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Elastic and Plastic Collisions Application physics momentum impulse mechanics
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7/18/2019 Elastic and Plastic Collisions Application physics momentum impulse mechanics
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v1 v2
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7/18/2019 Elastic and Plastic Collisions Application physics momentum impulse mechanics
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v1i = v
v2i = −2v
v1f = v1
v2f = v2
m1v1i + m2v2
i = m1v1
f + m2v2
f
⇒ mv − 2mX 2v = mv1 + 2mv2
⇒ v1 + 2v2 = −3v
v1i − v2i = v2f − v1f
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⇒ v − (−2v) = v2 − v1
⇒ v2 − v1 = 3v
⇒ 2v2 + v2 = 0
⇒ v2 = 0
⇒ v1 + 2v2 = −3v
⇒ v1 + 0 = −3v
⇒ v1 = −3v
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t = 2πr
v
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P 1 P 2 θ
7/18/2019 Elastic and Plastic Collisions Application physics momentum impulse mechanics
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P 2 = P 21 + P 2
2 + 2P 1P 2cosθ
⇒ P 2 = m2v2 + m2v2 + 2mvmvcosθ
⇒ P 2
= 2m2
v2
+ 2m2
v2
cosθ
P 2 = {2mv
2
}2 = m2v2
⇒ m2v2 = 2m2v2 + 2m2v2cosθ
⇒ cosθ =
−
1
2
= cos1200
⇒ θ = 1200
7/18/2019 Elastic and Plastic Collisions Application physics momentum impulse mechanics
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m1v1i + m2v2i = m1v1f + m2v2f
⇒ mv − mX 0 = mV + mV
⇒ V = v
2
⇒ K i = K = 12mv2
⇒ Kc = 1
2X 2mXV 2 = mX
v2
2
= mv2/4 = K
2
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θ
θ
θ
θ
θ
θ
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⇒ K i = K f = 1
2mv2 = 2
⇒ 1
2X 1Xv2 = 2
⇒ v2 = 4
⇒ v = 2 m/s
mv + mX 0 = 2mv
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v = v
2
K f = 12
2m(v)2 = 12
2mv2
2
⇒ v2
4 = 2
⇒ v = 2√
2 m/s
