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A 57.0-kg boy and his 43.0-kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backward with a velocity 3.00 m/s toward the west. Ignore friction.(a) Describe the subsequent motion of the girl. (b) How much potential energy in the girls body is converted into mechanical energy of the boy-girl system? (c) Is the momentum of the boy-girl system conserved in the pushing-apart process? If so, explain how that is possible considering (d) there are large forces acting and (e) there is no motion beforehand and plenty of motion afterward.

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akindelemf

Answer:

a) Since linear momentum IS conserved, the momentum of the girl is actually -4.7125*43 = -202.6 kg∙ m/s (opposite direction from the boy).

When this is added to the boy's momentum (+202.6), the result is zero.

This is expected since the initial momentum is also zero.  

b)  733.97 J  potential energy in the girls body is converted into mechanical energy of the boy-girl system

c) YES the momentum of the boy-girl system conserved in the pushing-apart process because the velocities (vectors) are opposite and the individual momentum therefore cancel leaving zero, the same as before the pushing.

Explanation:

b) ME = KE = ½ΣMV² = ½(57*3.0² + 43*4.7125²) = 733.97 J

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