Transcript Slide 1

Elastic Collisions
• Review Impulse/Momentum
• 3 cases
– Inelastic Collision x
– Elastic Collision
– Indeterminate x
• Elastic Collision – first principles
• Elastic Collision – Relative velocity equation
• Other Elastic examples
Review – Impulse/Momentum
• Collisions
F12 = force on
1 due to 2
1
2
• 𝐹12 = −𝐹21
𝑏𝑦 𝑎𝑐𝑡𝑖𝑜𝑛 − 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
• 𝐹12 ∆𝑡 = −𝐹21 ∆𝑡
𝐼𝑚𝑝𝑢𝑙𝑠𝑒1 = 𝐼𝑚𝑝𝑢𝑙𝑠𝑒2
• ∆𝑃1 = −∆𝑃2
(𝑚1 𝑣1′ − 𝑚1 𝑣1 ) = −(𝑚2 𝑣2′ − 𝑚2 𝑣2 )
• 𝑃1 + 𝑃2 = 𝑃1′ + 𝑃2 ′
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2 ′
Conservation of Momentum
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2 ′
• Too many variables – 3 possibilities
– Inelastic – eliminate variable
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣′ + 𝑚2 𝑣′
– Elastic - generate 2nd equation
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2 ′
1
1
1
1
𝑚1 𝑣1 2 + 𝑚2 𝑣2 2 = 𝑚1 𝑣1 ′2 + 𝑚2 𝑣2 ′2
2
2
2
2
– Indeterminate - need more info
Simple Elastic Collision
• Two carts with equal mass m
• One traveling at v1 m/s, other at 0 m/s
• Conserve momentum and energy:
𝑚𝑣1 + 𝑚0 = 𝑚𝑣1′ + 𝑚𝑣2′
(𝑃)
1
1
1
1
𝑚𝑣1 2 + 𝑚02 = 𝑚𝑣1 ′2 + 𝑚𝑣2 ′2
2
2
2
2
• Simplify:
𝑣1 = 𝑣1′ + 𝑣2′
𝑣1 2 = 𝑣1 ′2 + 𝑣2 ′2
(𝐾𝐸)
(𝑃)
(𝐾𝐸)
• Substitute 1st equation in 2nd equation:
𝑣2′ = 𝑣1 − 𝑣1′
2
′2
𝑣1 = 𝑣1 + 𝑣1 −
′ 2
𝑣1
(𝑃)
(𝐾𝐸 + 𝑃)
Simple Elastic Collision (cont)
• Previously:
𝑣2′ = 𝑣1 − 𝑣1′
𝑣1 2 = 𝑣1 ′2 + 𝑣1 − 𝑣1′
(𝑃)
2
(𝐾𝐸 + 𝑃)
• Expand 2nd equation and simplify:
𝑣1 2 = 𝑣1 ′2 + 𝑣1 2 − 2𝑣1 𝑣1′ + 𝑣1 ′2
(𝐾𝐸 + 𝑃)
0 = 2𝑣1 ′2 − 2𝑣1 𝑣1′
𝑣1 ′2 = 𝑣1 𝑣1′
• Solutions
𝑣1′ = 0
𝑣1′ = 𝑣1
→
→
𝑣2′ = 𝑣1
𝑣2′ = 0
1𝑠𝑡 𝑠𝑡𝑜𝑝𝑠, 2𝑛𝑑 𝑡𝑎𝑘𝑒𝑠 𝑜𝑟𝑖𝑔. 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 1𝑠𝑡
𝑡ℎ𝑒𝑦 𝑚𝑖𝑠𝑠𝑒𝑑
𝑑𝑢𝑚𝑏 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Relative velocity equation
• Can use as substitute for conservation of energy in
elastic collisions
• Much easier to solve!
• Derivable from conservation of momentum +
conservation of energy
Relative velocity equation
•
Conserve momentum and energy:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′
(𝑃)
1
1
1
1
𝑚1 𝑣1 2 + 𝑚2 𝑣2 2 = 𝑚1 𝑣1 ′2 + 𝑚2 𝑣2 ′2 (𝐾𝐸)
2
2
2
2
•
Rearrange both equations:
𝑚1 𝑣1 − 𝑣1′ = 𝑚2 𝑣2 ′ − 𝑣2
(𝑃)
𝑚1 𝑣1 2 − 𝑣1 ′2 = 𝑚2 𝑣2 ′2 − 𝑣2 2
•
•
Factor energy equation:
𝑚1 𝑣1 − 𝑣1′ = 𝑚2 𝑣2 ′ − 𝑣2
(𝑃)
𝑚1 𝑣1 − 𝑣1′ 𝑣1 + 𝑣1′ = 𝑚2 𝑣2 ′ − 𝑣2
𝑣2′ + 𝑣2
Eliminate what’s common to both:
𝑚1 𝑣1 − 𝑣1′ = 𝑚2 𝑣2 ′ − 𝑣2
𝑣1 + 𝑣1′ = 𝑣2′ + 𝑣2
•
(𝐾𝐸)
(𝑃)
(𝑟𝑒𝑙 𝑣𝑒𝑙)
Conservation of momentum and relative velocity equation:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′
𝑣1 − 𝑣2 = 𝑣2′ − 𝑣1′
(𝑃)
(𝑟𝑒𝑙 𝑣𝑒𝑙)
(𝐾𝐸)
Relative velocity equation
• Instead of solving this:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′ (𝑃)
1
1
1
1
𝑚1 𝑣1 2 + 𝑚2 𝑣2 2 = 𝑚1 𝑣1 ′2 + 𝑚2 𝑣2 ′2 (𝐾𝐸)
2
2
2
2
• You may substitute this
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′
𝑣1 − 𝑣2 = 𝑣2′ − 𝑣1′
(𝑃)
(𝑟𝑒𝑙 𝑣𝑒𝑙)
Which is a lot easier to solve!
Simple Elastic Collision (again)
• Two carts with equal mass m
• One traveling at v1 m/s, other at 0 m/s
• Conserve momentum and energy relative velocity equation:
𝑚𝑣1 + 𝑚0 = 𝑚𝑣1′ + 𝑚𝑣2′
𝑣1 − 𝑣2 = 𝑣2′ − 𝑣1′
• Simplify:
𝑣1 = 𝑣1′ + 𝑣2′
𝑣1 = 𝑣2′ − 𝑣1′
(𝑃)
𝑟𝑒𝑙 𝑣𝑒𝑙
𝑃
𝑟𝑒𝑙 𝑣𝑒𝑙
• Add both equations:
2𝑣1 = 2𝑣2′
𝑣2′ = 𝑣1
𝑣1′ = 0
Same result as before
2009 “Conserve-the-Wrong-Thing” Contest
• In collision, always conserve this:
m1v1 + m2v2 = m1v1’ + m2v2’
• If elastic, also conserve this:
½ m1v12 + ½ m2v22 = ½ m1v1’2 + ½ m2v2’2
Or use relative velocity equation
• Guaranteed way of losing 8 points in exam!
Elastic Example 1
• Conservation of momentum and relative velocity eqn:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′
𝑣1 − 𝑣2 = 𝑣2′ − 𝑣1′
(𝑃)
𝑟𝑒𝑙 𝑣𝑒𝑙
• Plug in numbers:
0.45 𝑘𝑔 3 𝑚 𝑠 + 0 = 0.45 𝑘𝑔 𝑣1′ + 0.9 𝑘𝑔 𝑣2′
3 𝑚 𝑠 − 0 = 𝑣2′ − 𝑣1′
(𝑃)
𝑟𝑒𝑙 𝑣𝑒𝑙
• Need to solve 2 simple equations:
1
2
Elastic Example 1 (cont)
• Previous page:
0.45 𝑘𝑔 3 𝑚 𝑠 = 0.45 𝑘𝑔 𝑣1′ + 0.9 𝑘𝑔 𝑣2′
3 𝑚 𝑠 = 𝑣2′ − 𝑣1′
(𝑃)
𝑟𝑒𝑙 𝑣𝑒𝑙
• Solution (my weird way):
0.45 𝑘𝑔 3 𝑚 𝑠 = 0.45 𝑘𝑔 𝑣1′ + 0.9 𝑘𝑔 𝑣2′
0.45 𝑘𝑔 3 𝑚 𝑠 = 0.45 𝑘𝑔 𝑣2′ − 0.45 𝑘𝑔 𝑣1′
• Add both equations:
2.7 𝑘𝑔 𝑚 𝑠 = 1.35 𝑘𝑔 𝑣2′
𝑣2′ = 2 𝑚 𝑠
1
• Substituting in rel. vel.
𝑣1′ = −1 𝑚 𝑠
Note negative sign – 1 bounces back!
2
Elastic Example 2
• Conservation of momentum and relative velocity eqn:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′
𝑣1 − 𝑣2 = 𝑣2′ − 𝑣1′
(𝑃)
𝑟𝑒𝑙 𝑣𝑒𝑙
• Plug in numbers:
0.06 𝑘𝑔 2.5 𝑚 𝑠 + 0.09 𝑘𝑔 1.15 𝑚 𝑠 = 0.06 𝑘𝑔 𝑣1′ + 0.09 𝑘𝑔 𝑣2′
2.5 𝑚 𝑠 − 1.15 𝑚 𝑠 = 𝑣2′ − 𝑣1′
1
𝑟𝑒𝑙 𝑣𝑒𝑙
2
(𝑃)
Elastic Example 2 (cont)
• Previous page:
0.06 𝑘𝑔 2.5 𝑚 𝑠 + 0.09 𝑘𝑔 1.15 𝑚 𝑠 = 0.06 𝑘𝑔 𝑣1′ + 0.09 𝑘𝑔 𝑣2′
1.35 𝑚 𝑠 = 𝑣2′ − 𝑣1′
• My devious tricks:
0.06 𝑘𝑔 2.5 𝑚 𝑠 + 0.09 𝑘𝑔 1.15 𝑚 𝑠 = 0.06 𝑘𝑔 𝑣1′ + 0.09 𝑘𝑔 𝑣2′
0.06 𝑘𝑔 1.35 𝑚 𝑠 = 0.06 𝑘𝑔 𝑣2′ − 0.06 𝑘𝑔 𝑣1′
• Add both equations:
0.06 𝑘𝑔 2.5 𝑚 𝑠 + 0.09 𝑘𝑔 1.15 𝑚 𝑠 + 0.06 𝑘𝑔 1.35 𝑚 𝑠 = 0.09 𝑘𝑔 𝑣2′ + 0.06 𝑘𝑔 𝑣2′
0. 3345 𝑘𝑔 𝑚 𝑠 = 0.15 𝑘𝑔 𝑣2′
𝑣2′ = 2.23 𝑚 𝑠
𝑣1′ = 0. 88 𝑚 𝑠
1
2
Elastic Example 3
• Conservation of momentum and
relative velocity eqn:
𝑚1 𝑣1 + 𝑚2 𝑣2 = 𝑚1 𝑣1′ + 𝑚2 𝑣2′
𝑣1 − 𝑣2 = 𝑣2′ − 𝑣1′
(𝑃)
𝑟𝑒𝑙 𝑣𝑒𝑙
• Plug in numbers:
450 𝑘𝑔 4.5 𝑚 𝑠 + 550 𝑘𝑔 3.7 𝑚 𝑠 = 450 𝑘𝑔 𝑣1′ + 550 𝑘𝑔 𝑣2′
4.5 𝑚 𝑠 − 3. 7 𝑚 𝑠 = 𝑣2′ − 𝑣1′
𝑟𝑒𝑙 𝑣𝑒𝑙
(𝑃)
Elastic Example 3 (cont)
• Previous page:
4060 𝑘𝑔 𝑚 𝑠 = 450 𝑘𝑔 𝑣1′ + 550 𝑘𝑔 𝑣2′
0.8 𝑚 𝑠 = 𝑣2′ − 𝑣1′
• Here we go again:
4060 𝑘𝑔 𝑚 𝑠 = 450 𝑘𝑔 𝑣1′ + 550 𝑘𝑔 𝑣2′
450 𝑘𝑔 0.8 𝑚 𝑠 = 450 𝑘𝑔 𝑣2′ − 450 𝑘𝑔 𝑣1′
• Adding
4060 𝑘𝑔 𝑚 𝑠 + 360 𝑚 𝑠 = 550 𝑘𝑔 𝑣2′ + 450 𝑘𝑔 𝑣2′
1000 𝑘𝑔 𝑣2′ = 4420 𝑘𝑔 𝑚 𝑠
𝑣2′ = 4.42 𝑚 𝑠
𝑣1′ =
3.62 𝑚 𝑠
1
2
Elastic Example 3 (cont)
• Momentum change of 1
∆𝑝1 = 𝑚1 𝑣1′ − 𝑚1 𝑣1
∆𝑝1 = 450 kg 3.62 𝑚 𝑠 − 450 kg 4.5 𝑚 𝑠
∆𝑝1 = −396 kg 𝑚 𝑠
• Momentum change of 2
∆𝑝2 = 𝑚2 𝑣2′ − 𝑚2 𝑣2
∆𝑝2 = 550 kg 4.42 𝑚 𝑠 − 550 kg 3.7 𝑚 𝑠
∆𝑝1 = +396 kg 𝑚 𝑠
What 1 lost, 2 gained.
Collisions in 2-dimensions
•
Conserve momentum in x and y direction
𝑚𝑣𝑎𝑥 + 𝑚𝑣𝑏𝑥 = 𝑚𝑣′𝑎𝑥 + 𝑚𝑣′𝑏𝑥
𝑚𝑣𝑎𝑦 + 𝑚𝑣𝑏𝑦 = 𝑚𝑣′𝑎𝑦 + 𝑚𝑣′𝑏𝑦
•
Cancelling m’s and filling in numbers
3 𝑚 𝑠 + 0 = 𝑣′𝑎 cos(45) + 𝑣 ′ 𝑏 cos(45)
0 + 0 = 𝑣′𝑎 sin(45) + −𝑣 ′ 𝑏 sin 45
𝑣′𝑏 = 𝑣′𝑎
•
y equation yields:
•
x equation yields: 2 𝑣′𝑎 cos(45) = 3 𝑚 𝑠
𝑣′𝑎 = 𝑣′𝑏 = 2.12 𝑚 𝑠
(𝑛𝑜𝑡𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑠𝑖𝑔𝑛)