PHYSICS 220 Lecture 11 11 Momentum and Impulse Textbook Sections 7.1 – 7.3 Lecture 11 Purdue University, Physics 220 1 Overview • Last Lecture – Conservative Forces • Work is independent of path • Define Potential Energy PE PE =mgy– gravity = m g y –PE spring = ½ k x 2 – Work – Energy Theorem ∑ – Power: P = W / Δt (Watt) Td nc WKU=Δ+Δ • Today – Impulse – Momentum – Linear Momentum Lecture 11 Purdue University, Physics 220 2 Key Ideas • Last Time: Work-Energy – Σ F = m a multiply both sides by Δx – Σ F Δx = m a Δx (note: a Δx = ½ Δv 2 ) – Σ F Δx = ½ m Δv 2 W K Dfi W k dKi ti E– Σ = Δ Define or an Kinetic nergy • This Time: Impulse-Momentum – Σ F = m a multiply both sides by Δt – Σ F Δt = m a Δt (note: a Δt = Δv) – Σ F Δt = m Δv – Σ I = Δp Define Impulse and Momentum Lecture 11 Purdue University, Physics 220 3 Impulse and Momentum • Define Momentum – Momentum is a VECTOR pmv= G G – The direction of the momentum is the same as the velocity Units are kg m / s– are • Define Impulse – Change in momentum requires Force acting over a period of time or Impulse – Δp=FΔtorI=Δpp = F t or I = Impulse: F Δt Vector Momentum transfer Lecture 11 Purdue University, Physics 220 4 Work: F r Δr Scalar Energy transfer Force and Momentum • If no force is acting on a particle, its velocity is constant – Newton’s First Law – Therefore, its momentum is also constant • If a net force does act on a particle, its velocity will change – Newton’s Second Law – It will have an acceleration – Its momentum will also change Lecture 11 5Purdue University, Physics 220 Force and Momentum • Assume the force and acceleration are constant Δ G G G vvv • Si t i ti l it th f b − == = Δ Δ G G fi mm m tt Fa • nce momen um s mass mes ve ocity, the orce can e related to the momentum • () Δ= − = − G G G G G fi fi tmFvvpp G • – This is the impulse theorem =Δ=Δ G Fpimpulse t Lecture 11 6Purdue University, Physics 220 Impulse Changes Momentum • Define Impulse I = FΔt Change in momentum requires Force acting– Force over a time or Impulse – Recall: mΔv = FΔt –Or Δp = FΔt Δp = I Impulse I = FΔt Momentum p = mv Lecture 11 Purdue University, Physics 220 7 1-D F vs t Plot Impulse is simply the area under the F (t) curve! Lecture 11 Purdue University, Physics 220 8 x Impulse and Variable Forces • The force does not need to be constant • The magnitude of the force grows rapidly from zero to a rapidly from zero to maximum value •The force then decreases to zero •Impulse = area under the force-time curve • Still =Δ G impulse p Lecture 11 Purdue University, Physics 220 9 Impulse • Impulse is equal to the area under the curve curve • The same value of the impulse can be be obtained by a large force with a short time or a small force with a long time • Applications include air bags Lecture 11 Purdue University, Physics 220 10 Impulse and Momentum I ≡ F ave Δt= p f -p i = Δp • For single object … • For collection of objects –F = 0⇒ momentum conserved (Δp= 0) collection of … –p total = Σp – Internal forces: forces between objects in system – External forces: all other forces – ΣF ext = 0 ⇒ total momentum conserved (Δp tot = 0) Lecture 11 Purdue University, Physics 220 11 –F ext = m total a Momentum of a System • To find the total momentum of a system of particles you , you need to add the momenta of all the individual particles in the system • == ∑∑ G G G total i i i mppv – The particles may be pieces of a solid object or individual particles associated with each ii other Lecture 11 12Purdue University, Physics 220 Quiz 1) A mass is attached to the bottom of a vertical spring. This causes the spring to stretch and the mass to move spring to stretch mass to move downward. Does the potential energy of the spring increase or decrease? Does the gravitational potential energy of the mass increase of decrease? mass increase of A) PE of spring decreases; PE of mass increases B) both decrease C) both increase D) PE of spring increases; PE of mass decreases PE of mass Lecture 11 Purdue University, Physics 220 13 Quiz 2) Which of the following statements correctly define a Conservative Force: A) A force is conservative when the work it does on a moving object is independent of the path of the motion is independent of the of the between the object's initial and final positions. B) A force is conservative when it does no net work on an force object moving around a closed path, starting and finishing at the same point. C) Both of the above statements are correct. D) Neither of the above statements is correct. Lecture 11 Purdue University, Physics 220 14 Newton’s Second Law F ma= ∑ G G IFtp=Δ=Δ ∑ G G New p F Δ ∑ G G formulation of Newton’s 2nd t = Δ Law Lecture 11 Purdue University, Physics 220 15 ILQ By what factor does an object’s kinetic energy change if its speed is doubled? By what factor does the magnitude of the momentum change? A) 4, 2 ), B) 2, 2 C) 0.5, 4 2 1 pmv K mv G G 2 == Lecture 11 Purdue University, Physics 220 16 Pushing Off… • Fred and Jane are on skates facing each other. Jane then pushes Fred with force F N2L Fred: F JF = m Fred a Fred Δv Fred = a Fred Δt = (F JF /m Fd ) Δt a = Δv/Δt Fred m Fred Δv Fred = F JF Δt N2L Jane: F FJ = m Jane a Jane Δv Jane = a Jane Δt = (F FJ /m Jane ) Δt m Jane Δv Jane = F FJ Δt N3L: For every action, there is an equal and opposite reaction. F FJ =-F JF Lecture 11 Purdue University, Physics 220 17 m Fred Δv Fred = -m Jane Δv Jane Pushing Off… Fred (75 kg) and Jane (50 kg) are on skates facing each other. Jane then pushes Fred with a constant force F = 45 N for a time Δt = 3 seconds. Who will be moving fastest at the end of the push? A) Fred B) Same C) Jane Fred Jane F = +45 N (positive direct.) I = +45 × 3 Ns = 135 Ns F = -45 N Newton’s 3 rd law I = -45 × 3 Ns = -135 Ns I = Δp = mv f –mv i I = Δp = mv f –mv i I/m = v f -v i v f = 135 N-s / 75 kg 18m/s I/m = v f -v i v f = -135 N-s / 50 kg 27m/s Lecture 11 Purdue University, Physics 220 18 = 1.8 m/s = -2.7 m/s Note: P fred + P jane = (1.8) 75 + (-2.7) 50 = 0! Momentum is Conserved • Define Momentum p = m v –m 1 Δv 1 = -m 2 Δv 2 Momentum is conserved – Δp 1 = -Δp 2 – Δp 1 + Δp 2 = 0 p 1f -p 1i +p 2f -p 2i =0 p +p =p +p – Σp initial = Σp final • Example: Jane pushes Fred so he is going 2.0 m/s. If Fred is twice as heavy as Jane how fast does 1f 2f 1i 2i as as Jane, Jane end up moving. – Σp = Σp initial final –0 = m Fred v Fred + m Jane v Jane – v Jane = -m Fred v Fred / m Jane = -4 m/s Lecture 11 Purdue University, Physics 220 19 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! Conservation in Lecture 11 Purdue University, Physics 220 20 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 p initial = m v initial = (1200 kg) x 40 m/s = 48,000 kg m/s North p final = m v final = (1200 kg) x 30 m/s = 36,000 kg m/s East North-South: p NS final – p NS initial = (0 – 48000) = -48,000 kg m/s East-West: p EW final –p EW initial = (36000 - 0) = +36,000 kg m/s Magnitude: 40 m/s Lecture 11 Purdue University, Physics 220 21 Sqrt(Δp NS 2 + Δp EW 2 ) = 60,000 kg m/s 30 m/s Soft vs Hard Impact 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 C) the same I = Δp Same change in momentum same In which case is the average force greater A) Floor B) Foam Δp = F Δt F = Δp/Δt Lecture 11 Purdue University, Physics 220 22 C) the same smaller Δt => large F Bouncy vs Sticky Ball 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 i h b i I hi h i h i d f h i loor w t out ounc ng. n whic case s t e magn tu e o t e mpu se given to the ball by the floor the biggest? A) Case 1 Δ mΔ mΔt FΔt I B) Case 2 C) The same p= v = ma t = F t = I Bouncing Ball Sticky Ball same Ball |I| = |Δp| | | |I| = |Δp| | |= mv final – m v initial = |m( v final -v initial )| = mv final – m v initial = |m(0 - v initial )| Lecture 11 Purdue University, Physics 220 23 = 2 m v = m v Bouncy vs Sticky Ball 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 In sticks floor . 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 21 I pmv mv=Δ = − G G G G Lecture 11 Purdue University, Physics 220 24 time Question Is it possible for a system of two objects to have zero total momentum while having a non-zero lki i ?total kinet c energy A) Yes The example of the ice skaters B) No 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 d i b h)none and gain a bunch) Lecture 11 Purdue University, Physics 220 25 Question Two objects are known to have the same kinetic energy. Do these two objects necessarily have the same momentum? A) Yes B) No Lecture 11 Purdue University, Physics 220 26 ILQ At the instant a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta Which object momenta. Which has the greater kinetic energy? Or are they the same A) The same Apply conservation of B) The gun C) The bullet momentum: P i =0 P f =M gun V gun +m bullet v bullet V gun =-(m bullet /M gun )v bullet 2 22 11 22 bullet bullet gun gun gun gun bullet bullet mm KMVM v K MM ⎛⎞ == = ⎜⎟ Lecture 11 Purdue University, Physics 220 27 gun gun ⎝⎠ Example • A fright train is being assembled in switching yard. Car 1 has a mass of m 1 =65 ×10 3 kg and moves at a velocity 080 / C2h f 92 10 3 kdv 01 =0.80 m s. Car 2 as a mass o m 2 = × kg and moves at a velocity v 02 =1.3 m/s and couples to it. Neglecting friction, find the common velocity v f of the cars ft th b l da er ey ecome coup e . m 2 m 1 v 02 v 01 initial m 2 m 1 v f final Lecture 11 Purdue University, Physics 220 28 Example • A fright train is being assembled in switching yard. Car 1 has a mass of m 1 =65 ×10 3 kg and moves at a velocity 080 / C2h f 92 10 3 kdv 01 =0.80 m s. Car 2 as a mass o m 2 = × kg and moves at a velocity v 02 =1.3 m/s and couples to it. Neglecting friction, find the common velocity v f of the cars ft th b l d Apply conservation of momentum: P i =m 1 v o1 +m 2 v 02 P f =(m 1 +m 2 )v f a er ey ecome coup e . (m 1 +m 2 )v f = m 1 v 01 +m 2 v 02 101 202 mv mv+ 12 33 (65 10 )(0 8 / ) (92 10 )(1 3 / ) f v mm kg m s kg m s == + ×+× Lecture 11 Purdue University, Physics 220 29 33 .8/ .3/ 1.1 / (65 10 92 10 ) ms kg kg = ×+× Example • Note the velocity of car 1 increases while the velocity of car 2 decreases. Th l ti d d l ti i b th• e acce eration an ece era on ar se ecause e 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) in elastic m 2 m 1 v 02 v 01 initial v f fl Lecture 11 Purdue University, Physics 220 30 m 2 m 1 final Summary of Concepts • Momentum p=mv p = mv – Momentum is VECTOR – Momentum is conserved (when ΣF = 0) (when • Σ mv initial = Σ mv final – If the vector sum of the external forces is zero ti dmomentum is conserve • Impulse I=FΔt I = F – Gives change in momentum I = Δp • Good Luck on EXAM Lecture 11 Purdue University, Physics 220 31 on Administrator Microsoft PowerPoint - Lecture11bw [Compatibility Mode]