Chapter 9: Linear Momentum

NEWTON’S LAWS OF MOTION! THE COURSE THEME: NEWTON’S LAWS OF MOTION! Chs. 4, 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.

In Chs. 4 & 5, 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 Ch. 9 discusses 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.

Sect. 9-1: Momentum & It’s Relation to Force Momentum: The momentum of an object is DEFINED as: (a vector || ) 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! Figure 8-1. Caption: Object of mass m: (a) falls a height h vertically; (b) is raised along an arbitrary two-dimensional path.

∑F = (dp/dt) = d(mv)/dt = m(dv/dt) = ma Momentum: The Most General Statement of Newton’s 2nd Law The total force acting on a mass = the time rate of change of the momentum. (The total force on a mass = the time derivative of the 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 = (dp/dt) = d(mv)/dt = m(dv/dt) = ma Actually, (1) was the form of N’s 2nd Law first written & used by Newton himself! (1)

∑F = (dp/dt) = m(dv/dt) = ma Newton’s 2nd Law (The Most General Form!) For m = constant, (1) becomes (as before): ∑F = (dp/dt) = m(dv/dt) = ma (1)

Example 9-1: Force of a Tennis Serve For a top player, a tennis ball may leave the racket l on the serve with a speed v2 = 55 m/s (~ 120 mi/h!). L The ball has mass m = 0.06 kg & is in contact with ll the racket for a time of about t = 4 ms (4  10-3 s) a. Estimate the average force Favg on the ball. b. Would this force be large enough to lift a 60-kg person? Use Newton’s 2nd Law: ∑F = p/t Ball’s initial momentum: p = 0. Ball’s final momentum: p = mv Average force: ∑F ≈ Fave = p/t Fave = p/t = (mv – 0)/t = [(0.06)(55) – 0]/(4  10-3 s) Fave = 800 N To lift a 60 kg person requires only F = mg = (60)(9.8) = 588 N !! Figure 9-1. Solution: The average force is the change in momentum divided by the time: 800 N. A 60-kg person weighs about 600 N, so this would be large enough.

Example 9-2: 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. (Ignore splashing back.) Calculate the average force exerted by the water on the car. Use Newton’s 2nd Law: ∑F = p/t Initial momentum: p = mv Final momentum p = 0 m = 1.5 kg each t = 1 s (water 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 Newton’s 3rd Law, the water exerts an equal & opposite force on the car! Figure 9-2. Solution: Every second, 1.5 kg of water hits the car and stops. The force is the change in momentum (final momentum is zero) divided by the time, or 30 N.

Exercise B (p 216) Water splashes back!

Section 9-2: Conservation of Momentum (Collisions) Experimental fact (provable from Newton’s Laws): For 2 colliding objects, (zero external force) the total momentum is conserved (constant) throughout the collision. The total (vector) momentum before the collision = the total (vector) momentum after the collision.  the Law of Conservation of Momentum

The Total Momentum Will Be Constant. Momentum Conservation can be derived from Newton’s Laws We are mainly interested in analyzing collisions between 2 masses, mA & mB. 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 Law, the internal forces are equal & opposite, The Total Momentum Will Be Constant. Figure 9-4. Caption: Collision of two objects. Their momenta before collision are pA and pB, and after collision are pA’ and pB’. At any moment during the collision each exerts a force on the other of equal magnitude but opposite direction.

For 2 masses in collision, we have, for the total momentum P (pi = momentum of mass i & Fi = force on mass i, i = A or B): The internal forces cancel, so (Fext = total external force) In the special case when the total external force is zero (Fext = 0), this is: So, when the total external force is zero, (dP/dt) = 0, or the total momentum of the 2 masses remains constant during the collision!! P = pA + pB = constant

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

 mAvA + mBvB = mA(vA) + mB (vB) Law of Conservation of Momentum (collisions): The total (vector) momentum before a collision = the total (vector) momentum after a 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  mAvA + mBvB = mA(vA) + mB (vB)

The vector sum of the momenta is a constant! Example: 2 billiard balls collide (zero external force) The vector sum of the momenta is a constant! Vector sum of momenta before = Vector sum of momenta after! Vector sum of momenta = constant!

Another brief 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: The force on A, due to B, for a small time t: FAB = pA/t = mA[(vA) - vA]/t The force on B, due to A, for the same 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)

 v = [(mAvA)/(mA + mB)] = 12 m/s Ex. 9-3: Railroad Cars Collide: Conservation of Momentum Simplest possible example!! Car A, mass mA = 10,000 kg, is 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 Before Collision Figure 9-5. Solution: Momentum is conserved; after the collision the cars have the same momentum. Therefore their common speed is 12.0 m/s. After Collision vA = 0, (vA) = (vB) = v mAvA+mBvB = (mA + m2B)v  v = [(mAvA)/(mA + mB)] = 12 m/s

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

Example: Rocket Propulsion Momentum Before Take Off = Momentum After Take Off 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 (dm/dt). Figure 9-6. Caption: (a) A rocket, containing fuel, at rest in some reference frame. (b) In the same reference frame, the rocket fires, and gases are expelled at high speed out the rear. The total vector momentum, P = pgas + procket, remains zero.

Example 9-4: 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 Shooting = Momentum After Shooting 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) Figure 9-7. Solution: Use conservation of momentum. The recoil velocity of the gun is -2.5 m/s. This gives: 0 = mB (vB)  + mR(vR)  (vR) = - 2.5 m/s (to the left, of course!)

Conceptual Example 9-5: Falling on or off a sled a. An empty sled is sliding on frictionless ice when l Susan drops vertically from a tree above onto the l sled. When she lands, does the sled speed up, l slow down, or keep the same speed? b. Later: Susan falls sideways off the sled. When l she drops off, does the sled speed up, slow l down, or keep the same speed? The sled will slow down. The sled keeps going at the same speed (as does Susan).

Another Treatment of Collisions 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 = 0 (1) Newton’s 2nd Law: (if no other forces act) F21 = m1a1 (2). F12 = m2a2 (3) Put (2) & (3) into (1)  m1a1 + m2a2 = 0 (4) Note v’s & F’s are vectors!!

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 = 0 (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 = 0 (5) The time derivative of m1v1 + m2v2 is = 0.  Calculus tells us that m1v1 + m2v2 = constant! (6) A Vector Equation!

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”!!

If m doesn’t depend on time: ∑F = [d(mv)/dt] Now consider one of the masses m & write: Newton’s 2nd Law: ∑F = ma Use the definition of the acceleration as the 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) W We did this for constant m. It can be shown that it is more general than this & that it is valid even if m changes with time.

∑F = (dp/dt) = d(mv)/dt = m(dv/dt) = ma Consider again the general statement of Newton’s 2nd Law: The total (net) force acting on a mass = the time rate of change in the mass’s momentum (the time derivative of the momentum). As we’ve said, (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 = (dp/dt) = d(mv)/dt = m(dv/dt) = ma (1)

ptot = p1 + p2 = (p1) & (p2) = constant Back to 2 moving masses interacting & isolated from the rest of the world. We found: d(p1 + p2)/dt = 0 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 is conserved! If, at some initial time, the 2 momenta are p1 & p2 & if at some final time they are (p1) & (p2), we can write: ptot = p1 + p2 = (p1) & (p2) = constant

Example: Archer An archer, m1 = 60 kg, v1 = 0, is on frictionless ice. He puts an arrow into his bow. m2 = 0.5 kg, v2 = 0. He shoots it horizontally to the right at (v2) = 50 m/s. What velocity (v1) does the have as a result? Momentum Conservation There are no external forces in the x-direction, so momentum is conserved in the x-direction The total momentum before shooting is =  The Total Momentum After Shooting is 0! Before Shooting the Arrow ptot = m1v1 + m2v2 = m1(0) + m2(0) = 0 Momentum is conserved! ptot = p1 + p2 = (p1) + (p1) = constant  After Shooting the Arrow m1(v1) + m2(v2) = 0 or, (v1) = - (m2/m1)(v2) (v1) = - 0.42 m/s (The minus means the archer slides to left!)