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PHYSICS 149: Lecture 18 • Chapter 7: Linear Momentum – 7.2 Momentum – 7.3 The Impulse-Momentum Theorem – 7.4 Conservation of Momentum Lecture 18 Purdue University, Physics 149 1 ILQ 1 Two objects are known to have the same kinetic energy. Do these two objects necessarily have the same momentum? A) Yes B) No Lecture 18 The example of the ice skaters starting from rest and pushing off of each other...the total momentum is zero because they travel in opposite directions but the kinetic energy is not zero (they start with none and gain a bunch) Purdue University, Physics 149 2 ILQ 2 An astronaut is taking a space walk when his tether breaks. breaks He has a big drill drill. How can he get back to the spaceship? A) Face away from the ship and throw the drill forward. B) Face the ship and throw the drill forward. C) Throw the drill towards the tether. Lecture 18 Purdue University, Physics 149 3 Momentum • Momentum is a vector quantity. quantity • Units: kg⋅m/s, N⋅s – Unit conversion: 1 kg⋅m/s = 1 N⋅s • Momentum is denoted by p. p • Momentum is sometimes referred to as “linear momentum” t ” to t distinguish di ti i h it ffrom angular l momentum. t Lecture 18 Purdue University, Physics 149 4 Momentum is Conserved • Momentum is “Conserved” meaning it can not be created nor destroyed – Can be transferred • Total Momentum does not change with time • Momentum is a VECTOR 3 Conservation Laws in one! Lecture 18 Purdue University, Physics 149 5 Momentum is Conserved • Define Momentum p = m v – – – – m1 Δv1 = -m2 Δv2 Δp1 = -Δp2 Δp1 + Δp2 = 0 Σpinitial = Σpfinal Momentum is conserved p1f-p1i+p2f-p2i=0 p1f+p2f=p1i+p2i • Example: Jane pushes Fred so he is going 2.0 m/s. If Fred is twice as heavy as Jane, Jane how fast does Jane end up moving? – Σpinitial = Σpfinal – 0 = mFred vFred + mJane vJane – vJane = -mFred vFred / mJane = -4 m/s Lecture 18 Purdue University, Physics 149 6 Momentum Transfer • During an interaction, momentum is transferred from one bodyy to another,, but the total momentum of the two is unchanged. G G G G G G Δp2 = −Δp1 or p1i + p2i = p1 f + p2 f • Consider a collision between two spaceships. G v1i G v1 f G v2 i G G G G G G F Δv1 = a1Δt = 12 Δt ⇒ Δp1 = m1Δv1 = F12 Δt m1 G G F21 = − F12 Lecture 18 G v2 f G G G G G G F21 Δv2 = a2 Δt = Δt ⇒ Δp2 = m2 Δv2 = F21Δt m2 G G G G G G ⇒ Δp2 = m2 Δv2 = F21Δt = (− F12 )Δt = −m1Δv1 = −Δp1 G G Thus, Δp2 = −Δp1 Purdue University, Physics 149 7 Impulse • Impulse is the product of (average) force and time interval during which the force acts. G G I = FΔt • The change g in momentum of an object j is equal q to the impulse (recall a collision between two spaceships). G G G Δp = I (= FΔt ) • Impulse is a vector quantity. quantity • Units: N⋅s, kg⋅m/s – Unit conversion: 1 N⋅s = 1 kg⋅m/s • Impulse is often denoted by II. Lecture 18 Purdue University, Physics 149 8 Impulse Changes Momentum • Define Impulse I = FΔt – Change in momentum requires Force acting over a time or Impulse – Recall: mΔv = FΔt – Or Δp = FΔt Δp = I Impulse I = FΔt Momentum p = mv Lecture 18 Purdue University, Physics 149 9 ILQ By what factor does an object’s kinetic energy change if its speed is doubled? By what factor does the momentum change? A)) 4,, 2 B) 2, 2 C) 0.5, 4 G G p = mv Lecture 18 1 2 K = mv 2 Purdue University, Physics 149 10 Impulse and Momentum I ≡ FaveΔt = pf - pi = Δp • For single object … – F = 0 ⇒ momentum conserved (Δp = 0) • For collection of objects … – ptotal = Σp – Internal forces: forces between objects in system – External forces: all other forces – ΣFext = 0 ⇒ total momentum conserved ((Δp ptot = 0)) – Fext = mtotal a Lecture 18 Purdue University, Physics 149 11 Conservation of Momentum • Momentum is a vector quantity, so both the magnitude and the direction of the total momentum at the beginning and end of the interaction must be the same. • The total momentum of a system y is the vector sum of the momenta of each object in the system. Internal interactions do not change the total momentum of a system. External interactions can change the total momentum of a system. • • • During a collision, the interaction often takes place during a sufficiently ffi i l short h amount off time i that h externall fforces may b be ignored; conservation of momentum may be used. Lecture 18 Purdue University, Physics 149 12 ILQ Two identical balls are dropped from the same height onto the floor. In each case they have velocity v downward just before hitting the floor. In case 1 the ball bounces back up, and in case 2 the ball sticks to the fl floor without ih b bouncing. i IIn which hi h case iis the h magnitude i d off the h iimpulse l given to the ball by the floor the biggest? A) Case 1 Δ mΔ = maΔt Δp=mΔv m Δt = FΔt = I B) Case 2 C) The same Lecture 18 Bouncing Ball Sticky Ball |I| = |Δp| |I| = |Δp| = |mv | final – m vinitial| = |mv | final – m vinitial| = |m( vfinal - vinitial)| = |m(0 - vinitial)| =2mv = mv Purdue University, Physics 149 13 ILQ Two identical balls are dropped from the same height onto the floor. In case 1 the ball bounces back up, and in case 2 the ball sticks to the floor without bouncing. bouncing In both cases of the above question, the direction of the impulse given to the ball by the floor is the same. What is this direction? A) Upward B) Downward G G G G I = Δp = mv2 − mv1 time Lecture 18 Purdue University, Physics 149 14 Momentum is a Vector A car with mass 1200 kg is driving north at 40 m/s, and turns east driving 30 m/s. What is the magnitude of the car’s change in momentum? A) 0 B) 12,000 C) 36,000 D) 48,000 E) 60,000 pinitial = m vinitial = (1200 kg) x 40 m/s = 48,000 kg m/s North pfinal = m vfinal = (1200 kg) x 30 m/s = 36,000 kg m/s East North-South: Pfinal – Pinitial = (0 – 48000) = -48,000 kg m/s East-West: Pfinal – Pinitial = (36000 - 0) = +36,000 kg m/s 30 m/s 40 m/s Magnitude: sqrt(P2North + P2East ) = 60,000 kg m/s Lecture 18 Purdue University, Physics 149 15 ILQ You drop an egg onto A) the floor B) a thick piece of foam rubber. In both cases, the egg does not bounce. In which case is the impulse greater? A) Floor B) Foam I = Δp C) the same Same change in momentum In which case is the average force greater A) Floor B)) Foam C) the same Lecture 18 Δp = F Δt F = Δp/Δt smaller Δt = large F Purdue University, Physics 149 16 Example: Time to Burn For a safe re-entry into the Earth's atmosphere the pilots of a space capsule must reduce their speed from 2.6 × 104 m/s to 1.1 × 104 m/s. The rocket engine g p produces a backward force on the capsule p of 1.8 × 105 N. The mass of the capsule is 3800 kg. For how long must they fire their engine? [Hint: Ignore the change in mass of the capsule due to the expulsion of exhaust gases.] Lecture 18 Purdue University, Physics 149 17 Example: Time to Burn vi=2.6×104 m/s vf=1.1×104 m/s F=1.8×105 N M=3800 M 3800 Kg Δt=? FΔ t = m Δ v m Δv m ( v f − v i ) Δt = = = F F 3800kg ( 2.6 − 1.1) × 104 m / s 1N × = 317 s = 5 2 1.8 × 10 N 1 Kgm / s Lecture 18 Purdue University, Physics 149 18 Example • A fright train is being assembled in switching yard. Car 1 has a mass of m1=65 ×103 kg and moves at a velocity v01=0.80 0 80 m/s. / Car C 2h has a mass off m2=92 92 ×10 103 kg k and d moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars after ft they th become b coupled. l d m2 m2 Lecture 18 v02 m1 m1 v01 vf Purdue University, Physics 149 initial final 19 Example • Note the velocity of car 1 increases while the velocity of car 2 decreases. • The Th acceleration l ti and dd deceleration l ti arise i b because th the cars exert internal forces on each other. • Momentum conservation allows us to determine the change in velocity without knowing what the internal forces are. • Note that often momentum is conserved but the kinetic energy is not! (Kinetic energy is conserved only in elastic collisions) m2 v02 m2 Lecture 18 m1 m1 v01 vf Purdue University, Physics 149 initial f l final 20 Example • A fright train is being assembled in switching yard. Car 1 has a mass of m1=65 ×103 kg and moves at a velocity v01=0.80 0 80 m/s. / Car C 2h has a mass off m2=92 92 ×10 103 kg k and d moves at a velocity v02=1.3 m/s and couples to it. Neglecting friction, find the common velocity vf of the cars after ft they th become b coupled. l d Apply conservation of momentum: Pi=m m1vo1+m m2v02 Pf=(m (m1+m m2)vf (m1+m2)vf = m1v01+m2v02 m1v01 + m2 v02 vf = = m1 + m2 (65 ×103 kg )(0.8 )(0 8m / s ) + (92 ×103 kg )(1.3 )(1 3m / s ) = 1.1m / s 3 3 (65 ×10 kg + 92 ×10 kg ) Lecture 18 Purdue University, Physics 149 21 ILQ At the instant a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta momenta. Which object has the greater kinetic energy? Or are they the same A) The same Apply conservation of B) The gun momentum: C) The bullet Pi=0 0 Pf=M MgunVgun+m mbulletvbullet Vgun=-(mbullet/Mgun)vbullet K gun ⎛ mbullet 1 1 2 = M gunVgun = M gun ⎜ ⎜M 2 2 ⎝ gun Lecture 18 2 ⎞ 2 mbullet K bullet ⎟⎟ vbullet = M gun ⎠ Purdue University, Physics 149 22 Impulse-Momentum Theorem • The “total” impulse on an object is equal to the change in the object’s momentum during the same time interval. G = Itotal • Example: – Today’s automobiles are designed with a number of safety features, all designed to make the impact last longer (Δt) and so applied force weaker (F). That way we can minimize injury upon collision. collision pi = mvi, pf = mvf = 0 Æ |Δp| = |pf – pi| = |–mvi| = |F|Δt The longer Δt, Δt the weaker |F| |F|. Lecture 18 Purdue University, Physics 149 23 Newton’s Second Law • The net force is the rate of change of momentum momentum. • This relation is valid even when mass is not constant (just like a rocket engine). g ) Lecture 18 Purdue University, Physics 149 24 Newton’s Second Law G G ∑ F = ma G G I = ∑ F Δt = Δp G G Δp ∑ F = Δt Lecture 18 New formulation of N2L Purdue University, Physics 149 25 ILQ • A 3.0 kg object is initially at rest. It then receives an impulse of magnitude 15 N⋅s N⋅s. After the impulse, the object has A) B) C) D) a momentum of magnitude 5.0 kg⋅m/s. a speed of 45 45.0 0 m/s. m/s a momentum of magnitude 15.0 kg⋅m/s. a speed of 7 7.5 5 m/s m/s. Lecture 18 Purdue University, Physics 149 26 ILQ • An 18-wheeler and a Volkswagen Beetle are rolling along with the same momentum momentum. If you exert the same force with the brakes to stop each one,, which takes a longer g time to bring g to rest? a) b)) c) d) 18-wheeler Volkswagen g beetle same for both impossible to say Lecture 18 Purdue University, Physics 149 27 Example • A pole-vaulter of mass 60.0 kg vaults to a height of 6.0 m before dropping to thick padding placed below to cushion her fall. (a) Find the speed with which she lands. – Gravity (conservative force) is the only force acting on her before g, so we mayy use the conservation of mechanical energy gy landing, (recall Chapter 6). – 0 + mgh = ½ mv2 + 0 Æ v = sqrt(2gh) = sqrt(2 ⋅ 9.8m/s2 ⋅ 6.0m) = 10.8 m/s (b) If the padding brings her to a stop in a time of 0.50 s, what is the average force on her body due to the padding d i that during h time i iinterval? l? – Use the impulse-momentum theorem – Δp = pf – pi = mvf – mvi = 60.0 kg [0 – (–10.8 m/s)] = 648 kg⋅m/s, upward F = Δp/Δt = 648 kg⋅m/s / 0.50 s = 1,296 N, upward Lecture 18 Purdue University, Physics 149 28