1. Ch. 9: Linear Momentum & Collisions

2. NEWTON’S LAWS OF MOTION!
THE COURSE THEME:
NEWTON’S LAWS OF MOTION!
Chs. 5 & 6: Motion analysis with Forces.
Chs. 7 & 8: Alternative analysis with Work & Energy.
Conservation of Energy: NOT a new law! We’ve seen that this is Newton’s Laws re-formulated or translated from Force Language to Energy Language.
NOW (Ch. 9): Another alternative analysis using the concept of (Linear) Momentum.
Conservation of (Linear) Momentum: NOT a new law!
We’ll see that this is just Newton’s Laws of Motion re-formulated or re-expressed (translated) from Force Language to (Linear) Momentum Language.

3. In Chs. 5 & 6, we expressed Newton’s Laws of Motion using the concepts of position, displacement, velocity, acceleration, force.
Newton’s Laws with Forces: General. In principle, could be used to solve any dynamics problem, But, often, they are very difficult to apply, especially to very complicated systems. So, alternate formulations have been developed. Often easier to apply.
In Chs. 7 & 8, we expressed Newton’s Laws using Work & Energy Language.
Newton’s Laws with Work & Energy: Also general. In principle, could be used to solve any dynamics problem, But, often, it’s more convenient to use still another formulation.
This is an approach that uses Momentum instead of Energy as the basic physical quantity.
Newton’s Laws in a different language (Momentum). Before we discuss these, we need to learn vocabulary in Momentum Language.

4. Momentum: The momentum of an object is DEFINED as:
p = mv (a vector || v)
SI Units: kgm/s = Ns
In 3 dimensions, momentum has 3 components:
px = mvx py = mvy pz = mvz
Newton called mv “quantity of motion”.
Question: How is the momentum of an object changed?
Answer: By the application of a force F!

5. Section 9.1: Linear Momentum
Consider an isolated system with 2 masses: m1 moves at velocity v1 & m2 moves at velocity v2. m1 feels a force F21 exerted on it by m2. m2 feels a force F12 exerted on it by m21. See figure  
NOTE: Misconception! The masses do NOT have to touch!
Newton’s 3rd Law: F21 = - F12
Or: F21 + F12 = (1)
Newton’s 2nd Law: (if no other forces act) F21 = m1a (2) F12 = m2a (3)
Put (2) & (3) into (1) 
m1a1 + m2a2 = (4)
Note
v’s & F’s are
vectors!!

6. A Vector Equation! 2 moving masses interacting.  
N’s 3rd Law: F21 + F12 = 0
N’s 2nd Law: F21 = m1a1. F12 = m2a2
Together: m1a1 + m2a2 = (4)
Acceleration definition: a ≡ (dv/dt)
The acceleration is the time derivative of the velocity
 (4) becomes: m1(dv1/dt) + m2(dv2/dt) = 0
Use simple calculus: d(m1v1)/dt + d(m2v2)/dt = 0
or d(m1v1 + m2v2)/dt = (5)
The time derivative of m1v1 + m2v2 is = 0.
 Calculus tells us that
m1v1 + m2v2 = constant! (6)
A Vector Equation!

7. With the definition of momentum: p1 = m1v1, p2 = m2v2 (6) becomes:
So, for 2 moving masses interacting & isolated from the rest of the world:
m1v1 + m2v2 = constant (6)
With the definition of momentum:
p1 = m1v1, p2 = m2v2
(6) becomes:
p1 + p2 = constant (7)
(7) says that, no matter how they interact & what motions they undergo, the vector sum of the momenta of otherwise isolated masses is ALWAYS THE SAME FOR ALL TIME!
Note: The plural of “momentum” is “momenta”, NOT “momentums”!!

8. Newton’s 2nd Law: ∑F = (dp/dt)
Consider now, one mass m & write:
Newton’s 2nd Law: ∑F = ma
Use the definition of acceleration as time derivative of the velocity: a ≡ (dv/dt).
Put this into Newton’s 2nd Law: ∑F = m(dv/dt)
If m doesn’t depend on time: ∑F = [d(mv)/dt]
Put the definition of momentum, p ≡ mv into
Newton’s 2nd Law: ∑F = (dp/dt)
Did this for constant m. Can be shown it’s more general & is valid even if m changes with time.

9. A general statement of Newton’s 2nd Law is: ∑F = (dp/dt) (1)
The total or net force acting on a mass = the time rate of change in the mass’s momentum.
(1) is more general than ∑F = ma because it allows for the mass m to change with time also!
Example, rocket motion! Later!
Note: if m is constant, (1) becomes:
∑F = (dp/dt) = d(mv)/dt = m(dv/dt) = ma

10. ptot = p1i + p2i = p1f + p2f = constant
Back to 2 moving masses interacting &
isolated from the rest of the world. We found: d(p1 + p2)/dt = or
p1 + p2 = constant (1)
This says that the total momentum of the
2 masses ptot = p1 + p2 = constant
Suppose due to the forces F21 & F12, p1 & p2
change with time. (1) tells us that, no matter
how they change individually, ptot = constant
So, the total VECTOR momentum
of the 2 masses isconserved!
If, at some initial time, the 2 momenta are p1i & p2i & if at some final time they are p1f & p2f, we can write:
ptot = p1i + p2i = p1f + p2f = constant

11. momentum before = momentum after!
Example: 2 billiard balls collide (zero external force)
momentum before = momentum after!
m1v1i
m2v2i
m1v1i + m2v2i =
m1v1f + m2v2f
the vector sum
is constant!
m1v1f
m2v2f

12. Example Initial Momentum = Final Momentum (1D)
v1i = 24 m/s
v2i = 0
vf = ??
m1v1i+m2v2i = (m1 + m2)vf v2i = 0, v1f = v2f = vf
 vf = [(m1v1)/(m1 + m2)] = 12 m/s

13. Example: Rocket Propulsion
Momentum Before = Momentum After
0 = Procket - Pgas

14. Example: Rifle Recoil Momentum Before = Momentum After
m1v1i+m2v2i = m1v1f + m2v2f
mB = 0.02 kg, mR = 5.0 kg, vB = 620 m/s
0 = mB vB + mRvR  vR = m/s (to left, of course!)
v1i = 0, v2i = 0 vR vB pR pB

15. Example 9.1: Archer
Archer, m1 = 60 kg, v1i = 0, stands on frictionless ice. Arrow, m2 = 0.5 kg, v2i = 0. Archer shoots arrow horizontally at
v2f = 50 m/s to the right. What velocity v2f does the archer have as a result?
Possible Approaches:
N’s 2nd Law in force form:
Can’t use: No information about F or a!
Energy approach: Can’t use:
No information about work, energy!
Momentum approach: Easily used!!!

16. Momentum is conserved! ptot = p1i + p2i = p1f + p2f = constant
m1 = 60 kg, v1i = 0, m2 = 0.5 kg, v2i = 0, v2f = 50 m/s, v2f = ?
Momentum
No external forces in x-direction, so arrow is isolated in the x-direction
Total momentum before shooting arrow is 0
 Total Momentum after shooting arrow is also 0!
Before Shooting Arrow
ptot = m1v1i + m2v21 = m1(0) + m2(0) = 0
Momentum is conserved! ptot = p1i + p2i = p1f + p2f = constant
 After Shooting Arrow
m1v1f + m2v2f = 0 or, v1f = - (m1/m1)v2f
v1f = m/s (Minus means archer slides to left!)