91Ó°ÊÓ

The mass of the girl is\({m_{\rm{g}}} = 20\;{\rm{kg}}\).

The mass of the boy is\({m_{\rm{b}}} = 40\;{\rm{kg}}\).

The speed of the boy is \({v_{\rm{b}}} = 1.0\;{\rm{m/s}}\).

The conservation of momentum principle states that the total momentum of an isolated system remains conserved. In this problem, the external force on the boy and girl is zero, so the momentum of the system that constitutes the boy and the girl will remain conserved.

Here, the initial momentum of the pair of children (girl and boy) is zero because they are in a rest position. On evaluating the final momentum of the girl from the conservation of momentum, it is obtained that her direction will be opposite, and her magnitude will be similar to the boy’s final momentum.

The relationship from conservation of momentum is given by:

\(\begin{aligned}{c}{P_1} = {P_2}\\{m_{\rm{g}}}{v_{\rm{g}}} = {m_{\rm{b}}}{v_{\rm{b}}}\end{aligned}\)

Here,\({v_{\rm{g}}}\)is the speed of the girl,\({P_1}\)is the momentum of the girl, and\({P_2}\)is the momentum of the boy.

On plugging the values in the above relation, you get:

\(\begin{aligned}{c}\left( {20\;{\rm{kg}}} \right){v_{\rm{g}}} = \left( {40\;{\rm{kg}}} \right)\left( {1\;{\rm{m/s}}} \right)\\{v_{\rm{g}}} = 2\;{\rm{m/s}}\end{aligned}\)

Thus, \({v_{\rm{g}}} = 2\;{\rm{m/s}}\) is the speed of the girl in the opposite direction.