What Is the Speed of the Second Puck After a Head-On Collision?

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Homework Help Overview

The discussion revolves around a head-on collision between two identical ice hockey pucks, focusing on the speeds before and after the collision. The original poster presents a scenario where one puck slows down while the other puck's speed needs to be determined. The context involves concepts of momentum conservation and elastic collisions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the conservation of momentum formula and question the correctness of the original poster's calculations. There is also a suggestion to simplify the equations by dropping the mass, given that the pucks are identical.

Discussion Status

Some participants have offered guidance on rearranging the momentum equation and have pointed out potential mistakes in the original calculations. There is an ongoing exploration of the implications of vector quantities in the context of the problem, particularly regarding the direction of velocities.

Contextual Notes

Participants note that the velocities should be treated as vectors, which may affect the calculations due to the direction of motion. There is a mention of confusion regarding the resulting speeds, indicating a need for clarity in the setup and assumptions of the problem.

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Homework Statement


ice hockey puck moving at speed V1 collides head on with a second identical puck moving towards it at speed V2. After the collision, the first puck slows down to speed v1 without changing direction.
a)After the collision what is the speed v2 of the second puck?
b) calculate the speed v2 of the second puck when the first puck had an initial speed of 18ms-1 that was changed to 2.0ms-1 by the collision, and the initial speed of the second puck was 12,0ms-1. both pucks have a mass of 0.16kg

The Attempt at a Solution



well the questions shows that the initial puck continues on its path at a lower speed to i guess this was an elastic collision

not sure if the math is right tho

m1V1+m2V2 = m1v1+m2v2
v2 = m1V1+m2V2/m1v1 + m2

when i factor in the given speeds and masses in the next part of the question i am not sure if it is right.

v2 = (0.16kg)(18.oms-1)+(0.16kg)(12.0ms-1)/(0.16kg)(2.0ms-1)+0.16kg
v2 = 15.16ms-1

is this right?
 
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Welcome to PF!

Hi lilmissbossy! I forgot to say earlier … Welcome to PF! :smile:
lilmissbossy said:
m1V1+m2V2 = m1v1+m2v2
v2 = m1V1+m2V2/m1v1 + m2

Nooo … v2 = (m1V1+m2V2 - m1v1)/m2.

Try again! :smile:
 
lilmissbossy said:

Homework Statement


ice hockey puck moving at speed V1 collides head on with a second identical puck moving towards it at speed V2. After the collision, the first puck slows down to speed v1 without changing direction.
a)After the collision what is the speed v2 of the second puck?
b) calculate the speed v2 of the second puck when the first puck had an initial speed of 18ms-1 that was changed to 2.0ms-1 by the collision, and the initial speed of the second puck was 12,0ms-1. both pucks have a mass of 0.16kg

The Attempt at a Solution



well the questions shows that the initial puck continues on its path at a lower speed to i guess this was an elastic collision

not sure if the math is right tho

m1V1+m2V2 = m1v1+m2v2
v2 = m1V1+m2V2/m1v1 + m2

when i factor in the given speeds and masses in the next part of the question i am not sure if it is right.

v2 = (0.16kg)(18.oms-1)+(0.16kg)(12.0ms-1)/(0.16kg)(2.0ms-1)+0.16kg
v2 = 15.16ms-1

is this right?

The hockey pucks are identical. You can drop the mass out the equations. Might make it simpler.
 


tiny-tim said:
Hi lilmissbossy! I forgot to say earlier … Welcome to PF! :smile:


Nooo … v2 = (m1V1+m2V2 - m1v1)/m2.

Try again! :smile:


Oh ok i see now i have made a mistake when rearranging the equation...
thats great i will factor in the values and it should make more sense:smile:
 
so the answer is 28 m/s that doesn't make much sence?
 
Hey guys, had the same problem.
Remember V is a vector quantity so if the pucks are heading toward each other one will be a negative quantity.

eg: (.16x18)+(.16x-12)=(.16x2)+(.16xV2)

so the answer is 4 not 28

Momentum Conservation
.96=.96
 

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