Momentum: Skater A & B on Frictionless Ice

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Homework Help Overview

The problem involves two skaters on frictionless ice, where one skater pushes the other, leading to questions about their final speeds based on the conservation of momentum. The scenario includes two parts: one where both skaters start at rest and another where they have an initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum, questioning the relevance of force in this context. There are attempts to set up equations based on momentum and inquiries about how to find the final velocities.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning assumptions about the forces involved. Some guidance on using conservation of momentum has been provided, but no consensus has been reached on the final approach or calculations.

Contextual Notes

There is a mention of gravity and acceleration, which some participants suggest may not be relevant to the problem setup on horizontal ice. The initial conditions for both parts of the problem are also being clarified.

runningirl
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Homework Statement



Two skaters are standing at rest on frictionless ice. Skater A, 65 kg, pushes skater B, 80 kg. Skater B ends up moving to the right at 2.0 m/s.

a) What is the final speed of skater A?

b) Repeat, but this time, assume both skaters are initially moving to the right at 0.50 m/s.

Homework Equations



-F(t1)=F(t2)

The Attempt at a Solution



-65(v-0)=80(v-2)

v=1.1 m/s for part a.

part b?
 
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hi runningirl! :smile:
runningirl said:
-F(t1)=F(t2)

nooo … force doesn't come into it :redface:

use conservation of momentum (for both parts) …

in collisions (or "reverse collisions" like this), momentum (also angular momentum, btw) is always conserved :smile:
 
tiny-tim said:
hi runningirl! :smile:


nooo … force doesn't come into it :redface:

use conservation of momentum (for both parts) …

in collisions (or "reverse collisions" like this), momentum (also angular momentum, btw) is always conserved :smile:

well, p=2.005(v)

how would find v?!

.36=(vf^2-vo^2)/2a
is acceleration just 9.8(2.005)?
 
hi runningirl! :smile:

(just got up :zzz: …)
runningirl said:
well, p=2.005(v)

how would find v?!

.36=(vf^2-vo^2)/2a
is acceleration just 9.8(2.005)?

where did 2.005 come from? :confused:

just write a conservation of momentum equation

total momentum after = total momentum before …

what do you get? :smile:

(and gravity (9.8) doesn't come into this … the ice is presumably horizontal! :wink:)
 

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