Download Momentum

Momentum
Physics
Linear Momentum
Physics Definition : Linear momentum
of an object of mass (m) moving with
a velocity (v) is defined as the product
of the mass and the velocity.
Momentum is represented by the
symbol (p).
Momentum Facts
• p = mv
• Momentum is a vector quantity!
• Velocity and momentum vectors point in the same direction.
• SI unit for momentum: kg·m/s (no special name).
• Momentum is a conserved quantity (this will be proven later).
• A net force is required to change a body’s momentum.
• Momentum is directly proportional to both mass and speed.
• Something big and slow could have the same momentum as
something small and fast.
The Equation:
p = mv
Bill Nye has something to say…
Let’s solve a problem
A 2250 kg pickup truck has a velocity of 25 m/s
to the east. What is the momentum?
p = 5.6 X 104 kg * m/s east
m = 2250 kg
v = 25 m/s east
p = mv
Relationships
• If you double the mass • Then double momentum
• If you double the
• Then double momentum
velocity
• Then momentum is zero
• If at rest (velocity is 0)
p = mv
http://www.youtube.com/watch?v=hTZI-kpppuw
Equivalent Momenta
Car: m = 1800 kg; v = 80 m /s
p = 1.44 ·105 kg · m /s
Bus: m = 9000 kg; v = 16 m /s
p = 1.44 ·105 kg · m /s
Train: m = 3.6 ·104 kg; v = 4 m /s
p = 1.44 ·105 kg · m /s
Due 1/11
Impulse and Momentum
Physics Definition: The product of the average
net force exerted on an object and the time
interval over which the force acts.
Impulse Equation
J=Ft
Example: A 50 N force is applied to a 100 kg boulder
for 3 s.
The impulse of this force is J = (50 N) (3 s) = 150 N ·s.
Note that we didn’t need to know the mass of the
object in the above example.
Impulse Units
J = F t shows why the SI unit for impulse is the Newton · second. There is no special name
for this unit, but it is equivalent to a kg · m /s.
proof: 1 N ·s = 1 (kg·m /s2) (s) = 1 kg·m /s
Fnet = m a shows this is equivalent to
a newton.
Therefore, impulse and momentum have the same units, which leads to a useful theorem.
Impulse - Momentum Theorem
The impulse due to all forces acting on an object (the net force) is equal to the change in
momentum of the object:
Fnet t =  p
We know the units on both sides of the equation are the same
(last slide), but let’s prove the theorem formally:
Fnet t = m a t = m( v / t)t = m v =  p
How are the velocities of ball (or an
object) before and after the collision
related to the force acting on it?
• N2L: describes how the velocity of a body is
changed by a net force acting on it.
– Ex: The changed in velocity of the ball must have
been caused by the force exerted by the tennis
racket on the ball.
– THE FORCE CHANGES OVER TIME.
Momentum/Impulse = Saves lives
Air Bags
Physics!
Impulse - Momentum Example
A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s who kicks it in the exact
opposite direction at 22 m/s with an average force of 1200 N. How long is his foot and the ball
in contact?
answer: We’ll use Fnet t =  p. Since the ball
changes direction,  p = m  v = m (vf - v0)
= 1.3 [22 - (-13)] = (1.3 kg) (35 m/s)
= 45.5 kg · m /s. Thus, t = 45.5 / 1200
= 0.0379 s, which is just under 40 ms.
During this contact time the ball compresses substantially and then decompresses. This
happens too quickly for us to see, though. This compression occurs in many cases, such as
hitting a baseball or golf ball.
Let’s solve another problem
A 1400 kg car moving westward with a velocity of
15 m/s collides with a utility pole and is brought
to rest in 0.3 seconds. Find the force exerted on
the car during the collision.
F = 7.0 X104 N east
F = (mv2 – mv1)/∆t
m = 1400 kg
v1 = 15 m/s west
v2 = 0 m/s west
t= 0.3 s
Due 1/13
Momentum is Conserved
Physics Definition: Law of conservation of
momentum- the total p of all objects
interacting with one another remains constant
regardless of the nature of the forces between
the object.
Explosions
• When two or more objects are
pushed apart by an internal force.
• The “explosive” force can be
provided by an actual explosion,
a spring, or a pair of magnets.
Collisions
• Collisions are when two or more objects run
into each other.
– They can stick together
– Spring back apart
Collisions
1. Elastic Collisions Bounces off
2. Inelastic Collisions Sticks together
Elastic Collisions
• No permanent deformation
• Energy and momentum is conserved
• Ex: Billiard ball colliding with another
Keep direction in mind…
Inelastic Collision
•
•
•
•
Objects sticks together
Momentum is conserved
Energy is not conserved
Ex: a ball of putty hitting and sticking to
another ball
• Ex: Two railroad cars colliding and coupling
together
Partially Elastic Collision
• There is some deformation
• They do not stick together
• Automobile collision is a good example(
mathematically, a partially elastic collision is
handled the same way as an elastic collision
except that energy is not conserved.
Conservation of Momentum Equation
m1v1 +m2v2 = m1v1’ + m2v2’
• Note: regardless of the type of collision, you
will most likely use momentum to solve it!
• Each object has a velocity before and after
the collision but their masses remain the
same.
Inelastic Collisions
In any case where objects
stick together they will
both have the same
velocity and will end up
sharing the momentum of
the first object.
Let’s Calculate Now
• A 4kg block with an initial velocity of 10 m/s
that is colliding with an 6 kg block that is
stationary. After the collision, the 6 kg block is
seen to be moving 8 m/s. Your job is to
determine the velocity of the 4 kg block after
the collision.
4 kg
6 kg
4 kg
V1 = 10 m/s
V2 = 0 m/s
V1 ‘ = ? m/s
6 kg
V2 ‘ = 8 m/s
1st let’s start with the collision
equation
•
•
•
•
•
•
•
m1v1 +m2v2 = m1v1’ + m2v2’
m1= 4 kg
m2 = 6 kg
4(10) + 6(0) = 4(v1’) + 6(8)
v1 = 10 m/s
40 + 0 = 4(v1’) + 48
-8 = 4(v1’)
v2 = 0 m/s
-2 m/s = v1’
v1’ = ?
v2’ = 8 m/s
Another one….
• A 4 kg block with an initial velocity of 10 m/s is
colliding with an 6 kg block that is stationary.
After the collision, both blocks are stuck
together and are moving together. Your job is
to determine the velocity of the linked blocks
after the collision.
4 kg
6 kg
V1 = 10 m/s
V2 = 0 m/s
4 kg
6 kg
V‘ = ? m/s
Again start with the collision equation
•
•
•
•
•
•
m1v1 +m2v2 = (m1+ m2)v’
m1= 4 kg
m2 = 6 kg
4(10) + 6(0) = 4 + 6(v’)
v1 = 10 m/s
40 + 0 = 10(v’)
v2 = 0 m/s
+ 4 m/s = v’
v’ = ?