1. Chapter 9: Linear Momentum

2. THE COURSE THEME: NEWTON’S LAWS OF MOTION! Chs. 4 & 5: Motion analysis with Forces. Ch. 6: Alternative analysis with Work & Energy. –Work-Energy Theorem & Conservation of Mechanical Energy: NOT new laws! We’ve seen that they are Newton’s Laws re-formulated or translated from Force Language to Energy Language. NOW (Ch. 7): 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 re-formulated or re- expressed (translated) from Force & Acceleration Language to Force & (Linear) Momentum Language.

3. In Chs. 4 & 5, we expressed Newton’s Laws of Motion using the position, displacement, velocity, acceleration, & force concepts. Newton’s Laws with Forces & Accelerations: Very 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 which are often easier to apply. In Ch. 6, we expressed Newton’s Laws in Work & Energy Language. Newton’s Laws with Work & Energy: Very general. In principle, could be used to solve any dynamics problem, But, often (especially in collision problems) it’s more convenient to use still another formulation. The Ch. 7 formulation uses Momentum & Force as the basic physical quantities. Newton’s Laws in Momentum & Force Language Before we discuss these, we need to learn the vocabulary of this language.

4. Sect. 7-1: Momentum & It’s Relation to Force Momentum: The momentum of an object is DEFINED as: (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. Momentum: The most general statement of Newton’s 2 nd Law is: (  p/  t) (1) –The total force acting on an object = time rate of change of momentum. (1) is more general than ∑F = ma because it allows for the mass m to change with time also! Example, rocket motion! –Note: if m is constant, (1) becomes: ∑F = (  p/  t) = [  (mv)/  t] = m(  v/  t) = ma

6. Newton’s 2 nd Law (General Form!) ∑F =  p/  t (1) m = constant. Initial p 0 = mv 0. Final momentum p = mv. (1) becomes: ∑F =  p/  t = m(v-v 0 )/  t = m(  v/  t) = ma (as before)

7. Example 7-1: Force of a tennis serve For a top player, a tennis ball may leave the racket on the serve with a speed v 2 = 55 m/s (about 120 mi/h). The ball has mass m = 0.06 kg & is in contact with the racket for a time of about  t = 4 ms (4 x 10 -3 s), Estimate the average force F avg on the ball. Would this force be large enough to lift a 60-kg person?

8. Example 7-2: Washing a Car, Momentum Change & Force Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s & is aimed at the side of a car, which stops it. (That is, we ignore any splashing back.) Calculate the force exerted by the water on the car. Newton’s 2 nd Law: ∑F =  p/  t Initial p = mv, Final p = 0, m = 1.5 kg each  t = 1 s (water instantaneously stops before splashing back)  p = 0 – mv F = (  p/  t) = [(0 -mv)/  t] = - 30 N This is the force the car exerts on the water. By N’s 3 rd Law, the water exerts an equal & opposite force on the car!

9. Conceptual Exercise B (p 169) Water splashes back!

10. Section 7-2: Conservation of Momentum (Collisions!!) An Experimental fact (provable from Newton’s Laws): For 2 colliding objects, (zero external force) the total momentum is conserved (constant) throughout the collision. That is, The total (vector) momentum before the collision = the total (vector) momentum after the collision.  Law of Conservation of Momentum

11. Law of Conservation of Momentum The total (vector) momentum before the collision = the total (vector) momentum after the collision.  p total = p A + p B = (p A )+ (p B ) = constant or:  p total =  p A +  p B = 0 p A = m A v A, p B = m B v B, Initial momenta (p A ) = m A (v A ), (p B ) = m B (v B ), Final momenta or:  m A v A + m B v B = m A (v A ) + m B (v B )

12. Momentum Conservation can be derived from Newton’s Laws. We are mainly interested in analyzing collisions between 2 masses, say m A & m B. We assume that a collision takes a short enough time that external forces can usually be ignored, so that all that matters is the internal forces between the 2 masses during the collision. Since, by Newton’s 3 rd Law the internal forces are equal & opposite The Total Momentum Will Be Constant.

13. If masses m A & m B collide, N’s 2 nd Law (in terms of momentum) holds for each: ∑F A = (  p A /  t) & ∑F B = (  p B /  t). p A & p B, = momenta of m A & m B ∑F A & ∑F B = total forces on m A & m B, including both internal + external forces. Define the total momentum: P = p A + p B and add the N’s 2 nd Law equations: (  P/  t) = (  p A /  t) + (  p B /  t) = ∑F A + ∑F B. By N’s 3 rd Law, internal forces cancel & the right side = ∑F ext = total external force. So (  P/  t) = (  p A /  t) + (  p B /  t) = ∑F ext So, in the special case when the total external force is zero (F ext = 0), this is: (  P/  t) = 0 = (  p A /  t) + (  p B /  t) or  P =  p A +  p B = 0 So, when the total external force is zero, (ΔP/Δt) = 0, or the total momentum of the 2 masses remains constant during the collision! P = p A + p B = constant = (p A ) + (p B ) a or m A v A + m B v B = m A (v A ) + m B (v B ) Momentum Conservation in Collisions A Proof, using Newton’s Laws of Motion.

14. Newton’s 2 nd Law: Force on A, due to B, small  t: F AB =  p A /  t = m A [(v A ) - v A ]/  t Force on B, due to A, small  t: F BA =  p B /  t = m B [(v B ) - v B ]/  t Newton’s 3 rd Law: F AB = - F BA = F  m A [(v A ) - v A ]/  t = - m B [(v B ) - v B ]/  t or: m A v A + m B v B = m A (v A ) + m B (v B )  Proven! So, for Collisions: m A v A + m B v B = m A (v A ) + m B (v B ) Another Proof, using Newton’s 2 nd & 3 rd Laws Two masses, m A & m B in collision: Internal forces: F AB = - F BA by Newton’s 3 rd Law

15. This is known as the Law of Conservation of Linear Momentum “When the total external force on a system of masses is zero, the total momentum of the system remains constant.” Equivalently, “The total momentum of an isolated system remains constant.”

16. Collision: m A v A + m B v B = m A (v A ) + m B (v B ) (1) Another simple Proof, using Newton’s Laws of Motion: Newton’s 2 nd Law for each mass: Force on A, due to B, small  t: F AB = (  p A /  t) = m A [(v A ) - v A ]/  t Force on B, due to A, small  t: F BA =  p B /  t = m B [(v B ) - v B ]/  t Newton’s 3 rd Law: F BA = - F 1AB  m A [(v A ) - v A ]/  t = m B [(v B ) - v B ]/  t or m A v A +m B v B = m A (v A ) + m B (v B )  Proven!

17. The vector sum is constant! Momentum before = Momentum after! Example: 2 pool balls collide (zero external force)

18. Simplest possible example!! Car A, mass m A = 10,000 kg, traveling at speed v A = 24 m/s strikes car B (same mass, m B = 10,000 kg), initially at rest (v B = 0). The cars lock together after the collision. Calculate their speed v immediately after the collision. Ex. 7-3: Railroad Cars Collide: Momentum Conserved Conservation of Momentum in 1dimension Initial Momentum = Final Momentum v A = 0, (v A ) = (v B ) = v so, m A v A +m B v B = (m A + m 2B )v  v = [(m A v A )/(m A + m B )] = 12 m/s

19. Momentum Before = Momentum After m A v A + m B v B = m A (v A ) + m B (v B ) Initially: m A explodes, breaking up into m B & m C. So: 0 = m B v B + m C v C Given 2 masses & 1 velocity, can calulate the other velocity Example: An explosion as a “collision”! A A B B C C

20. Momentum Before = Momentum After Momentum conservation works for a rocket if we consider the rocket & its fuel to be one system, & we account for the mass loss of the rocket (Δm/Δt). Example: Rocket Propulsion

21. Example 7-4: Rifle recoil Calculate the recoil velocity of a rifle, mass m R = 5 kg, that shoots a bullet, mass m B = 0.02 kg, at speed v B = 620 m/s. This gives: 0 = m B (v B ) + m R (v R )  (v R ) = - 2.5 m/s (to the left, of course!) Momentum Before = Momentum After Momentum conservation works here if we consider rifle & bullet as one system m B = 0.02 kg, m R = 5.0 kg (v B ) = 620 m/s Conservation of Momentum m A v A + m B v B = m R (v R ) + m B (v B )

22. Example: An Archer A man (m M = 60 kg, v M = 0) stands on frictionless ice. He shoots an arrow (m A = 0.5 kg, v A = 0) horizontally & it leaves the bow at (v A ) = 50 m/s to the right. What velocity (v M ) does he have as a result? m M = 60 kg, v M = 0, m A = 0.5 kg, v A = 0 (v A ) = 50 m/s, (v M ) = ? The total momentum before the arrow is shot is 0 & momentum is conserved so The total Momentum after the arrow is shot is also 0! p A + p M = 0 = (p A ) + (p M ) or: m A v A + m M v M = m A (v A ) + m M (v M ) m A (0) + m M (0) = m A (v A ) + m M (v M ) 0 = m A (v A ) + m M (v M ) or: (v M ) = - (m A /m M )(v A ) (v M ) = - 0.42 m/s (The minus means that the man slides to the left!)