Conservation of Momentum in Skating Question

In summary, the conversation discusses two different solutions to finding the distance traveled by a pair of figure skaters after a collision. The first solution uses the inelastic collision formula and the impulse formula, while the second solution combines the work equations. After plugging in the values, the first solution calculates the distance as 1.44m, while the second solution calculates it as 1.0548m. The conversation ends with the realization that the second solution was incorrect due to not considering the final velocity as zero. Therefore, the more accurate solution is 1.44m.
  • #1
Ketchup1
2
0
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?
 
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  • #2
Ketchup1 said:
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
Oh! That works out! I didn't think of that for some reason
Thank you very much :D
The answer works out perfectly.
 

Related to Conservation of Momentum in Skating Question

1. What is conservation of momentum in skating?

Conservation of momentum in skating refers to the principle that the total amount of momentum in a closed system remains constant, unless acted upon by an external force. In skating, this means that the momentum of the skater and their surroundings will not change unless an external force, such as friction or air resistance, is applied.

2. How does conservation of momentum apply to skating?

In skating, conservation of momentum applies to any situation where a skater is moving on a frictionless surface, such as ice. This means that the skater's momentum will remain constant as they move, and any changes in direction or speed will be due to external forces acting on them, such as pushing off with their skates or air resistance.

3. Why is conservation of momentum important in skating?

Conservation of momentum is important in skating because it helps explain the physics behind the movements and actions of skaters. Understanding this principle can also help skaters improve their technique and conserve their energy while on the ice.

4. How does friction affect conservation of momentum in skating?

Friction is an external force that can affect conservation of momentum in skating. It can slow down a skater's momentum and make it more difficult for them to move smoothly and efficiently. Skaters must learn how to minimize the effects of friction in order to maintain their momentum and perform at their best.

5. Can conservation of momentum be violated in skating?

No, conservation of momentum cannot be violated in skating or any other physical system. This principle is based on the laws of physics and has been proven to hold true in all observed situations. While external forces can affect the overall momentum of a skater, they cannot violate the principle of conservation of momentum.

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