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

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.

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)

Review of Work/Energy 𝑣 𝑥 2 = 𝑣 𝑥0 2 +2 𝑎 𝑥 𝑥 Start with: 𝑣 𝑥 2 = 𝑣 𝑥0 2 +2 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 0 2 +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

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

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

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

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

Impulse/Momentum Example 3 A 0.145 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= 0.145 𝑘𝑔∙52 𝑚 𝑠 − 0.145 𝑘𝑔∙−39 𝑚 𝑠 =13.195 𝑘𝑔 𝑚/𝑠 Impulse 𝐹 ∆𝑡=𝐼𝑚𝑝𝑢𝑙𝑠𝑒=∆𝑝=13.195 𝑘𝑔 𝑚/𝑠 Force on ball (note positive) 𝐹= 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 ∆𝑡 = 13.195 𝑘𝑔 𝑚 𝑠 .003 𝑠 =4400 𝑁 Force on bat (note negative) 𝐹 𝑐𝑎𝑟−𝑤𝑎𝑡𝑒𝑟 =− 𝐹 𝑤𝑎𝑡𝑒𝑟−𝑐𝑎𝑟 =−4400 𝑁

Conservation of 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

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 + 2 𝑘𝑔 0 𝑚 𝑠 = .023 𝑘𝑔 170 𝑚 𝑠 + 2 𝑘𝑔 𝑣 𝑏𝑙𝑜𝑐𝑘 ′ 𝑣 𝑏𝑙𝑜𝑐𝑘 =0.69 𝑚 𝑠

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 222 𝑣 ′ = 𝑚 4 𝑣 4 + 𝑚 218 𝑣 218 222 𝑢∙420 𝑚 𝑠 =4 𝑢∙ 𝑣 4 +218 𝑢∙350 𝑚 𝑠 𝑣 4 =4200 𝑚 𝑠

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