1. Ch. 9: Linear Momentum & Collisions

2. 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: p x = mv x p y = mv y p z = mv z 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: m 1 moves at velocity v 1 & m 2 moves at velocity v 2. m 1 feels a force F 21 exerted on it by m 2. m 2 feels a force F 12 exerted on it by m 21. See figure   NOTE: Misconception! The masses do NOT have to touch! Newton’s 3 rd Law: F 21 = - F 12 Or: F 21 + F 12 = 0 (1) Newton’s 2 nd Law: (if no other forces act) F 21 = m 1 a 1 (2). F 12 = m 2 a 2 (3) Put (2) & (3) into (1)  m 1 a 1 + m 2 a 2 = 0 (4) Note v’s & F’s are vectors!!

6. 2 moving masses interacting.   N’s 3 rd Law: F 21 + F 12 = 0 N’s 2 nd Law: F 21 = m 1 a 1. F 12 = m 2 a 2 Together: m 1 a 1 + m 2 a 2 = 0 (4) Acceleration definition: a ≡ (dv/dt) The acceleration is the time derivative of the velocity  (4) becomes: m 1 (dv 1 /dt) + m 2 (dv 2 /dt) = 0 Use simple calculus: d(m 1 v 1 )/dt + d(m 2 v 2 )/dt = 0 or d(m 1 v 1 + m 2 v 2 )/dt = 0 (5) The time derivative of m 1 v 1 + m 2 v 2 is = 0.  Calculus tells us that m 1 v 1 + m 2 v 2 = constant! (6) A Vector Equation!

7. So, for 2 moving masses interacting & isolated from the rest of the world: m 1 v 1 + m 2 v 2 = constant (6) With the definition of momentum: p 1 = m 1 v 1, p 2 = m 2 v 2 (6) becomes: p 1 + p 2 = 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. Consider now, one mass m & write: Newton’s 2 nd Law: ∑F = ma Use the definition of acceleration as time derivative of the velocity: a ≡ (dv/dt). Put this into Newton’s 2 nd 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 2 nd 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 2 nd 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. Back to 2 moving masses interacting & isolated from the rest of the world. We found: d(p 1 + p 2 )/dt = 0 or p 1 + p 2 = constant (1) This says that the total momentum of the 2 masses p tot = p 1 + p 2 = constant Suppose due to the forces F 21 & F 12, p 1 & p 2 change with time. (1) tells us that, no matter how they change individually, p tot = constant So, the total VECTOR momentum of the 2 masses isconserved! If, at some initial time, the 2 momenta are p 1i & p 2i & if at some final time they are p 1f & p 2f, we can write: p tot = p 1i + p 2i = p 1f + p 2f = constant

11. Example: 2 billiard balls collide (zero external force) m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f the vector sum is constant! momentum before = momentum after! m 1 v 1i m 2 v 2i m 1 v 1f m 2 v 2f

12. Example Initial Momentum = Final Momentum (1D) m 1 v 1i +m 2 v 2i = (m 1 + m 2 )v f v 2i = 0, v 1f = v 2f = v f  v f = [(m 1 v 1 )/(m 1 + m 2 )] = 12 m/s v 2i = 0 v 1i = 24 m/s v f = ??

13. Example: Rocket Propulsion Momentum Before = Momentum After 0 = P rocket - P gas

14. Example: Rifle Recoil Momentum Before = Momentum After m 1 v 1i +m 2 v 2i = m 1 v 1f + m 2 v 2f m B = 0.02 kg, m R = 5.0 kg, v B = 620 m/s 0 = m B v B + m R v R  v R = - 2.5 m/s (to left, of course!) v 1i = 0, v 2i = 0 vRvR pRpR vBvB pBpB

15. Example 9.1: Archer Archer, m 1 = 60 kg, v 1i = 0, stands on frictionless ice. Arrow, m 2 = 0.5 kg, v 2i = 0. Archer shoots arrow horizontally at v 2f = 50 m/s to the right. What velocity v 2f does the archer have as a result? Possible Approaches: N’s 2 nd 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. m 1 = 60 kg, v 1i = 0, m 2 = 0.5 kg, v 2i = 0, v 2f = 50 m/s, v 2f = ? 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 p tot = m 1 v 1i + m 2 v 21 = m 1 (0) + m 2 (0) = 0 Momentum is conserved! p tot = p 1i + p 2i = p 1f + p 2f = constant  After Shooting Arrow m 1 v 1f + m 2 v 2f = 0 or, v 1f = - (m 1 /m 1 )v 2f v 1f = - 0.42 m/s (Minus means archer slides to left!)