QQuestionPhysics
QuestionPhysics
An 80.0 kg man is standing on a frictionless ice surface when he throws a 2.00 kg book at 10.0 m/s. With what velocity does the man move across the ice?
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Answer
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Step 1:: Write down the principle of conservation of momentum.
The total momentum of a closed system remains constant before and after an interaction, assuming no external forces act on the system. In this case, the man and the book form a closed system.
Step 2:: Express the principle of conservation of momentum mathematically.
m_{man} v_{man,i} + m_{book} v_{book,i} = m_{man} v_{man,f} + m_{book} v_{book,f}
Let m\_man be the mass of the man, m\_book be the mass of the book, v\_man,i be the initial velocity of the man, v\_book,i be the initial velocity of the book, v\_man,f be the final velocity of the man, and v\_book,f be the final velocity of the book. Then, the principle of conservation of momentum can be written as:
Step 3:: Identify the known and unknown quantities.
We know that m\_man = 80.0 kg, m\_book = 2.00 kg, v\_book,i = 10.0 m/s, and v\_book,f = 0 m/s (since the book comes to a stop after hitting the ice). We want to find v\_man,f, the final velocity of the man.
Step 4:: Solve for the unknown quantity.
v_{man,f} = \frac{{20.0~{\text{kg}} \times {\text{m/s}}}}{{80.0~{\text{kg}}}} = 0.25~{\text{m/s}}
Substitute the known quantities into the equation from Step 2: Simplify the equation: Divide both sides by 80.0 kg:
Final Answer
The man moves across the ice with a velocity of 0.25 m/s.
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