Question

In: Physics

A 62.0-kg boy and his 36.0-kg sister, both wearing roller blades, face each other at rest....

A 62.0-kg boy and his 36.0-kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backward with a velocity 3.10 m/s toward the west. Ignore friction.

(a) Describe the subsequent motion of the girl.

She will move west, in the same direction as her brother, with a speed 3.10 m/s.She will move east, directly away from her brother with a speed 6.89 m/s.    She will move east, directly away from her brother with a speed 5.34 m/s.She will move east, directly away from her brother with a speed 1.55 m/s.


(b) How much potential energy in the girl's body is converted into mechanical energy of the boy–girl system?
J

(c) Is the momentum of the boy–girl system conserved in the pushing-apart process?

YesNo    


(d) If so, explain how that is possible considering there are large forces acting. (If momentum is not conserved, enter "Momentum is not conserved.")

This answer has not been graded yet.



(e) If so, explain how that is possible considering there is no motion beforehand and plenty of motion afterward. (If momentum is not conserved, enter "Momentum is not conserved.")

Solutions

Expert Solution

a) She will move east, directly away from her brother with a speed 5.34 m/s

As the boy gets momentum in West direction, the girl gets the same amount of momentum in East direction.

let
m1 = 62 kg, m2 = 36 kg
v1 = 3.10 m/s

use, m2*v2 = m1*v1

v2 = m1*v1/m2

= 62*3.1/36

= 5.34 m/s

b) mechanical energy of the boy–girl system = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

= (1/2)*62*3.1^2 + (1/2)*36*5.34^2

= 811 J

c) Yes

d) Here the girl exert force on boy and the boy also exertes same amount of force in the opposite direction according newton's third law.

these two forces are internal forces. When extenral force acting on a system is zero, the total momentum of the system is conserved.

so, internal forces do not affect the total momentum of the system.


e) let +x be the East

initial momentum = 0


final momentum = m1*v1 + m2*v2

= 62*(-3.1) + 36*(5.34)

= 0

final momentum is also zero.

so, momentum is conserved.


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