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Lo.Lee.Ta.
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Please help me! Is this problem worked correctly...? If not, why?
1. In a pool ball game, consider the collision of two .5kg balls. One ball is at rest initially, the other approaches with a speed vi= 10m/s. The collision is not head-on, and after collision, ball #1 moves with a speed of 6.0m/s in a direction 60deg away from the initial line of motion. What is the speed and direction of ball#2?
Okay, so I know this is Elastic Collision, so momentum and energy are conserved.
Pi = Pf
Pi(x) =Pf(x)
Pi(y) = Pf(y)
m(1,i)(v) + m(2,i)(v) = m(1,f)(v) + m(2,f)(v)
P(x): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(cos(θ))(vf)
P(y): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(sin(θ))(vf)
Now I'm doing y/x to figure out the θ.
y/x = 5/5 = (1.5 + .5sinθ(vf))/(1.5 + .5cos(vf))
1 = tanθ
θ = 45deg
Put the 45deg back into x-comp. momentum:
(.5)(10) + (.5)(0) = (.5)cos(60)(6) + (.5)(cos45)(vf)
vf = 9.90m/s
...I'm thinking this is incorrect, though. It seems the vf is too high...
Please help me figure this problem out! Thank you! :)
1. In a pool ball game, consider the collision of two .5kg balls. One ball is at rest initially, the other approaches with a speed vi= 10m/s. The collision is not head-on, and after collision, ball #1 moves with a speed of 6.0m/s in a direction 60deg away from the initial line of motion. What is the speed and direction of ball#2?
Okay, so I know this is Elastic Collision, so momentum and energy are conserved.
Pi = Pf
Pi(x) =Pf(x)
Pi(y) = Pf(y)
m(1,i)(v) + m(2,i)(v) = m(1,f)(v) + m(2,f)(v)
P(x): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(cos(θ))(vf)
P(y): (.5)(10) + (.5)(0) = (.5)(cos(60))(6) + (.5)(sin(θ))(vf)
Now I'm doing y/x to figure out the θ.
y/x = 5/5 = (1.5 + .5sinθ(vf))/(1.5 + .5cos(vf))
1 = tanθ
θ = 45deg
Put the 45deg back into x-comp. momentum:
(.5)(10) + (.5)(0) = (.5)cos(60)(6) + (.5)(cos45)(vf)
vf = 9.90m/s
...I'm thinking this is incorrect, though. It seems the vf is too high...
Please help me figure this problem out! Thank you! :)