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Last time, we examined how an external impulse changes the momentum of an object. Now, we will look at how multiple objects interact in a closed system (no external forces)

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● In any closed system (no external force) total momentum is conserved. p initial = p final ● When we have more than one object, p 1initial + p 2initial = p 1final + p 2final OR m 1 v 1initial + m 2 v 2initial = m 1 v 1final + m 2 v 2final Conservation of momentum is a much easier way to deal with collisions than Newton’s Laws.

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Example: Two-particle collision m1m1 v 1i m2m2 v 2i velocities just before interaction (collision) velocities just after interaction (collision) v 1f v 2f Total linear momentum in a system is conserved! p after = p before m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f

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A 60.0-kg astronaut is on a space walk when her tether line breaks. She throws her 10.0-kg oxygen tank away from the shuttle with a speed of 12.0 m/s to propel herself back to the shuttle. What is her velocity? 70 before v initial = 0 after 10 12.0 m/s 60 v 1final = ? Example: Explosion / Ejection

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A 60.0-kg astronaut is on a space walk when her tether line breaks. She throws her 10.0-kg oxygen tank away from the shuttle with a speed of 12.0 m/s to propel herself back to the shuttle. What is her velocity? p before = p after 70 before v = 0 after 10 12.0 m/s 60 v 1final = ? 0 = m 1 v 1final + m 2 v 2final 0 = 60.0 v 1 + 10.0 (12.0) v 1 = − 2.0 m/s moving in the negative direction means toward shuttle Example: Explosion / Ejection

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Similar Scenarios hot gas ejected at very high speed p before = p after 0 = m 1 v 1 + m 2 v 2 m 1 v 1 = - m 2 v 2 Gas ejected from a rocket Recoil of a gun Throwing something off a boat etc.

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There are two fish in the sea. A 6 kg fish and a 2 kg fish. The big fish swallows the small one. What is its velocity immediately after lunch? a.the big fish swims at 1 m/s toward and swallows the small fish that is at rest. 1 m/s before lunch after lunch v = ? Collision Problem – We do

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There are two fish in the sea. A 6 kg fish and a 2 kg fish. The big fish swallows the small one. What is its velocity immediately after lunch? a.the big fish swims at 1 m/s toward and swallows the small fish that is at rest. Net external force is zero. Momentum is conserved. p before lunch = p after lunch momentum is vector, direction matters; choose positive direction in the direction of big fish. + m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f 1 m/s before lunch after lunch (6 kg)(1 m/s) + (2 kg)(0 m/s) = (6kg + 2 kg) v 6 kg m/s = (8 kg) v v = 0.75 m/s in the direction of the large fish before lunch v = ? Collision Problem – We do

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b. Suppose the small fish is not at rest but is swimming toward the large fish at 2 m/s. Collision Problem – You do

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p before lunch = p after lunch + Mu 1 + mu 2 = (M+m)v 1 m/s before lunch after lunch v = 0.25 m/s in the direction of the large fish before lunch v = ? b. Suppose the small fish is not at rest but is swimming toward the large fish at 2 m/s. - 2 m/s (6 ) (1 ) + (2 ) (—2 ) = (6 + 2 ) v 6 — 4 = 8 v The negative momentum of the small fish is very effective in slowing the large fish. Collision Problem – You do

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p before lunch = p after lunch + Mu 1 + mu 2 = (M+m)v 1 m/s before lunch after lunch v = 0 m/s v = ? c. Small fish swims toward the large fish at 3 m/s. - 3 m/s (6 ) (1 ) + (2 ) (—3 ) = (6 + 2 ) v 6 — 6 = (8 ) v fish have equal and opposite momenta. Zero momentum before lunch is equal to zero momentum after lunch, and both fish come to a halt. Additional problem for at-home practice

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p before lunch = p after lunch + Mu 1 + mu 2 = (M+m)v 1 m/s before lunch after lunch v = — 0.25 m/s v = ? d. Small fish swims toward the large fish at 4 m/s. - 4 m/s (6 ) (1 ) + (2 ) (—4) = (6 + 2 ) v 6 — 8 = 8 v The minus sign tells us that after lunch the two-fish system moves in a direction opposite to the large fish’s direction before lunch. Additional problem for at-home practice

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What if the objects travel at angles to each other???

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Resolve vectors into components Apply conservation of momentum to the x direction and the y direction. p xi = p xf AND p yi = p yf

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A red ball traveling with a speed of 2 m/s along the x-axis hits the eight ball. After the collision, the red ball travels with a speed of 1.6 m/s in a direction 37 o below the positive x-axis. The two balls have equal mass. At what angle will the eight ball fall in the side pocket? What is the speed of the blue (8 th ) ball after collision. before collision: after collision: 8 8 v 1i v 2i = 0 θ2θ2 37 0 the point of collision v 2f v 1f

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A red ball traveling with a speed of 2 m/s along the x-axis hits the eight ball. After the collision, the red ball travels with a speed of 1.6 m/s in a direction 37 o below the positive x-axis. The two balls have equal mass. At what angle will the eight ball fall in the side pocket? What is the speed of the blue (8 th ) ball after collision. before collision: after collision: 8 8 v 1i v 2i = 0 θ2θ2 37 0 the point of collision v 2f v 1f p before = p after in x – direction m v 1i + 0 = mv 1f cos 37 0 + m v 2f cos 2 v 2f cos 2 = v 1i - v 1i cos 37 0 = 0.72 m/s (1) in y – direction 0 = - m v 1f sin 37 0 + m v 2f sin θ 2 v 2f sin θ 2 = v 1f sin 37 0 = 0.96 m/s (2) direction of v 2: tan θ 2 = 1.33 θ 2 = 53 0 (2) → v 2 = 0.96 / sin 53 0 v 2 = 1.2 m/s

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PN J S N P J S

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You Do

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CONSERVATION OF MOMENTUM. When two particles collide they exert equal and opposite impulses on each other. It follows that for the two particles, the.

CONSERVATION OF MOMENTUM. When two particles collide they exert equal and opposite impulses on each other. It follows that for the two particles, the.

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