A 53.0-kg boy and his 37.0-kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backward with a velocity 2.70 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 2.70 m/s.She will move east, directly away from her brother with a speed 3.87 m/s. She will move east, directly away from her brother with a speed 1.35 m/s.She will move east, directly away from her brother with a speed 5.22 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.")
Given that
The mass of the boy mb =53kg
The mass of the girl is mg =37kg
Final velocity of the boy is vfb =-2.70m/s
a)
Now from the law of conservation of momentum
mbvib+mgvig=mbvfb+mgvfg
0 =53kg*(-2.70)+37(vfg)
Now vfg is given by =143.1/37=3.867m/s
Therefore the speed of the girl is vgf =3.867m/s
b)
Now the potential energy in the girls body is the total kineitc energy of boy and girl
KE of the gril =(1/2)mgvg2 =0.5*37(3,867)2 =276.72J
KE of the boy =(1/2)mbvb2 =0.5*53(-2.70)2 =193.185J
Total KE =469.905J Which is nearly equal to =470J
c)
Here the momentum remains conserved because the according to the law of conservation of momentum when there is no external force acting on the system, the momntum remains conserved.
d)
If the external force acts on the system the momentum does not remains conserved
e)
if we consider the there is no motion beforehand and plenty of motion afterward also the momentum is not conserved.
Get Answers For Free
Most questions answered within 1 hours.