1. Momentum Review Energy, Intro Momentum Impulse-Momentum
Examples of Impulse-Momentum
Conservation of Momentum
Inelastic, Elastic, and Indeterminate Collisions
Indeterminate Collision Examples

2. Energy vs. Momentum Energy Momentum
Useful with conservative forces – where work only depends on difference of 2 endpoints.
Momentum
Useful in collisions – where objects exert equal and opposite forces by action-reaction.

3. Vectors vs. Scalars Chapter 2 – kinematics VECTOR
Chapter 3 – 2-D kinematics VECTOR
Chapter 4 – Newton’s Laws VECTOR
Chapter 5 – Circular Motion VECTOR
Chapter 6 - Energy SCALAR
Chapter 7 - Momentum VECTOR
(we only do in one dimension)

4. Review of Work/Energy 𝑣 𝑥 2 = 𝑣 𝑥0 2 +2 𝑎 𝑥 𝑥
Start with:
𝑣 𝑥 2 = 𝑣 𝑥 𝑎 𝑥 𝑥
Along arbitrary path
Each term product of vectors, hence scalar
Multiply by ½m:
1 2 𝑚 𝑣 2 = 1 2 𝑚 𝑣 𝐹∙𝑥 cos⁡(𝜃)
Scalar energy
Scalar work
Important Points
Work – force in direction of distance same as distance in direction of force
Multiple –sum of works same as work of sums, can neglect some works
PE – some works can be treated as change in PE

5. Intro to Impulse/Momentum
Start with
𝑣= 𝑣 0 +𝑎𝑡
vector expression
(back to x and y components)
Multiply by scalar m:
𝑚𝑣= 𝑚𝑣 0 +𝑚𝑎𝑡
𝑚𝑣= 𝑚𝑣 0 +𝐹𝑡
vector momentum
vector impulse
Notes
Momentum mv increased or decreased by addition of impulse Ft
Symbol p = mv, units kg-m/s
Momentum is vector. (but we only do in one dimension)
momentum
momentum
impulse

6. Use of Momentum Collisions 𝐹 12 =− 𝐹 21 𝑏𝑦 𝑎𝑐𝑡𝑖𝑜𝑛−𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
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

7. Impulse/Momentum Example 1
Tennis ball leaves racquet at speed 55 m/s. Mass is .06 kg and contact time 4 ms (4 x 10-3 s). Estimate force.
Change in momentum
∆𝑝=𝑚 𝑣 ′ −𝑚𝑣=(.06 𝑘𝑔∙55 𝑚 𝑠 )−(0)=3.3 𝑘𝑔 𝑚/𝑠
Impulse
𝐹 ∆𝑡=𝐼𝑚𝑝𝑢𝑙𝑠𝑒=∆𝑝=3.3 𝑘𝑔 𝑚/𝑠
𝐹= 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 ∆𝑡 = 3.3 𝑘𝑔 𝑚 𝑠 4∙ 10 −3 𝑠 =825 𝑁

8. Impulse/Momentum Example 2
Water leaves hose at rate 1.5 kg/s with speed 20 m/s. Aimed directly at side of car which stops it. Estimate force of a) car on water, b) water on car.
Change in momentum each second
∆𝑝=𝑚 𝑣 ′ −𝑚𝑣= 0 −1.5 𝑘𝑔∙20 𝑚 𝑠 =−30 𝑘𝑔 𝑚/𝑠
Impulse (note negative)
𝐹 ∆𝑡=𝐼𝑚𝑝𝑢𝑙𝑠𝑒=∆𝑝=−30 𝑘𝑔 𝑚/𝑠
Force on water (note negative)
𝐹= 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 ∆𝑡 = −30 𝑘𝑔 𝑚 𝑠 1 𝑠 =−30 𝑁
Force on car (note positive)
𝐹 𝑐𝑎𝑟−𝑤𝑎𝑡𝑒𝑟 =− 𝐹 𝑤𝑎𝑡𝑒𝑟−𝑐𝑎𝑟 =+30 𝑁

9. Impulse/Momentum Example 3
A kg baseball pitched at 39 m/s is hit on a horizontal line drive straight back toward the pitcher at 52 m/s. If the contact time between the bat and ball is 3 ms, calculate the average force on ball and bat during the hit.
-39 m/s
52 m/s
Change in momentum
∆𝑝=𝑚 𝑣 ′ −𝑚𝑣= 𝑘𝑔∙52 𝑚 𝑠 − 𝑘𝑔∙−39 𝑚 𝑠 = 𝑘𝑔 𝑚/𝑠
Impulse
𝐹 ∆𝑡=𝐼𝑚𝑝𝑢𝑙𝑠𝑒=∆𝑝= 𝑘𝑔 𝑚/𝑠
Force on ball (note positive)
𝐹= 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 ∆𝑡 = 𝑘𝑔 𝑚 𝑠 𝑠 =4400 𝑁
Force on bat (note negative)
𝐹 𝑐𝑎𝑟−𝑤𝑎𝑡𝑒𝑟 =− 𝐹 𝑤𝑎𝑡𝑒𝑟−𝑐𝑎𝑟 =−4400 𝑁

10. Conservation of 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

11. 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

12. Case 3 - Indeterminate Conservation of momentum
𝑚 𝑏𝑢𝑙 𝑣 𝑏𝑢𝑙 + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 𝑏𝑙𝑜𝑐𝑘 = 𝑚 𝑏𝑢𝑙 𝑣 𝑏𝑢𝑙 ′ + 𝑚 𝑏𝑙𝑜𝑐𝑘 𝑣 𝑏𝑙𝑜𝑐𝑘 ′
.023 𝑘𝑔 230 𝑚 𝑠 𝑘𝑔 0 𝑚 𝑠 = 𝑘𝑔 170 𝑚 𝑠 𝑘𝑔 𝑣 𝑏𝑙𝑜𝑐𝑘 ′
𝑣 𝑏𝑙𝑜𝑐𝑘 =0.69 𝑚 𝑠

13. A little nuclear physics (Indeterminate)
Could convert “u” to kg, but conversion would just cancel in every term!
Conservation of momentum – original nucleus on left, products on right
𝑚 𝑣 ′ = 𝑚 4 𝑣 4 + 𝑚 218 𝑣 218
222 𝑢∙420 𝑚 𝑠 =4 𝑢∙ 𝑣 𝑢∙350 𝑚 𝑠
𝑣 4 =4200 𝑚 𝑠

14. 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 (tomorrow)
𝑚 1 𝑣 𝑚 2 𝑣 2 = 𝑚 1 𝑣′ + 𝑚 2 𝑣′
Elastic - generate 2nd equation (next week)
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