Calculating Energy Loss in a Hockey Puck Collision

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In a collision between two identical hockey pucks on a frictionless rink, the first puck with a mass of 0.1 kg and an initial velocity of 3.8 m/s loses energy after striking a stationary puck. To calculate the energy lost, one can use the formula E=1/2mv² for both before and after the collision. The discussion clarifies that the focus is on the energy lost by the first puck specifically, rather than the entire system. The conservation of momentum is highlighted, indicating that while momentum is conserved, energy can be lost in inelastic collisions. The key takeaway is to calculate the difference in kinetic energy before and after the collision to determine the fraction of energy lost.
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One hockey puck of mass 0.1 kg strikes an identical stationary puck on a frictionless ice rink. If the first puck had a velocity V(0)=3.8m/s before the collision and v(1)=0.8 m/s after in the same direction, what fraction of the energy was lost?
Where do I even start to find the answer??
 
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If the rink is frictionless, then no energy should be lost. This is a momentum problem, and momentum is a conserved quantity. Is this your question, or are you trying to find the fraction of energy lost in the first puck as opposed to the fraction of energy lost of the system?
 
I would assume the question is asking how much energy was lost by the first puck.

That's a simple matter of taking the energy before and after via E=1/2mv2

The difference would be the energy lost by the first puck.
 
If you are indeed talking about the energy lost by the first puck then dantose is correct.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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