Conservation of Momentum in Skating Question

  • Thread starter Ketchup1
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Hello, I've recently came across a question that I have two solutions for. Though I don't see how either of them can be wrong, I'm getting two different answers :( Can someone explain me why this is happening (maybe I'm doing something wrong?) and which solution is better to use.

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/s2
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)vf2 - 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?
 

Answers and Replies

  • #2
gneill
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Hello, I've recently came across a question that I have two solutions for. Though I don't see how either of them can be wrong, I'm getting two different answers :( Can someone explain me why this is happening (maybe I'm doing something wrong?) and which solution is better to use.

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/s2
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)vf2 - 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?
Hi Ketchup1, Welcome to Physics Forums.

I think something's gone wrong in your second solution. To find the distance traveled via the work done keep in mind that the initial velocity is that of the post-collision pair, and the final velocity is zero (they come to rest when their traveling is done).
 
  • #3
2
0
Oh! That works out! I didn't think of that for some reason
Thank you very much :D
The answer works out perfectly.
 

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