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Chapter 9: Linear Momentum

Chapter 9: Linear Momentum. 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 &

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Chapter 9: Linear Momentum

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  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:NOTa 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 anydynamics 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. 7formulation usesMomentum & Forceas 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 isDEFINEDas: (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. Momentum: • The most general statement of Newton’s 2nd 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 2nd Law(General Form!) ∑F = p/t (1) m = constant. Initial p0 = mv0. Final momentum p = mv. (1) becomes: ∑F = p/t = m(v-v0)/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 speedv2 = 55 m/s(about 120 mi/h).The ball has massm = 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 forceFavgon the ball. Would this force be large enough to lift a60-kgperson?

  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 2nd 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 3rd 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.  ptotal = pA + pB = (pA)+ (pB) = constant or:ptotal = pA + pB = 0 pA = mAvA, pB = mBvB, Initial momenta (pA)= mA(vA), (pB)= mB(vB), Final momenta or:  mAvA + mBvB = mA(vA) + mB (vB)

  12. Momentum Conservation can be derived from Newton’s Laws. We are mainly interested in analyzing collisions between 2 masses, say mA & mB. 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 3rd Lawthe internal forces are equal & opposite The Total Momentum Will Be Constant.

  13. Momentum Conservation in Collisions A Proof,using Newton’s Laws of Motion. If masses mA & mB collide, N’s 2nd Law(in terms of momentum) holds for each: ∑FA = (pA/t) &∑FB = (pB/t). pA & pB, = momenta of mA & mB∑FA&∑FB = total forces onmA & mB, including both internal + external forces. Define the total momentum: P = pA + pB and add the N’s 2nd Lawequations: (P/t) = (pA/t) + (pB/t) = ∑FA+ ∑FB. By N’s 3rd Law, internal forces cancel & the right side = ∑Fext= total external force. So (P/t) = (pA/t) + (pB/t) = ∑Fext So, in the special case when the total external force is zero (Fext = 0), this is: (P/t) = 0 = (pA/t) + (pB/t) or P = pA + pB = 0 So, whenthe total external force is zero, (ΔP/Δt) = 0, or the total momentum of the 2 masses remains constant during the collision! P = pA + pB = constant = (pA)+ (pB)a or mAvA + mBvB = mA(vA) + mB (vB)

  14. Another Proof, using Newton’s 2nd & 3rd Laws Two masses, mA & mB in collision: Internal forces: FAB = - FBA by Newton’s 3rd Law Newton’s 2nd Law: Force on A, due to B, small t: FAB = pA/t = mA[(vA)- vA]/t Force on B, due to A, small t: FBA = pB/t = mB[(vB)- vB]/ t Newton’s 3rd Law:FAB = - FBA = F  mA[(vA)- vA]/t = - mB[(vB)- vB]/t or: mAvA + mBvB = mA(vA)+ mB (vB) Proven! So,for Collisions:mAvA + mBvB = mA(vA)+ mB (vB)

  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:mAvA+ mBvB = mA (vA)+ mB (vB) (1) • Another simple Proof, using Newton’s Laws of Motion: Newton’s 2nd Law for each mass: Force on A, due to B, small t: FAB = (pA/t) = mA[(vA)- vA]/t Force on B, due to A, small t: FBA = pB/t = mB[(vB)- vB]/ t Newton’s 3rd Law: FBA = - F1AB  mA[(vA)- vA]/t = mB[(vB)- vB]/t or mAvA+mBvB = mA(vA)+ mB (vB) Proven!

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

  18. Ex. 7-3: Railroad Cars Collide: Momentum Conserved Simplest possible example!! Car A, mass mA = 10,000 kg, traveling at speed vA = 24 m/s strikes car B(same mass, mB = 10,000 kg), initially at rest (vB = 0).The cars lock together after the collision. Calculate their speed v immediately after the collision. Conservation of Momentum in 1dimension Initial Momentum = Final Momentum vA = 0, (vA) = (vB) = v so, mAvA+mBvB = (mA + m2B)v  v = [(mAvA)/(mA + mB)] = 12 m/s

  19. Example: An explosion as a “collision”! A Momentum Before = Momentum After mAvA + mBvB = mA(vA)+ mB (vB) Initially: mA explodes, breaking up into mB & mC. So: 0 = mBvB + mCvC Given 2 masses & 1 velocity, can calulate the other velocity A B C C B

  20. Example: Rocket Propulsion 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).

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

  22. A man (mM = 60 kg, vM = 0) stands on frictionless ice. He shoots an arrow (mA = 0.5 kg, vA = 0) horizontally & it leaves the bow at (vA) = 50 m/s to the right. What velocity (vM)does he have as a result? mM = 60 kg, vM = 0, mA = 0.5 kg, vA = 0 (vA) = 50 m/s, (vM) = ? • The total momentum before the arrow is shot is 0 &momentum is conserved so The total Momentumafter the arrow is shot is also0! pA + pM = 0 = (pA)+ (pM) or: mAvA + mMvM = mA(vA) + mM(vM) mA(0) + mM(0) = mA(vA) + mM(vM) 0 = mA(vA) + mM(vM) or:(vM) = - (mA/mM)(vA) (vM) = - 0.42 m/s (The minus means that the man slides to the left!) Example: An Archer

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