Asked by Elisa
A 45.0 kg girl is standing on a 163 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.45 m/s relative to the plank.
(a) What is her velocity relative to the surface of ice?
(b) What is the velocity of the plank relative to the surface of ice?
(a) What is her velocity relative to the surface of ice?
(b) What is the velocity of the plank relative to the surface of ice?
Answers
Answered by
Damon
If the girl is walking at 1.45 m / s in the x direction relative to the plank and the plank is moving at velocity V in the x direction then her total velocity is 1.45 + V in the x direction. (note - V better come out negative, in the negative x direction in part b)
There are no external horizontal forces on this system.
Therefore there will be no change in the initial horizontal momentum, which is zero.
final momentum:
0 = 45 * (1.45+V) + 163 V
0 = 65.25 + 45 V + 163 V
208 V = - 65.25
V = -0.314 m/s
by the way that means the girl is going at 1.45 -.314 = 1.14 relative to the ice and as a check in the ice coordinate system
0 = -0.314 * 163 + 1.14 * 45 ???
= -51.2 + 51.3 close enough
There are no external horizontal forces on this system.
Therefore there will be no change in the initial horizontal momentum, which is zero.
final momentum:
0 = 45 * (1.45+V) + 163 V
0 = 65.25 + 45 V + 163 V
208 V = - 65.25
V = -0.314 m/s
by the way that means the girl is going at 1.45 -.314 = 1.14 relative to the ice and as a check in the ice coordinate system
0 = -0.314 * 163 + 1.14 * 45 ???
= -51.2 + 51.3 close enough
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