Momentum: Skater A & B on Frictionless Ice

[runningirl]

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?

 

[tiny-tim]

hi runningirl!

-F(t1)=F(t2)

nooo … force doesn't come into it

use conservation of momentum (for both parts) …

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

 

[runningirl]

hi runningirl!


nooo … force doesn't come into it

use conservation of momentum (for both parts) …

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

well, p=2.005(v)

how would find v?!

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

 

[tiny-tim]

hi runningirl!

(just got up :zzz: …)

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?

just write a conservation of momentum equation …

total momentum after = total momentum before …

what do you get? ​

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