Conservation of Momentum - Elastic Collision

AI Thread Summary
In the discussion on the conservation of momentum during an elastic collision, a 20kg curling stone moving at 1.5m/s collides with a stationary 0.16kg hockey puck, which then moves at 2.5m/s post-collision. The momentum conservation equation used is mstonevixstone + mpuckvixpuck = mstonevfxstone + mpuckvfxpuck. The calculated final velocity of the curling stone is 1.48m/s. The solution appears to be correct, and no further questions were raised. This demonstrates the application of momentum conservation principles in collision scenarios.
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Homework Statement


A curling stone with a mass of 20kg slides across the ice at 1.5m/s. It collides head on with a stationary 0.16kg hockey puck. After the collision, the puck's speed is 2.5m/s. Assume the motion occurs in the horizontal direction. What is the stone's final velocity? Follow the steps below to answer the question.


Homework Equations


P=mv
Pi=Pf



The Attempt at a Solution


mstonevixstone + mpuckvixpuck = mstonevfxstone + mpuckvfxpuck

20*1.5 + .16*0 = 20vfxstone + .16*2.5
vfxstone = 1.48 m/s
 
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Looks okay to me. What's your question?
 
Haha. Well if it's correct, I suppose I have none. Thanks!
 

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Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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