Q3.5 Actions and reactions (S)

A girl of mass 40 kg and a sleigh of mass 10 kg rest on the smooth surface of a frozen lake, a distance 15 m apart. By means of a rope the girl exerts a force of 5 N on the sleigh, drawing it towards her. Find

  1. the acceleration of the sleigh
  2. the acceleration of the girl
  3. the distance they each have moved when they meet up.

In what respects do your results depend upon the particular value of the force that the girl exerts?

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Hint

The third law is the key: Key Point 2.3.

In the last part you will stand a better chance if you have remembered Guideline 0.6.

 

Solution

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Solution

Let F=5 N represent the force that the girl exerts on the sleigh. By the 3rd law this is also the magnitude of the force that the sleigh exerts on her (through the rope: we will see this a bit more clearly when we have examined the idea of tension in S2.7).

Graphic - No title - girlsleigh
  1. [1. and 2.] It is easy to get lost with signs here. It is clear that the sleigh accelerates to the right, while the girl accelerates to the left. So let us represent the rightward acceleration of the sleigh by a1 and the leftward acceleration of the girl by a2. Both of these quantities are positive. Using m1 and m2 for the masses (of sleigh and girl) we have, by Newton’s 2nd law (applied to each)
    \[ m_1a_1 =F \hspace{1cm}\mbox{\rm and}\hspace{1cm} m_2a_2 =F \]
    Thus
    \[ a_1 =\frac{F}{m_1} = \frac{1}{2} \, ms^{-2} \hspace{1cm}\mbox{\rm and}\hspace{1cm} a_2 =\frac{F}{m_2}=\frac{1}{8} \, ms^{-2} \]
  2. [3.] Suppose it takes a time t for the girl and the sleigh to meet up. In this time the sleigh, starting from rest, will move a distance (to the right)
    \[ d_1 = \frac{1}{2} a_1 t^2 \]
    while the girl will move a distance (to the left)
    \[ d_2 = \frac{1}{2} a_2 t^2 \]
    It follows that
    \[ \frac{d_1}{d_2} = \frac{a_1}{a_2} =\frac{F/m_1}{F/m_2} =\frac{m_2}{m_1} \]
    Note that F cancels out from our argument at this point. But we know that d1+d2=d where d=15 m is the initial distance between the girl and the sleigh. Substituting d2=d-d1 we find
    \[ \frac{m_2}{m_1} = \frac{d_1}{d_2} = \frac{d_1}{d-d_1} \]
    which we can reorganise to give
    \[ d_1 =\frac{dm_2}{m_1+m_2} =12\,m \]
    In the same way we find
    \[ d_2 =\frac{dm_1}{m_1+m_2} =3\,m \]
    Notice that, in accord with experience, the two bodies move through distances that are inversely proportional to their masses.

While the specific strength of the force applied does enter the accelerations we calculated in the first part of the question, the result for the distances that the girl and the sleigh move do not depend on the magnitude of this force (which, as noted, cancels from the argument). This suggests that a more general principle is at work here. There is indeed; it is linear momentum conservation, expressed in the behaviour of the Centre of Mass. More of this in S4.2.