Why is the girl and the sled accelerating in opposite directions?

[dwdoyle8854]

Homework Statement

A 40kg girl and an 8.4 kg sled are on the frictionless ice of afrozen lake, 15m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope. Whatare the aceleration magnitudes of a) the sled and b) the girl? c) How far from the firls initial position do they meet?

Homework Equations

f = ma
Δx=V₀t + 1/2 at²

The Attempt at a Solution

So this is easy enough until part c) where the solution manual and I disagree.
The solution manual states "The accelerations of the sled and girl are in opposite directions."
and i just can't imagine why.

In parts a) and b) i find the magnitudes of the accelerations for the girl and sled respectively to be
0.62 m/s² = a_sled
0.13 m/s² = a_girl

to solve for when the collision i guess happens i set
x_{final, girl}=x_{final, sled}

and solving for time:
t=√( (2x_{0, girl}/(a_sled - a_girl) )

the solution manual however has:
t=√( (2x_{0, girl}/(a_sled + a_girl) )
which is because they state the girl is accelerating in the opposite direction of the sled.
But the girl is pulling on the sled, and everything is frictionless. How could she be accelerating in the opposite direction of the sled if she is pulling it?

freebody diagram of girl, x-axis

<--5.2N--()--5.2N--> by Newtons 3rd law.

 

[SammyS]

Homework Statement

A 40kg girl and an 8.4 kg sled are on the frictionless ice of afrozen lake, 15m apart but connected by a rope of negligible mass. The girl exerts a horizontal 5.2 N force on the rope. Whatare the aceleration magnitudes of a) the sled and b) the girl? c) How far from the firls initial position do they meet?

Homework Equations

f = ma
Δx=V₀t + 1/2 at²

The Attempt at a Solution

So this is easy enough until part c) where the solution manual and I disagree.
The solution manual states "The accelerations of the sled and girl are in opposite directions."
and i just can't imagine why.

In parts a) and b) i find the magnitudes of the accelerations for the girl and sled respectively to be
0.62 m/s² = a_sled
0.13 m/s² = a_girl

to solve for when the collision i guess happens i set
x_{final, girl}=x_{final, sled}

and solving for time:
t=√( (2x_{0, girl}/(a_sled - a_girl) )

the solution manual however has:
t=√( (2x_{0, girl}/(a_sled + a_girl) )
which is because they state the girl is accelerating in the opposite direction of the sled.
But the girl is pulling on the sled, and everything is frictionless. How could she be accelerating in the opposite direction of the sled if she is pulling it?

freebody diagram of girl, x-axis

<--5.2N--()--5.2N--> by Newtons 3rd law.

Conservation of Momentum.

The accelerations are in opposite directions, but towards each other.

 

[dwdoyle8854]

Conservation of Momentum.

The accelerations are in opposite directions, but towards each other.

okay, so


(sled)--------(girl)

(sled)--5.2N--> <--5.2N--(girl)

is this a better diagram?

So, i guess the little scenario I have in my head isn't what the question proposes. The girl is pulling the thing, but moving toward the sled? She isn't walking along pulling the sled behind her?

Well, i guess that solves it. Cheers.

 

[tms]

The girl is pulling the thing, but moving toward the sled? She isn't walking along pulling the sled behind her?

Since the ice is frictionless, she could not walk on it (this is idealized ice).

 

[CWatters]

So, i guess the little scenario I have in my head isn't what the question proposes. The girl is pulling the thing, but moving toward the sled? She isn't walking along pulling the sled behind her?

That's correct. She's pulling it towards her and the sled it pulling her towards it.

It's impossible for her to walk along pulling the sled behind her because the ice is described as frictionless.