1. Wednesday(10/28): Introduction to Momentum (Ch.9). The Impulse-

Momentum Relation, Elastic and Inelastic Collisions, Recoil. HW: Read Ch.9,

pages 235-262 and Solve prob. 1, 5, 9, 13, 21, and 23 on pages 264-6.

2. Thursday(10/29): Collisions in Two Dimensions. Finding Center of Mass.

HW: Solve prob. 27, 33, 39, 41, 43, 51, and 52 on pages 266-9.

3. Friday(10/30): LAB on Center of Mass. HW: Process lab data.

4. Monday(11/02): Post-Lab discussion. HW: Process lab data and write

lab report (due Tuesday).

5. Tuesday(11/03): Review Ch.9. HW: Complete Review handout.

6. Wednesday(11/04): TEST on Ch.9 - Momentum. HW: Go to web-site

for notes on Ch.10 & 11 - Rotational Mechanics.

Very Important: If you have any questions or miss a class, see me

before school (8:00 - 8:30 AM), during Lunch, or after school. Best to

send an email to persin@nova.edu.

1. The momentum of an object is the product of the object's mass and

its velocity. We use lowercase p for momentum and this gives the

formula p = mv .

2. Since mass is a scalar and velocity is a vector, momentum is a vector

because the product of a scalar and a vector is a vector.

3. So, anything we have done with vectors previously can be done with

momentum. It can be broken into horizontal and vertical components,

and several momentum vectors can produce a single resultant.

4. The Pythagorean theorem and SOHCAHTOA are still used to solve

momentum problems.

5. The unit for momentum in MKS is the kg m/s , but when working in

CGS we use g cm/s.

6. The change in momentum, Δp , of an object is equal to the impulse

that acts that acts on it. This is known as the Impulse-Momentum

Relation. This relation can also be derived from Newton's 2nd Law,

and the definition of acceleration. Use F = ma with a = Δv/Δt .

7. Impulse is the product of the force acting on an object and the

time in which it acts. This gives us the formula for impulse which

is J = F·Δt. The unit of impulse in MKS is the Newton-second or Ns.

Impulse can also be derived using Calculus with J = ∫F(t)dt = Δp,

same as area under a curve.

8. The Law of Conservation of Momentum states that "In a closed,

isolated system, the total momentum of the system does not change".

9. This law means that in all interactions between isolated objects,

momentum is conserved, or,  p_i = p_f  .

10. Objects can transfer momentum during collisions, but the total

momentum in the system before the collision must equal the total

momentum after the collision, again,  p_i = p_f  .

11. During a collision, the change in momentum of the first object is

equal to and opposite the change in momentum of the second object.

12. Collisions can be classified as elastic or inelastic based on

whether or not energy is also conserved. For the elastic collision we

have m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' . If the collision is inelastic we

have m₁v₁ + m₂v₂ = (m₁ + m₂)v_1,2 .

13. In a perfectly elastic collision, both momentum and kinetic energy

are conserved, p_i = p_f   and   E_i = E_f

14. In a perfectly inelastic collision, two objects stick together and

move as one mass after the collision. This means that momentum is

conserved but kinetic energy is not conserved.  p_i = p_f   and   E_i > E_f

15. In an inelastic collision, kinetic energy is converted into internal

elastic potential energy when the objects deform. Some kinetic energy

is also converted to sound and thermal energy.

16. There is a special case called "recoil" in which, for example, two

objects are pushed against opposite ends of a spring. Clearly the

momentum before they are released is zero, so according to the

conservation law the momentum after they are released also must

be zero.

17. Recoil then is supported by the equation 0 = m₁v₁ + m₂v₂ ,

which requires that v₁ and v₂ must have opposite directions.

18. An interesting aspect of the relationship between momentum and

kinetic energy is that if the mass, m, and the velocity, v, of an object

is known, then we automatically know its momentum and kinetic

energy. Recall the p = mv and K = ½mv² , therefore we can express

the objects kinetic energy as K = ½pv .

19. Realize that in the macro world, few collisions are perfectly elastic

or perfectly inelastic.

And still, we need these steps to solve any problem in Physics:

(i) read the problem and identify the given variables

(ii) determine what you are asked to solve for

(iii) find the correct vector formula to use

(iv) use Algebra, Trigonometry, and/or Calculus to isolate the unknown

(v) substitute-in the given information and simplify.