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)

● 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.

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

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 m/s 60 v 1final = ? Example: Explosion / Ejection

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 m/s 60 v 1final = ? 0 = m 1 v 1final + m 2 v 2final 0 = 60.0 v (12.0) v 1 = − 2.0 m/s moving in the negative direction means toward shuttle Example: Explosion / Ejection

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.

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

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

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

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

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

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

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

 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

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 θ2θ the point of collision v 2f v 1f

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 θ2θ the point of collision v 2f v 1f p before = p after in x – direction m v 1i + 0 = mv 1f cos 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 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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You Do