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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.

User Joe Buckle
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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.

Step-by-step explanation:

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

User BerriJ
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