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## Unit 8: Linear Momentum

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**What is implied?**Unit 8: Linear Momentum Work on Unit 8 HW asap Demos: Collision carts and track U7b: All students get perfect Score - answers online Impulsive Force Model**Agenda**• Review Dynamics for exam • Operational Definitions • momentum • units • Momentum Motion Map • Elastic and Inelastic Collisions • Conservation of Momentum • Mid Term Review Impulsive Force Particle**Scratch off Practice**• The University of New England was founded in… • A) 1953 • B) 1978 • C) 1996 • D) 1939 • E) 1831 • Add 147 to the above answer to determine when UNE was first established. • A) 1978 • B) 1953 • C) 1996 • D) 1831 • E) 1939 Impulsive Force Particle**Operational Definitions**Linear momentum: p=mv Units (kg.m/s) Elastic Collision ~ “No-Stick” Inelastic Collision ~ “Stick” Impulsive Force Particle**Clicker ?**t=0 1 t=0 1 2 3 4 +x 3 4 60 • Ekf = Eki, pf ≠ pi • Ekf = Eki, pf = pi • Ekf ≠ Eki, pf ≠ pi • Ekf ≠ Eki, pf = pi • During elastic collisions… Impulsive Force Particle**Clicker ?**t=0 1 t=0 1 2 3 4 +x 60 • Ekf = Eki, pf ≠ pi • Ekf = Eki, pf = pi • Ekf ≠ Eki, pf ≠ pi • Ekf ≠ Eki, pf = pi • During inelastic collisions… Impulsive Force Particle**Elastic Collision**• Kinetic energy is conserved (Eki = Ekf) • Linear momentumis conserved (pf=pi=p) Ekf (J) pf (kg.m/s) 1 1 pi (kg.m/s) Eki (J) 0 0 1 1 Ekf(J) = 0.95 Eki(J) pf(kg.m/s) = 0.95 p i(kg.m/s) Impulsive Force Particle**Inelastic Collision**• Kinetic energy is NOT conserved (Eki ≠ Ekf) • Linear momentum is ALWAYS conserved (pf=pi=p) Ekf (J) pf (kg.m/s) 1 1 pi (kg.m/s) Eki (J) 0 0 1 1 Ekf(J) = 0.5 Eki(J) pf(kg.m/s) = 0.95 pi(kg.m/s) Impulsive Force Particle**Elastic Collision Motion Map**• Insert masses at positions proportional to those in the collision. No motion => no change in position. Time increases from the top down. t=0 1 t=0 1 2 3 4 +x 3 4 Impulsive Force Particle**Elastic Collision MM1**• Insert velocity vectors at each block’s position. Identify the interaction point with a dashed circle. Define initial and final momenta for each object’s mass and velocity. t=0 p2i=0 1 t=0 1 2 3 4 p1i p2f 3 p1f=0 4 Impulsive Force Particle**Elastic Collision MM2**• Insert total initial and final momentum as vector sums of either the initial or final momenta. p1i + p2i = pi = p p1f +p2f = pf = p t=0 + = + = p2i=0 1 t=0 1 2 3 4 p1i p2f 3 p1f=0 4 Impulsive Force Particle**Elastic Collision MM3**• Predictive Power: Since pf = p2f = pi = p1i • => m2v2f = m1v1i • since m1=m2 => v2f = v1i p1i + p2i = pi = p p1f +p2f = pf = p t=0 + = + = p2i=0 1 t=0 1 2 3 4 p1i p2f 3 p1f=0 4 Impulsive Force Particle**Alternative Representations**p p Before After p1 p2 p1 p2 • Area of velocity vs. mass plot (same area). p p v v Before After m1 m2 m1 m2 • Bar graph to monitor before, after, and total momentum. Impulsive Force Particle**Inelastic Collision MM1**• Insert velocity vectors at each block’s position. Identify the interaction point with a dashed circle. Define initial and final momenta for each particle based on masses and velocities. t=0 p2i=0 1 t=0 1 2 3 4 p1i p1&2f Impulsive Force Particle**Inelastic Collision MM2**• Insert total initial and final momentum as vector sums of either the initial or final momenta. p1i + p2i = pi = p p1&2f = pi = p t=0 + = = p2i=0 1 t=0 1 2 3 4 p1i p1&2f Impulsive Force Particle**Inelastic Collision MM3**• Predictive Power: Since pf = p1&2f = pi = p1i • => (m1+m2)v1&2f = m1v1i • since m1=m2 => v1&2f = (1/2)v1i p1i + p2i = pi = p p1&2f = pi = p t=0 + = = t=0 1 2 3 4 p1i p1&2f Impulsive Force Particle**Alternative Representations**p p Before After p1 p2 p1&2 • Area of velocity vs. mass plot (same area!). p p v v Before After m1 m2 m1+m2 • Bar graph to monitor before, after, and total momentum. Impulsive Force Particle**Work out problem**• Create an inelastic MM for m1=m striking m2=3m. • Predict v1&2f if v1i = +4m/s, v2i = 0 t=0 p1i t=0 2 Impulsive Force Particle**Inelastic Collision MM5**• Predictive Power: Since pf = p1&2f = pi = p1i • => (m1+m2)v1&2f = m1v1i • m1=m, m2=3m => v1&2f=(1/4)v1i =(1/4)(+4m/s)=+1m/s t=0 p1i + p2i = pi = p p1&2f = pi = p p2i=0 + = = 1 p1i p1&2f t=0 3 4 1 2 Impulsive Force Particle**Alternative Representations**p p Before After p1 p2 p1&2 • Area of velocity vs. mass plot (same area!!!). p p v v Before After m1 m2 m1+m2 • Bar Graph to represent initial, final, and total momentum Impulsive Force Particle**Clicker ?**2 t=0 1 p1i p2i=0 60 • 2m rebounds, m moves slowly • 2m stops, m moves at medium pace • 2m continues slow, m moves quickly • None of the above • If mass “2m” collides ELASTICALLY with mass “m” then… Impulsive Force Particle**Elastic Collision MM4**• Create an elastic MM for m1=2m striking m2=m. • Predict v1f and v2f if v1i = +3m/s, v2i=0. • Very challenging because we have two unknowns so that we need two equations. • Solved generically. Details are provide below but will not be discussed. Impulsive Force Particle**Elastic Collision Prediction**• Because momentum is conserved, pf=pi: m1v1f + m2v2f = m1v1i • Because the collision is elastic, Ekf = Eki (1/2)m1v1f2 + (1/2)m2v2f2 = (1/2)m1v1i2 • Two equations with two unknowns: (v1f and v2f given m1 = 2m, m2 = m, v1i = +3m/s). • To simplify let v1i = v, v2f = v2, v1f = v1 then: m1v1+m2v2=m1v • Solve for v1: v1 = (m1v-m2v2)/m1 Impulsive Force Particle**Crunch Time 1**• Square v1 and multiply by m1… v12= [(m1v-m2v2)/m1]2 = v2-2vv2m2/m1+v22m22/m12 => m1v12=m1v2-2vv2m2+v22m22/m1 • Dropping (1/2) in Ek: m1v12 + m2v22 = m1v2 • Substitute m1v12 into Ek … m1v2-2vv2m2+v22m22/m1 + m2v22 = m1v2 • Cancel m1v2 on both sides and divide by v2… -2vm2 + v2m22/m1 + m2v2 = 0 Impulsive Force Particle**Crunch Time 2**• Divide by m2 and isolate v2… v2(m2/m1 + 1) = 2v • Solve for v2… v2 = 2v/(1+m2/m1) = v2f • Solve for v1… v1 = v-m2v2/m1 = v - 2m2v/m1(1+m2/m1) = v(m1+m2-2m2)/(m1+m2) = v(m1-m2)/(m1+m2) v1 = v(1-m2/m1)/(1+m2/m1) = v1f Impulsive Force Particle**m1 > m2 => m1 Plows on**• Since m1=2m, m2=m and v1i=3m/s • v1f = v(1-m2/m1)/(1+m2/m1) = +1m/s • v2f = 2v/(1+m2/m1) = +4m/s p1f+p2f = pf p1i + p2i = pi t=0 + = + = p2i=0 1 2 3 4 3 4 t=0 1 4 p1i p1f p2f Impulsive Force Particle**Alternative Representations**p p Before After p1 p2 p1 p2 • Area of velocity vs. mass plot (same area!!!). p p v v Before After m1 m2 m1 m2 • Bar graphs to represent initial, final and total momentum Impulsive Force Particle**Clicker ?**2 t=0 1 p2i=0 p1i 60 • m rebounds, 2m moves slowly • m stops, 2m moves at medium pace • m continues slow, 2m moves quickly • None of the above • If mass “m” collides ELASTICALLY with mass “2m” then… Impulsive Force Particle**m1 < m2 => m1 bounces back!**• If m1=m, m2=2m and v1i=3m/s • v1f = v(1-m2/m1)/(1+m2/m1) = -1m/s • v2f = 2v/(1+m2/m1) = +2m/s p1i + p2i = pi p1f+p2f = pf t=0 + = + = p2i=0 p1i 1 2 3 4 t=0 2 p1f p2f 4 3 Before After p1 p2 p1 p2 Impulsive Force Particle**Unit 8a: Momentum ctd.**HW U8a due after Thanksgiving Demos: Collision carts and two air pucks Impulsive Force Model**Inelastic Collision Motion Map**• Insert masses and positions proportional to those in the collision. No motion => no change in position. Time increases from the top down. t=0 1 t=0 1 2 3 4 +x Impulsive Force Particle**Agenda**• Refined Operational Definitions: • Elastic Collision • Inelastic Collision • Explosions • Two objects in motion and colliding • Two-Dimensional Collisions Impulsive Force Particle**Operational Definitions**Linear momentum: p=mv Units (kg.m/s) Elastic Collision: Eki = Ekf Inelastic Collision: Eki ≠ Ekf Impulsive Force Particle**What is your opinion?**• True • False Explosions are examples of elastic collisions because kinetic energy before and after the explosion is conserved. Impulsive Force Particle**Explosions 1**• MMM an explosion of an object into two pieces, one with mass m, the other with mass 2m. What is the pi? Therefore what must be pf? Is this by definition an elastic or inelastic event? t=0 Before After p1 p2 p1 p2 Impulsive Force Particle**Explosions 2**• Since Eki = 0 and Ekf≠0 => Inelastic “event” • Predictive Power: Since pf = p2f + pif = pi = 0 • => +m2v2f - m1v1f = 0 • since m2=2m1 => v2f = (1/2)v1f p1i + p2i = pi t=0 + = p2i=0 p1i=0 1 p1f p2f 3 4 2 3 4 p1f + p2f = pf + = Before p After p p1 p2 p1 p2 Impulsive Force Particle**2 Moving Objects 1**• MMM the situation when two carts of equal mass m collide elastically, but in opposite directions and different velocities. Look at the individual and total momenta we observe that momentum is transferred. t=0 t=0 1 2 1 Impulsive Force Particle**2 Moving Objects 1**• Prediction: Since momenta are transferred: • p1i = p2f and p2i = p1f • And since m1=m2=m • v2f = v1i and v1f = v2i Rigorous solution requires conservation of kinetic energy, lots of algebra. t=0 1 2 1 t=0 p1i + p2i = pi p1i p2i 4 3 2 3 4 p1f+ p2f = pf p2f p1f Before After p p p1 p2 p1 p2 Impulsive Force Particle**Two Dimensional Collisions**• Motivation: A 1000 kg eastbound car traveling 30m/s runs a red light and center-punches a 10,000 kg truck traveling southbound at 4 m/s. The two collide inelastically. • Draw a top view of the incident and qualitatively estimate where the two vehicles end up, assuming no interactions with any other obstacles. Impulsive Force Particle**Clicker ?**A B q N W E S C 60 D • A • B • C • D • Choose the correct path, either A, B, C, or D AND explain why the paths you did not choose are incorrect. Impulsive Force Particle**Because p = constant…**A B q N W E S C D • … only “C” can be correct since it is the only one with momenta both in eastern and southern directions. • But what is pf and q? Impulsive Force Particle**Use Vector Tool: Sp=0**• pf is the total momentum after the collision. What is it comprised of? • The momenta of the car (pc) and truck (pt). What are these in terms of pf? q N W E S pf Impulsive Force Particle**Seek Resultant Vector**• The components of total momentum! • pcar = mV(east) = 30,000kgm/s (east) • ptruck = Mv(south) = 40,000kgm/s(south) • pf = ?, q = ? • pf = 50,000 kgm/s • q = tan-1(4/3) = 53° • What is vf =? • vf = pf/(m+M) = 4.5m/s pcar q ptruck pf Impulsive Force Particle**Arbitrary q Impact**• Gets messy, though understandable through tip to tail vector addition. • For example, two billiard balls. • Easy to solve if q + f = 90° f p1f p2f p1i q f q pi pf Impulsive Force Particle**Arbitrary q Impact**• p1f = pfcosf and p2f = pfcosq • COM: pf = p1f + p2f = pi • p1fy = p1fsinf = -p2fy = -p2fsinq • The vertical components of the final momenta must cancel. f p1f p2f p1i q f q pi pf Impulsive Force Particle**Clicker ?**A B C 60 0 / 54 • A • B • C • Two identical billiard balls collide; what will be their post-collision paths? Impulsive Force Particle**Unit 8b: Impulse Force**HW U8b due after Thanksgiving Demos: Force sensor and motion detector with collision cart, eggs and blanket, small and big balls. Impulsive Force Model**Agenda**• Newton’s Third Law • Impulsive Force and change of velocity • Examples Impulsive Force Particle**Clicker ?**Pow! 60 • VW Bug • Mac Truck • Both feel same force • No Clue • Which feels the greater force, a VW Bug colliding at 60 mph with the Mac Truck or the Mac Truck colliding at 30 mph with the VW Bug? Impulsive Force Particle**Clicker ?**Pow! 60 • VW Bug • Mac Truck • Both interact over same Dt • Texting buddies • Which vehicle interacts over a longer time during the collision? Impulsive Force Particle