What is the Final Speed of the Stationary Snooker Ball After Collision?

[ZxcvbnM2000]

Homework Statement

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?

Homework Equations

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

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 :)

 

[anathema]

it is important to always double-check your calculations and assumptions to ensure accuracy. Your solution appears to be correct, but it is always a good idea to have someone else review your work or to run through the calculations again yourself to ensure there are no errors. Additionally, it would be helpful to label your equations and show your work step by step to make it easier for others to follow along and understand your solution. Overall, your approach using conservation of momentum and the law of restitution is correct and your final answer of 0.57 m/s seems reasonable. Great job!