Question
Asked Dec 18, 2019
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A 45.0 - kg girl is standing on a 150. - 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 1.50 m/s to the right relative to the plank.
(a) What is her velocity relative to the surface of the ice? (b)
What is the velocity of the plank relative to the surface of the ice?

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Expert Answer

Step 1

Part A:

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Given Info: Mass of girl is mc = 45.0 kg, Mass of plank is mp = 15.0 kg, Initial velocity of the girl and plank is VGI and vpi = 0 and velocity of girl relative to plank is VGP = 1.50 m/s Explanation: The expression for velocity is + V pI VcI = VcP VGI is the Velocity of the girl relative to the ice VGP is the Velocity of the girl relative to the Plank vpi is the Velocity of the Plank relative to the ice Apply conservation of momentum to girl – plank system (тсvсI + mpupІ); — (тҫvсI + трург); mg is the mass of the girl. is the mass of the plank. тр

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Step 2
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Use the initial velocity to above equation to solve for equation for UPI ) VGI vPI = тр Rewrite the equation in terms vGI UCP UGI тс 1+ тр Substitute 1.50m/s for vcP, 45.0 kg for mc and 150.0 kg for mp. 1.50 m/s UGI 45.0kg 1+ 150.0 kg = 1.154 m/s = 1.15 m/s Conclusion: Thus the velocity of the girl relative to ice is 1.15 m/s.

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