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Chapter 8 Momentum, Impulse and Collisions 8.1 Momentum and Impulse In the previous two chapters we have reformulated the Newton’s second law in terms of conservation of energies (kinetic, potential, internal). We can also express it as ! ⃗ i = d⃗p (8.1) F dt i where ⃗p is a new physical quantity known as momentum. In this course we define it as ⃗p ≡ m⃗v. (8.2) Note, however, that in context of the so-called Hamiltonian mechanics (which is the starting point in both quantum mechanics and statistical mechanics) it is defined as " # ∂E x, dx dt ⃗p = . (8.3) ) ∂( dx dt For example if then % $ %2 $ 1 d⃗x 1 d⃗x = m + kx2 E ⃗x, dt 2 dt 2 ⃗p = " x# ∂E ⃗x, d⃗ dt x ∂( d⃗ ) dt which is in agreement with (8.2). 96 =m d⃗x dt (8.4) (8.5) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 97 The units of momentum are given by [Momentum] = [Mass] × [V elocity] (8.6) and for example in SI units 1 N · 1 s = 1 kg · m/s2 . (8.7) Alternatively one can write units of momentum as [Momentum] = [Mass] × [V elocity] = [Mass] × [Acceleration] × [T ime] = [F orce] × [T ime] (8.8) where the latter form suggests that when some force is applied to a system for some time then, it may effect the momentum. In fact if we define impulse (of the time-independent net force) as ! ⃗J = ⃗ i ∆t F (8.9) i then from (8.1) we get ⃗J = ⃗pf − ⃗pi . When the net force is time dependent the impulse is defined as & tf ! ⃗J = ⃗ i dt F ti (8.10) (8.11) i and from (8.1) we get ⃗J = & tf ti ! i ⃗ i dt = F & tf ti d⃗p dt = ⃗pf − ⃗pi dt (8.12) which is the same as equation (8.10). Note also that if the average force is defined as ' tf ( ⃗ i Fi dt ⃗ avg ≡ ti (8.13) F tf − ti then the impulse is just similarly to (8.9). ⃗J = F ⃗ avg (tf − ti ) (8.14) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 98 Similarly to the energy conservation which is fundamentally due to timeshift symmetry of physics laws, the momentum conservation is due to spaceshift symmetry. For this reason the conservation of energy expresses changes caused by force in time ! ⃗J = ⃗ i dt = ⃗pf − ⃗pi F (8.15) i and the conservation of momentum expresses the changes caused by forces in space ! ⃗ i · d⃗l = Kf − Ki . W = F (8.16) i The fact that the two expressions look so much alike might be surprising at first but this is what lead people to eventually discover a more fundamental and unified conserved quantity known as energy-momentum tensor as well as other conserved quantities such as electric charge. In fact discovery of new symmetries and developing theories based on these symmetries is what physicists did for the most part of the 20th century. Example 8.1 . We can now go back to the example 6.5 where we considered a race of two iceboats on a frictionless frozen lake. The boats have ⃗ masses m and 2m, and the wind exerts the same constant horizontal force F on each boat. The boats start from rest and cross the finish like a distance s away. Which boat crosses the final line with greater momentum. Although it is true that the final kinetic energy of both boats will be the CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 99 same, K1 = K2 1 1 (m) v12 = (2m) v22 2 2 and the final velocities where not the same v1 √ = 2. v2 (8.17) (8.18) and thus momenta are related by v1 √ p1 = = 2. p2 v2 (8.19) This is due to the fact that the same forces were acting for different periods of time. Using the impulse-momentum theorem we can conclude that F ∆t1 = mv1 F ∆t2 = mv2 (8.20) and thus ∆t1 v1 √ = = 2. (8.21) ∆t2 v2 Example 8.2. You throw a ball with a mass of 0.40 kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20 m/s. (a) Find the impulse of the net force on the ball during its collision with the wall. (b) If the ball is in contact with the wall for 0.010 s, find the average horizontal force that the wall exerts on the ball during impact. The momentum-energy theorem implies ⃗J = ⃗pf − ⃗pi . (8.22) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 100 Assuming that the x-axis is chosen in the direction of motion of the ball before collision we get ⃗J = m (⃗vf − ⃗vi ) ) * ) * = (0.40 kg) −30 m/sî − 20 m/sî = (−20 N·s) î. (8.23) Now if the ball is in contact with the wall for ∆t = 0.010 s then ⃗ ⃗ avg = J = (−20 N·s) î = 2000 Nî F ∆t 0.010 s 8.2 (8.24) Conservation of Momentum When a given system is analyzed it often useful to distinguish two types of forces: • internal forces (forces exerted on each other by objects inside the system) • external forces (forces exerted on the system by objects outside of the system) When there are no external forces the systems is said to be closed or isolated. For isolated systems one can write down the impulse-momentum theorem for each object separately ⃗ A on B = d⃗pB F dt d⃗ ⃗ B on A = pA F dt (8.25) but because of the Newton’s third law ⃗ A on B = −F ⃗ B on A F (8.26) we get d⃗pA d⃗pB =− . dt dt If we now define the total momentum of all particles as ⃗ = ⃗pA + ⃗pB P (8.27) (8.28) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 101 then we get the law of conservation of total momentum ⃗ dP = 0. dt This is true for an arbitrary collection of particles, i.e. ⃗ = ⃗pA + ⃗pB + ⃗pC + ... = mA ⃗vA + mB ⃗vB + mC ⃗vC + ... P (8.29) (8.30) given that there are no external (but only internal) forces acting on these particles. Note that since (8.29) is a vector equation it must be satisfied along all three components dPx = d(mA VAx +mB VdtBx +mC VCx +...) = 0 dt dPy d(mA VAy +mB VBy +mC VCy +...) = =0 dt dt dPz = d(mA VAz +mB VdtBz +mC VCz +...) = 0 (8.31) dt and for arbitrary time interval Pix = Pf x Piy = Pf y Piz = Pf z ⇒ ⇒ ⇒ mA VAix + mB VBix + mC VCix + ... = mA VAf x + mB VBf x + mC VCf x + ... mA VAiy + mB VBiy + mC VCiy + ... = mA VAf y + mB VBf y + mC VCf y + ... mA VAiz + mB VBiz + mC VCiz + ... = mA VAf z + mB VBf z + mC VCf z(8.32) + ... Example 8.4. A marksman holds a rifle of mass mR = 3.00 kg loosely, so it can recoil freely. He fires a bullet of mass mB = 5.00 g horizontally with a velocity relative to the ground of vBx = 300 m/s. What is the recoil velocity vRx of the rifle? What are the final momentum and kinetic energy of the bullet and rifle? CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 102 Conservation of momentum implies mB ⃗vBi + mR ⃗vRi = mB ⃗vBf + mR ⃗vRf 0 = mB vBx + mR vRx mB vRx = − vBx mR 5.00 × 10−3 kg (300 m/s) vRx = − 3.00 kg vRx = −0.500 m/s. (8.33) So the final momenta and kinetic energies are ⃗pR = (3.00 kg) (−0.500 m/s) î = −1.50 kg · m/s î " # ⃗pB = 5.00 × 10−3 kg (300 m/s) î = 1.50 kg · m/s 1 KR = (3.00 kg) (−0.500 m/s)2 = 0.375 J 2 # 1" 5.00 × 10−3 kg (300 m/s)2 = 225 J. KB = 2 (8.34) Example 8.6. Consider two batting robots on a frictionless surface. Robot A, with mass 20 kg, initially moves at 2.0 m/s parallel to the x-axis. It collides with robot B, which has mass 12 kg and is initially at rest. After the collision, robot A moves at 1.0 m/s in a direction that makes an angle α = 30◦ with its initial direction. What is the final velocity of robot B. CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 103 Conservation of momentum implies mA ⃗vA1 = mA ⃗vA2 + mB ⃗vB2 (8.35) or along x-axis mA vA1 = mA vA2 cos(α) + mB vB2x mA (vA1 − vA2 cos(α)) vB2x = mB (20 kg) vB2x = (2.0 m/s − 1.0 m/s cos(30◦ )) = 1.89 m/s (12 kg) (8.36) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 104 and along y-axis 0 = −mA vA2 sin(α) + mB vB2y mA vB2y = − vA2 sin(α) mB (20 kg) (1.0 m/s) sin(30◦) = −0.83 m/s vB2y = − (12 kg) (8.37) and so ⃗vB2 = 1.89 m/s î − 0.83 m/s ĵ. 8.3 (8.38) Inelastic Collisions In this section we will discuss instantaneous events - we call collisions which suddenly change the kinetic energies of objects. • If forces between colliding objects are conservative, then the total kinetic energy right before and right after collision is the same and the the collision is called elastic. ∆K = 0 (8.39) • If the forces are non-conservative, then the total kinetic energy is not conserved and the collision is called inelastic (or completely inelastic is objects stick together). ∆K ̸= 0 (8.40) The key point is that although the (kinetic) energy might not be conserved, the momentum is still conserved for both elastic and inelastic collisions: ⃗ =0 ∆P (8.41) Consider a completely inelastic two-bodies collision, i.e. ⃗vA2 = ⃗vB2 = ⃗v2 (8.42) then conservation of momentum implies mA ⃗vA1 + mB ⃗vB1 = (mA + mB ) ⃗v2 or ⃗v2 = mA ⃗vA1 + mB ⃗vB1 . (mA + mB ) (8.43) (8.44) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 105 For example if object B was originally at rest, then v2 = mA vA1 (mA + mB ) (8.45) and then it is easy to show that after the (completely inelastic) collision the total kinetic energies are 1 2 mA vA1 2 $ %2 1 1 mA 1 m2A 1 2 2 mA v2 + mB v2 = (mA + mB ) vA1 = (8.46) v2 = 2 2 2 (mA + mB ) 2 (mA + mB ) A1 K1 = K2 and so K2 mA = <1 K1 (mA + mB ) (8.47) or the total kinetic energy after the collision is lower than before collision K 2 < K1 . (8.48) Example 8.8. Consider a ballistic pendulum, a simple system of measuring speed of a bullet. CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 106 A bullet of mass mB makes a completely inelastic collision with a block of wood of mass mW , which is suspended like a pendulum. After impact, the block swings up to a maximum height y. In terms of y, mB , and mW , what is the initial speed v1 of the bullet. During the inelastic collision of the bullet and the block of wood the momentum is conserved and so mB v1 = (mB + mW )v2 ⇒ v1 = mB v2 mB + mW (8.49) During the swing the the total energy is conserved and so 1 (mB + mW ) v22 = (mB + mW ) gy 2 ⇒ v2 = + 2gy (8.50) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS By combining the two equations we get √ mB 2gy . v1 = mB + mW 107 (8.51) Example 8.9. A 1000 − kg car traveling north at 15 m/s collides with a 2000 − kg truck traveling east at 10 m/s. The occupants, wearing seat belts, are uninjured, but the two vehicles move away from the impact point as one. The insurance adjustor asks you to find the velocity of the wreckage just after the impact. What is your answer? Conservation of momentum implies ) * mC vC î + mT vT ĵ = (mC + mT ) v cos θ î + v sin θ ĵ (8.52) and thus we have two equations mC vC = (mC + mT ) v cos θ mT vT = (mC + mT ) v sin θ with two unknowns v and θ. With solutions mT vT tan θ = mC VC % $ 2000 kg · 15 m/s ≈ 37◦ θ = arctan 1000 kg · 10 m/s (8.53) (8.54) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 108 and mT vT sin θ (mC + mT ) 2000 kg · 15 m/s = 8.3 m/s. v = sin (37◦ ) (1000 kg + 2000 kg) v = 8.4 (8.55) Elastic Collisions In elastic collisions the total kinetic energy before and after collisions is unchanged and so one can use both conservation of energy and momenta ∆K = 0 ⃗ = 0 ∆P (8.56) Example 8.12. Consider an elastic collision of two pucks (masses mA = 0.500 kg and mB = 0.300 kg) on a frictionless air-hockey table. Puck A has an initial velocity of 4.00 m/s in the positive x-direction and a final velocity of 2.00 m/s in an unknown direction α. Puck B is initially at rest. Find the final speed vB2 of puck B and the angles α and β. From conservation of energy or 2 2 2 mA vA1 mA vA2 mB vB2 = + 2 2 2 , 2 2 mA (vA1 − vA2 ) vB2 = mB (8.57) (8.58) CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 109 and from conservation of momentum mA ⃗vA1 = mA ⃗vA2 + mB ⃗vB2 (8.59) mA vA1 = mA vA2 cos α + mB vB2 cos β 0 = mA vA2 sin α − mB vB2 sin β. (8.60) or along each axis From (8.60) we get mA (vA1 − vA2 cos α) mB mA vA2 sin α vB2 sin β = mB vB2 cos β = and by combining with (8.58) we get 2 2 2 vB2 = vB2 cos2 β + vB2 sin2 β $ %2 2 2 # mA (vA1 − vA2 ) mA " 2 = (vA1 − vA2 cos α)2 + vA2 sin2 α mB mB % $ # " 2 # mB " 2 2 2 2 vA1 − vA2 = vA1 − 2vA1 vA2 cos α + vA2 cos2 α + vA2 sin2 α m $ A% # mB " 2 2 2 2 vA1 − vA2 = vA1 − 2vA1 vA2 cos α + vA2 mA ) * mB 2 2 2 2 (vA1 − vA2 ) + (vA1 + vA2 ) mA cos α = = 36.9◦ 2vA1 vA2 and mA (vA1 − vA2 cos α) m v $ B B2 $ ! mB " 2 2 %% 2 +v 2 v −vA2 )+(vA1 A2 ) mA ( A1 mA vA1 − vA2 2vA1 vA2 cos β = 2 −v 2 mA (vA1 A2 ) mB mB , 2 2 mA − mB vA1 − vA2 cos β = = 26.6◦ . 2vA1 mB mA cos β = CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 8.5 110 Center of Mass The conservation of momentum can be restated in terms of center of mass defined as the mass-weighted average position of particle ( m1⃗r1 + m2⃗r2 + m3⃗r3 + ... mi⃗ri ⃗rcm ≡ . = (i m1 + m2 + m3 + ... i mi Then the total momentum can be written as ⃗ = m1⃗v1 + m2⃗v2 + m3⃗v3 + ... = P d = (m1⃗r1 + m2⃗r2 + m3⃗r3 + ...) = dt $ % d m1⃗r1 + m2⃗r2 + m3⃗r3 + ... = (m1 + m2 + m3 + ...) = dt m1 + m2 + m3 + ... d⃗rcm = (m1 + m2 + m3 + ...) dt or simply ⃗ = M⃗vcm P where M= ! mi i is the total mass and ⃗vcm = d⃗rcm dt is the velocity of the center of mass. Example 8.14. James (mJ = 90.0 kg) and Ramon (mR = 60.0 kg) are 20 m apart on a frozen pond. Midway between them is a mug of their favorite beverage. They pull on the ends of a light rope stretched between them. When James has moved 6.0 m towards the mug, how far and in what direction has Ramon moved? CHAPTER 8. MOMENTUM, IMPULSE AND COLLISIONS 111 From the conservation of the total momentum Pix = Pf x 0 = M⃗vcm ricm = rf cm If we set the origin to be in the location of the mug, then xiJ = −10 m xiR = 10 m xf j = −10 m + 6 m = −4 m and so the initial (and final) location of the center of mass is xf J mJ + xf R mR xiJ mJ + xiR mR = mJ + mR mJ + mR or xf R = xiJ mJ + xiR mR − xf J mJ (−10 m) (90.0 kg) + (10 m) (60.0 kg) − (−4 m) (90.0 kg) = = 1.0 m mR (60.0 kg) If there are external forces acting on the objects, then the total momentum changes as ! ⃗ ⃗ ext = dP = d (M⃗vcm ) = M⃗acm . F dt dt Example. Will the center of mass continue on the same parabolic trajectory even after one of the fragments hits the ground? Why or why not?