kg·m/s) ÷ (7 kg)
v = 3 m/s
pafter = 21 kg·m/s
Conservation of Momentum

A 50-kg clown is shot out of a 250-kg cannon
at a speed of 20 m/s. What is the recoil speed
of the cannon?
BEFORE
AFTER
Clown:
m = 50 kg
v = 0 m/s
p=0
Clown:
p = 1000 kg·m/s
m = 50 kg
v = 20 m/s
Cannon:
m = 250 kg
v = 0 m/s
p=0
Cannon: p = -1000 kg·m/s
m = 250 kg
v = ? m/s
pbefore = 0
pafter = 0
Conservation of Momentum

So…now we can solve for velocity.
WORK:
GIVEN:
p = -1000 kg·m/s v = p ÷ m
m = 250 kg
v = (-1000 kg·m/s)÷(250
v=?
kg)
p
m v
v = - 4 m/s
(4 m/s backwards)
JET CAR CHALLENGE
CHALLENGE:
Construct a car that will travel as far as
possible (at least 3 meters) using only the
following materials.




scissors
tape
4 plastic lids
2 skewers



2 straws
1 balloon
1 tray
How do each of Newton’s Laws apply?