Inelastic Collisions Review Impulse/Momentum 3 cases Indeterminate Inelastic Collision examples Ballistic Pendulum Other Inelastic examples

“Gravity” movie Newton’s Laws (conservation of momentum) 90 minute orbital time (different orbits?) CO2 fire extinguisher thruster Reentry (she’s going fast!)

Review – Impulse/Momentum Collisions 𝐹 12 =− 𝐹 21 𝑏𝑦 𝑎𝑐𝑡𝑖𝑜𝑛−𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝐹 12 ∆𝑡=− 𝐹 21 ∆𝑡 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 1 = 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 2 ∆ 𝑃 1 =−∆ 𝑃 2 (𝑚 1 𝑣 1 ′ − 𝑚 1 𝑣 1 )= −(𝑚 2 𝑣 2 ′ − 𝑚 2 𝑣 2 ) 𝑃 1 + 𝑃 2 = 𝑃 1 ′ + 𝑃 2 ′ 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣 1 ′ + 𝑚 2 𝑣 2 ′ F12 = force on 1 due to 2 1 2

Conservation of Momentum 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣 1 ′ + 𝑚 2 𝑣 2 ′ Too many variables – 3 possibilities Inelastic – eliminate variable 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣′ + 𝑚 2 𝑣′ Elastic - generate 2nd equation 1 2 𝑚 1 𝑣 1 2 + 1 2 𝑚 2 𝑣 2 2 = 1 2 𝑚 1 𝑣 1 ′ 2 + 1 2 𝑚 2 𝑣 2 ′ 2 Indeterminate - need more info

Inelastic Example 1 Conservation of momentum Inelastic 10,000 kg railroad car traveling at 24 m/s strikes a second 10,000 kg car at rest. They couple together, what is final speed? Conservation of momentum 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣 1 ′ + 𝑚 2 𝑣 2 ′ Inelastic 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣′ + 𝑚 2 𝑣′ 10,000 𝑘𝑔∙ 24 𝑚 𝑠 +0=(10,000 𝑘𝑔+10,000 𝑘𝑔) 𝑣′ 𝑣 ′ =12 𝑚/𝑠

Inelastic Example 1 (cont) Make the moving car 20,000 kg 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣′ + 𝑚 2 𝑣′ 20,000 𝑘𝑔∙ 24 𝑚 𝑠 +0=(20,000 𝑘𝑔+10,000 𝑘𝑔) 𝑣′ 𝑣 ′ =16 𝑚/𝑠 Make the stationary car 20,000 kg 10,000 𝑘𝑔∙ 24 𝑚 𝑠 +0=(10,000 𝑘𝑔+20,000 𝑘𝑔) 𝑣′ 𝑣 ′ =8 𝑚/𝑠

Inelastic Example 1 (cont) What happened to the energy? Kinetic energy before 1 2 𝑚 1 𝑣 1 2 + 1 2 𝑚 2 𝑣 2 2 = 1 2 10,000𝑘𝑔 24 𝑚 𝑠 2 =2.88∙ 10 6 𝐽 Kinetic energy after 1 2 (𝑚 1 + 𝑚 2 ) 𝑣′ 2 = 1 2 20,000𝑘𝑔 12 𝑚 𝑠 2 =1.44∙ 10 6 𝐽 Lost in heat

𝑣 1 ′ =−2.5 𝑚/𝑠 in opposite direction Inelastic Example 2 Calculate the recoil velocity of 5 kg rifle that shoots a 0.02 kg bullet at 620 m/s. Conservation of momentum 𝑚 1 𝑣 1 + 𝑚 2 𝑣 2 = 𝑚 1 𝑣 1 ′ + 𝑚 2 𝑣 2 ′ Inelastic collision in reverse. 5 𝑘𝑔∙0 𝑚 𝑠 +.02 𝑘𝑔∙0 𝑚 𝑠 =0=5 𝑘𝑔 ∙𝑣 ′ 1 + .02 𝑘𝑔∙620𝑚/𝑠 𝑣 1 ′ =−2.5 𝑚/𝑠 in opposite direction

Ballistic Pendulum Sequence Solve Backwards! Conservation momentum Conservation of energy Solve Backwards! Step 1: Conservation momentum: 𝑚𝑣= 𝑚+𝑀 𝑣 ′ 𝑣= 𝑚+𝑀 𝑚 𝑣′ Step 2: Conservation of energy: 1 2 𝑚+𝑀 𝑣′ 2 = 𝑚+𝑀 𝑔ℎ 𝑣 ′ = 2𝑔ℎ Combining: 𝑣= 𝑚+𝑀 𝑚 2𝑔ℎ

Ballistic Pendulum Example For h2 𝑣 2 = 𝑚+𝑀 𝑚 2𝑔 ℎ 2 𝑣 1 = 𝑚+𝑀 𝑚 2𝑔 ℎ 1 Combining 𝑣 2 𝑣 1 = 𝑚+𝑀 𝑚 2𝑔 ℎ 2 𝑚+𝑀 𝑚 2𝑔 ℎ 1 = ℎ 2 ℎ 1

Example 1 1. Conservation momentum 2. Conservation energy 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 ′ = 𝑚 𝑏𝑢𝑙 𝑣 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 𝑏𝑙𝑜𝑐𝑘 .029 𝑘𝑔+1.4 𝑘𝑔 𝑣 ′ =.029 𝑘𝑔 510 𝑚 𝑠 +0 𝑣′=10.3 𝑚/𝑠 2. Conservation energy 1 2 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣′ 2 = 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑔ℎ ℎ= 𝑣′ 2 2𝑔 =5.47 𝑚

Example 2 1. Conservation of momentum 2. Work-energy 0− 1 2 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 ′ 2 = −𝜇 𝑘 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑔 𝑥 𝑣 ′ = 2𝜇 𝑘 𝑔 𝑥 𝑣 ′ = 2∙0.25∙9.8 𝑚 𝑠 2 ∙9.5 𝑚 =6.82 𝑚 𝑠 1. Conservation of momentum 𝑚 𝑏𝑢𝑙 𝑣 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 𝑏𝑙𝑜𝑐𝑘 = 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 ′ 𝑣 𝑏𝑢𝑙= 𝑚 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑚 𝑏𝑢𝑙 𝑣 ′ =375 𝑚 𝑠

Inelastic car collision Work-energy 0− 1 2 𝑚 𝑆𝐶 + 𝑚 𝑆𝑈𝑉 𝑣 ′ 2 = −𝜇 𝑘 𝑚 𝑆𝐶 + 𝑚 𝑆𝑈𝑉 𝑔 𝑥 𝑣 ′ = 2𝜇 𝑘 𝑔 𝑥 𝑣 ′ = 2∙0.8∙9.8 𝑚 𝑠 2 ∙2.8 𝑚 =6.62 𝑚 𝑠 Conservation of momentum 𝑚 𝑆𝐶 𝑣 𝑆𝐶 + 𝑚 𝑆𝑈𝑉 𝑣 𝑆𝑈𝑉 = 𝑚 𝑆𝐶 + 𝑚 𝑆𝑈𝑉 𝑣 ′ 𝑣 𝑆𝐶= 𝑚 𝑆𝐶 + 𝑚 𝑆𝑈𝑉 𝑚 𝑆𝐶 𝑣 ′ = 920 𝑘𝑔+2300 𝑘𝑔 920 𝑘𝑔 6.62 𝑚 𝑠 =23.2 𝑚/𝑠