- #1

Ketchup1

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## Homework Statement

During a free dance program in figure skating, Phrank (m = 71kg) glides at a 2.1m/s to a stationary Phyllis (m= 52kg) and hangs on. How far will the pair slide after the "collision" if coefficient of kinetic friction (μk) between their skates and the ice is 0.052?

## Homework Equations

m1v1 1+m2v2 2=(m1+m2)vf (Inelastic collision, since Phrank grabs on)

Solution 1:

F⋅Δt=Δp =m(vf−vi) (Impulse)

Δd = 1/2(vi+vf)Δt

Solution 2:

W=F⋅d

Wnet=ΔK

## The Attempt at a Solution

__Solution 1:__

Use the inelastic collision formula to solve for the final velocity (vf).

m1v1 + m2v2 = (m1+m2)vf

(71kg)(2.1m/s) + (52kg)(0) = (71kg + 52kg)vf

vf = 1.2122m/s

Plug that into the impulse forumla.

F⋅Δt=Δp =m(vf−vi)

Δt = Δp/F

Δt = m(vf - vi)/μk⋅m⋅g (masses cancel out)

Δt = vf - vi/μk⋅g

Δt = (0 - 2.1m/s)/0.052⋅9.81 m/s

^{2}

Δt = 2.3763s

Plug the time into the the second kinematics equation.

Δd = 1/2(vi+vf)Δt

Δd = 1/2(2.1m/s + 1.2122m/s)(2.3763s)

Δd = 1.44 m

__Solution 2:__

Use the inelastic collision formula to solve for the final velocity (vf).

m1v1 + m2v2 = (m1+m2)vf

(71kg)(2.1m/s) + (52kg)(0) = (71kg + 52kg)vf

vf = 1.2122m/s

Combine the two work equations.

W=F⋅d

Wnet=ΔK

F⋅d=ΔK

Δd = ΔK/F

Δd = 1/2(m1 + m2)vf

^{2}- 1/2(m1)(v1)

^{2}/ μk⋅m⋅g

Its messy, so I'll skip plugging in all the values.

Δd = -66.185/-62.74476

Δd = 1.0548m

**Both solutions seem reasonable to me, but why am I getting two different answers? Which solution is more accurate?**