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A $$65.0-kg$$ boy and his $$40.0-kg$$ sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backward with velocity $$2.90 m/s$$ toward the west. Ignore friction.
Describe the subsequent motion of the girl.

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Solution
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If we consider the boy and the girl as a system then there is no any external force in horizontal direction. The force with which the girl pushes the boy and the reaction force on the girl by the boy (Newton’s third law) are internal forces; hence the momentum of the system remains conserved. As the reaction force on the girl is in opposite direction the girl will be pushed backward and hence will move in backward direction.
Let the mass of the boy is m1 = 65.0 kg and that of the girl is m2 = 40.0 kg.
the velocity with which the boy is pushed is v1 = 2.90 m/s and the velocity of the girl after the push is v2.
As both are initially at rest, initial momentum of the system will be zero. Momentum of the system after push will be
Momentum of boy + momentum of girl = m1v1 + m2v2 According to law of conservation of momentum in horizontal direction
Final momentum = initial momentum Or m1v1 + m2v2 = 0 Gives 1 1 2 2 65.0 * 2.90 4.71 40 m v v m       m/s Negative sign means that the velocity of the girl is in negative direction. Hence after push the girl will move towards east with a velocity of 4.71 m/s.

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