Fraction of Energy Lost in Inelastic Collision | Hockey Puck Problem

In summary: To calculate the fraction of energy lost, we need to compare the initial energy before collision with the final energy after collision. We can do this by dividing the lost energy by the initial energy:Fraction of energy lost = (lost energy)/(initial energy) = (.48J)/(1.444J) = 0.33 or 33%In summary, we can calculate the fraction of energy lost in a collision by comparing the lost energy to the initial energy. In this case, 33% of the initial energy was lost when one hockey puck of mass 0.2 kg with an initial velocity of 3.8 m/s collided with an identical stationary puck on a frictionless ice rink.
  • #1
sirfinklstin
15
0

Homework Statement



One hockey puck of mass 0.2 kg strikes an identical stationary puck on a frictionless ice rink. If the first puck had a velocity V0 = 3.8 m/s before the collision and V1 = 0.8 m/s after in the same direction, what fraction of the energy was lost?

Homework Equations


Conservation of momentum m1v1 + m2v2 = m3v3 + m4v4
Work energy theorem k = (1/2) mv^2 or (mv)^2/2m


The Attempt at a Solution


3.8 + 0 = .8 + 3
(I left out mass becuase all objects have the same mass.)

I am pretty much stuck becuase I do not know the final velocity of V2, and if there was energy lost then V2 shouldn't equal 3 m/s.
 
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  • #2
Welcome to PF!

Hi sirfinklstin! Welcome to PF! :wink:
sirfinklstin said:
3.8 + 0 = .8 + 3

I am pretty much stuck becuase … if there was energy lost then V2 shouldn't equal 3 m/s.

Why not?? :confused:

What is your energy equation? :smile:
 
  • #3
u almost solved the problem...
v1= 3.8
v2=0...bcoz stationary given
v3=0.8
and v4 u have to find out...final velocity of other mass...
which u got 3

now jus find difff between initial and final energy
wats d problm??
 
  • #4
Ok, thanks for the help with that, i thought a loss of energy would mean less than 3 m/s for V4, here is what I have now -

puck 1 before collision
(1/2)(.2)(3.8)^2
1.444

puck 2 before collision
(1/2)(.2)(3.8)^2
0

puck 1 after collision
(1/2)(.2)(.8)^2
.064

puck 2 after collision
(1/2)(.2)(3)^2
.9

loss of energy = .48J

is this correct?
 
  • #5
looks good! just convert that into a fraction...
 
  • #6
(48/100)j ?
 
  • #7
"what fraction of the energy was lost?"
 
  • #8
So, how to calculate fraction of the energy was lost? I'm lost :(
 

Related to Fraction of Energy Lost in Inelastic Collision | Hockey Puck Problem

1. What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. This means that after the collision, the objects involved will have less total kinetic energy than before the collision.

2. How is the fraction of energy lost in an inelastic collision calculated?

The fraction of energy lost in an inelastic collision is calculated by dividing the change in kinetic energy by the initial kinetic energy. This can be represented by the equation: (KEinitial - KEfinal) / KEinitial.

3. Why is the fraction of energy lost important in the hockey puck problem?

In the hockey puck problem, the fraction of energy lost is important because it tells us how much of the initial kinetic energy of the puck is converted into other forms of energy, such as heat and sound. This information can help us understand the overall efficiency of the collision and make predictions about the motion of the puck.

4. How does the mass of the hockey puck affect the fraction of energy lost?

The mass of the hockey puck does not directly affect the fraction of energy lost in an inelastic collision. However, a heavier puck may result in a larger change in kinetic energy, which could result in a higher fraction of energy lost.

5. Can the fraction of energy lost be greater than 1 in an inelastic collision?

No, the fraction of energy lost cannot be greater than 1 in an inelastic collision. This would indicate that the final kinetic energy is greater than the initial kinetic energy, which goes against the principle of conservation of energy. The fraction of energy lost is always a decimal value between 0 and 1.

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