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

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

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

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

5. 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 𝑚 2 𝑣 2 = 𝑚 1 𝑣 1 ′ + 𝑚 2 𝑣 2 ′
Inelastic
𝑚 1 𝑣 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

6. Inelastic Example 1 (cont)
Make the moving car 20,000 kg
𝑚 1 𝑣 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

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

8. 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 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 ′ 𝑘𝑔∙620𝑚/𝑠
𝑣 1 ′ =−2.5 𝑚/𝑠 in opposite direction

9. 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

10. 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 = ℎ 2 ℎ 1

11. 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

12. Example 2 1. Conservation of momentum 2. Work-energy
0− 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

13. Inelastic car collision
Work-energy
0− 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 ′ = 𝑘𝑔+2300 𝑘𝑔 920 𝑘𝑔 𝑚 𝑠 =23.2 𝑚/𝑠