Not Head-On Collision

ZxcvbnM2000

1. The problem statement, all variables and given/known data

One snooker ball (initially traveling at 0.6 m s-1) hits another of the same mass (0.4 kg) which was initially at rest. They collide with a coefficient of restitution of 0.9. The balls do not collide head-on, and so they come away at different angles. If the final speed of the incident ball is 0.3 m s-1, what is the final speed of the other (originally stationary) ball?

2. Relevant equations

conservation of momentum on x,y axis
law of restitution on the line of impact


3. The attempt at a solution

Momentum is conserved on the y axis ( perpendicular to the line of impact )

so u₁cosφ=V₁cosθ + V₂ 1)

Momentum is conserved on the x axis ( the line of impact )

u₁sinφ= V₁sinθ 2)


Applying the law of restitution on the line of impact :

e= ( V₂-V₁cosθ)/(u₁cosφ) 3)



So if we re-arrange a bit we get :

0.6cosφ = 0.3cosθ + V₂

0.54cosφ = -0.3cosθ + V₂

Αnd if we subtract these two we get :

cosφ = 10cosθ 4)


Νow if we plug 4) to 0.6cosφ = 0.3cosθ + V₂ we get :

5.7cosθ=V₂

But we also know that from 1) & 3) V₂ is also equal to = ((e+cosφ)u₁)/2 so

i found that cosθ=0.1 and i then i plugged it to 5.7cosθ=V₂ and found that V₂ is 0.57 m/s .

Is my solution correct ?


Thank you :)