1. Momentum
Physics

2. 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).
It describes an objects motion

3. Momentum Facts p = m v 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.

4. p = mv The Equation: P = kg m/s M = kg = how much matter in an object
V = m/s = speed in a given direction

5. Bill Nye has something to say…

6. p = mv 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 = m = v =
5.6 X 104 kg * m/s east
2250 kg
p = mv
25 m/s east

7. p = mv Relationships If you double the mass If you double the velocity
If at rest (velocity is 0)
Then double momentum
Then double momentum
Then momentum is zero
p = mv

8. 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

9. Due 1/11

10. 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.
Newton wrote his three laws of motion in terms of momentum, which he called the quantity of motion.

11. Impulse Equation
J = F t
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.

12. { Impulse Units proof: 1 N · s = 1 (kg · m /s2) (s) = 1 kg · m /s
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.

13. 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

14. 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.

15. Momentum/Impulse = Saves lives
Air Bags
Physics!

16. 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 / = 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.

17. 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 = m = v1 = v2 = t=
7.0 X104 N east
F = (mv2 – mv1)/∆t
1400 kg
15 m/s west
0 m/s west
0.3 s

18. Due 1/13

19. 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.

20. 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.

21. Collisions
Collisions are when two or more objects run into each other.
They can stick together
Spring back apart

22. Collisions 1. Elastic Collisions - Bounces off
2. Inelastic Collisions - Sticks together

23. Elastic Collisions No permanent deformation
Energy and momentum is conserved
Ex: Billiard ball colliding with another

24. Keep direction in mind…

25. 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

26. 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.

27. 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.

28. 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.

29. 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
6 kg
V1 = 10 m/s
V1 ‘ = ? m/s
V2 ‘ = 8 m/s
V2 = 0 m/s

30. 1st let’s start with the collision equation
m1v1 +m2v2 = m1v1’ + m2v2’
m1= 4 kg
m2 = 6 kg
v1 = 10 m/s
v2 = 0 m/s
v1’ = ?
v2’ = 8 m/s
4(10) + 6(0) = 4(v1’) + 6(8)
= 4(v1’) + 48
-8 = 4(v1’)
-2 m/s = v1’

31. 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
4 kg
6 kg
V‘ = ? m/s
V1 = 10 m/s
V2 = 0 m/s

32. Again start with the collision equation
m1v1 +m2v2 = (m1+ m2)v’
m1= 4 kg
m2 = 6 kg
v1 = 10 m/s
v2 = 0 m/s
v’ = ?
4(10) + 6(0) = 4 + 6(v’)
= 10(v’)
+ 4 m/s = v’
So the combine will be moving at 4 m/s in the original direction of the first object