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6 Overview. Momentum and its Conservation Collisions and Impulse Homework: RQ: 2, 3, 4, 5, 8, 16, 18, 19. Ex: 14, 24, 31, 35, 47, 51. Problems: 1, 3, 6. How do we measure amount of motion?. velocity (how fast) mass (how much) with equal importance mass x velocity = “momentum”. Momentum.
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6 Overview • Momentum and its Conservation • Collisions and Impulse • Homework: • RQ: 2, 3, 4, 5, 8, 16, 18, 19. • Ex: 14, 24, 31, 35, 47, 51. • Problems: 1, 3, 6.
How do we measure amount of motion? • velocity (how fast) • mass (how much) • with equal importance • mass x velocity = “momentum”
Momentum • Momentum = mv • SI Unit: kg·m/s • Ex. 1000kg car moves at 8m/s. • mv = (1000kg)(8m/s) = 8,000 kg·m/s
Motion and Its Conservation Not Conserved Not Conserved Not Conserved Conserved Conserved stops smooth and level rough
Impulse and Momentum • Impulse = Ft • consequence of Newton’s 2nd law that Ft = change in momentum • F = ma • Ft = mat • Ft = (m)(at) • Ft = (m)(Dv) • N·s = kg·m/s
Impulse Example • A braking force of 4000N acts for 0.75s on a 1000kg car moving at 5.0m/s. • impulse = Ft = (4000N)(0.75s) = -3000 N·s • Ft = mDv = -3000 kg·m/s • initial momentum = mvi = (1000)(5) = 5000 • final momentum = mvi + mDv • = 5000 kg·m/s + (-3000 kg·m/s) • = 2000 kg·m/s
Collisions • ‘brief’ interaction (between masses) • Types: • Inelastic (heat, sound, etc. are generated). Ex. Almost all collisions are inelastic. • Elastic (no heat, sound, etc. is created). Ex. Two magnets ‘collide’ without touching.
Conservation of Momentum • two objects “bump” into one another • forces are opposite by Newton’s 3rd Law • e.g., object 1: -Ft, object 2: +Ft • net impulse to System = -Ft + Ft = 0 • net change in momentum due to the bump is zero. • If no other net-force acts, Then total momentum of objects is same
Effect of Interactions • bigger mass moves slower • smaller mass moves faster • momentums are equal and opposite • momentum of system is unchanged
2 Object Momentum Conservation • momentum before = momentum after • (m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final • When can we use this equation? • When net force due to all other objects acting on 1 and 2 is zero. • Or, very soon after collision ends
when can momentum conservation be used? • objects are on a level, frictionless surface (e.g. physics lab) • collision forces much bigger than friction (often) • Ex. Rail-car collision force much bigger than wheel friction
Ex. completely inelastic collision • equal mass rail cars collideand connect together • m1 = m2 = m • i) v1 = 10 v2 = 0 • f) v1 = v2 = vfinal = ? • (m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final • m(10) + m(0) = mvfinal + mvfinal . • 10m = 2mvfinal • 5 = vfinal
Ex. 2m versus m • m1 = 6kg collides with m2 = 3kg. • i) v1 = 5m/s v2 = 0 • f) v1 = v2 = vfinal = ? (complete inelastic) • (m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final • (6kg)(5m/s) + (3kg)(0) = (6kg)(vf)+(3kg)(vf) • 30kgm/s = (9kg)(vf) • vf = (30/9)m/s = 3.3m/s
Ex. 1500kg hits 1000kg • m1 = 1500kg collides with m2 = 1000kg. • i) v1 = 10m/s v2 = 0 • f) v1 = v2 = vfinal = ? (complete inelastic) • try it yourself first • (m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final • (1500kg)(10m/s) + (1000kg)(0) = (1500kg)(vf)+(1000kg)(vf) • 15,000kgm/s = (2500kg)(vf) • vf = (15,000/2500)m/s = 6.0m/s
Two Dimensional Collisions • Vector System Momentum: conserved in many situations (at least for a short time) Example: pool game, cue ball strikes another ball
Bouncing Motion • bouncing creates more force than stop • bouncing is two actions: stop & start is more action than stop alone. • : D(mv) = final mom. – initial mom. • Stop: D(mv) = 0 – mv = – mv Bounce: D(mv) = –mv – mv = – 2mv • change in momentum is double for elastic bounce, thus the impluse is double
Summary • mv = momentum • Impulse = Ft = D(mv) • momentum is conserved in collisions when all other forces are small compared to the collision forces • momentum is a vector and can be conserved in one, two, or three dimensions. • bouncing causes greater force than a stop
(cont.) Ex. 1500kg hits 1000kg • impulse received by m2 = change in momentum of m2 = m2(Dv) • Ft = m2(Dv) = (1000kg)(6m/s – 0m/s) • Ft = m2(Dv) = 6000kgm/s • F = (6000kgm/s)/t • if t = 1s: F = (6000kgm/s)/1s = 6,000N • if t = 0.01s: F = (6000kgm/s)/0.01s = 600,000N • cars have crumple zones to increase the collision time & reduce the force
Whenever an interaction occurs in a system, forces occur in equal and opposite pairs. Which of the following do not always occur in equal and opposite pairs? 1. Impulses.2. Accelerations.3. Momentum changes.4. All of these occur in equal and opposite pairs.5. None of these do.
1. Impulses.2. Accelerations.3. Momentum changes.4. All of these occur in equal and opposite pairs.5. None of these do. Whenever an interaction occurs in a system, forces occur in equal and opposite pairs. Which of the following do not always occur in equal and opposite pairs?
An ice sailcraft is stalled on a frozen lake on a windless day. A large fan blows air into the sail. If the wind produced by the fan strikes and bounces backward from the sail, the sailcraft will move 1. to the left (backward).2. to the right (forward).3. not at all.
1. to the left (backward).2. to the right (forward).3. not at all. An ice sailcraft is stalled on a frozen lake on a windless day. A large fan blows air into the sail. If the wind produced by the fan strikes and bounces backward from the sail, the sailcraft will move
Practicing Physics • p31 #4 • p32 #5 • p33 #1