A) A 46.1 kg girl is standing on a 169 kg plank. The plank, originally at...

A) A 46.1 kg girl is standing on a 169 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 1.61 m/s relative to the plank. What is her velocity relative to the ice surface?

B) What is the velocity of the plank relative to the ice surface?

Solutions

Expert Solution

The velocity of the girl relative to the ice is equal to the sum of the velocity of the

girl relative to the plank and the velocity of the plank relative to the ice.

From the law of conservation of momentum, we have

Then, the velocity of the plank relative to the ice is,

Hence, the velocity of the girl relative to the ice is,

$v_{gi} = 1.61\textup{\mathrm{} m/s} -\left ( \frac{m_{g}}{m_{p}} \right )v_{gi}$

After solve, we get

(B)

The velocity of the plank relative to the ice is,

thus, the velocity is equal to 0.344 m/s and is directed opposite to the motion of the girl.

genius_generous answered 1 year ago

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