Momentum of Skiers in a Completely Inelastic Collision

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Homework Help Overview

The discussion revolves around a physics problem involving the momentum of two skiers in a completely inelastic collision. The original poster describes a scenario where a skier descends from a height and collides with another stationary skier, prompting questions about calculating final velocities and energy transformations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to find the speed of the first skier using potential and kinetic energy relationships. There are attempts to apply conservation of momentum principles, and some participants express confusion about calculating velocity from height.

Discussion Status

There is ongoing exploration of energy conservation principles and momentum calculations. Some participants have offered guidance on using potential energy to find the skier's speed, while others are seeking clarification on their calculations and the relevance of kinetic energy in this context.

Contextual Notes

Participants note that the collision is completely inelastic, which implies that only momentum is conserved during the collision. There is also mention of the original poster's uncertainty regarding the velocity calculations and the appropriateness of their energy calculations.

HarleyM
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Homework Statement


In a movie stunt, a 65 kg skier starts from a rest position at the top of a hill 30m high. She slides down the hill to the bottom, where she collides with a 45 kg stationary skier. The collision is completely inelastic. Find the final velocity of the skiers.


Homework Equations


M1V1 +M2V2= (M1+M2)V
ET'=ET'
ET'=(M1+m2)gh ( ithink this is relevant?)



The Attempt at a Solution



ET'= (65+45)(9.8)(30)
ET'= 32,340

32,340= 1/2mv2



Sq RT[(32,340)(2)/(65)] = velocity of skier going down hill?

=31.5 m/s (100 km/h + seems kinda fast for a skier)
 
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I think you need to first find the speed of the 65kg skier from their starting height. You can then calculate momentum and go from there.
If the collision is completely inelastic then ONLY momentum is conserved in the collision.
 
How would I calculate the velocity of the skier? This is where I am confused.. Thats why I used the ET' calculation..

I know you don't want to give me the answer.. but any hints would be immensely appreciated.. I mean I know the mass and height of the skier, no idea how to calculate their velocity though..

thanks for your time and response!
 
the skier gains KE as a result of 'falling' a distance of 30m.
You need to use PE = KE to calculate the speed of the 65kg skier.
Then it is a 'straight forward' momentum before = momentum after calculation.
Have a go and if you get stuck let me know where it goes wrong.
 
Last edited:
technician said:
the skier gains KE as a result of 'falling' a distance of 30m.
You need to use PE = KE to calculate the speed of the 65kg skier.
Then it is a 'straight forward' momentum before = momentum after calculation.
Have a go and if you get stuck let me know where it goes wrong.


THANKS again!

So PE=KE
mgh=1/2mv2
√(2)(9.8)(30)= V
V= 24.25 m/s


Ek = 1/2mv2
=1/2(65)(24.3)2
=19,190.9 J (initial energy) ( don't think I have to do this)

PT=PT'

M1V1+M2V2=(M1+M2)V'
(65)(24.3)+(45)(0)=(110)V'
V'=(65*24.3)/(110)
V'=14.36 m/s

Ek= 1/2(M1+M2)V'2
=1/2(110)(14.36)2
=11 341.53


how does the final velocity of both look? As that's what I am most concerned about.

Again thank you so much for your help!
 
That's it... you did not need to calculate KE in the question but well done.
 
Thank you so much ! Is there a rating system at PF?
 
Don't know !... not bothered...
 

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