Solve for Man's Velocity on Ice After Throwing Book at 20 m/s

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An 80 kg man throws a 4 kg book at 20 m/s while standing on a frictionless ice surface. To find the man's velocity after the throw, the conservation of linear momentum is applied. Initially, both the man and the book are at rest, resulting in a total momentum of zero. After the throw, the equation 0 = (80 kg * V) + (4 kg * 20 m/s) is used to calculate the man's velocity. The final result shows that the man moves at -1 m/s across the ice.
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Velocity question?? Need help..

An 80.0 kg man is standing on a frictionless ice surface when he throws a 4.00 kg book at 20.0 m/s. WIth what velocity does the man move across the ice?


Would I be using Mass= weight/gravity?

Astronaut weight = 80.0 kg
Astronaut v = ? m/s

Book weight= 4.00 kg
Book v= 20.0 m/s
 
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terpsgirl said:
An 80.0 kg man is standing on a frictionless ice surface when he throws a 4.00 kg book at 20.0 m/s. WIth what velocity does the man move across the ice?


Would I be using Mass= weight/gravity?

Astronaut weight = 80.0 kg
Astronaut v = ? m/s

Book weight= 4.00 kg
Book v= 20.0 m/s

Apply the law of linear momentum.
Before the throw: Man mass = 80kg, velocity = 0, Book mass = 4kg, velocity = 0.
After the throw: Man mass = 80kg, velocity = V. Book mass = 4kg, velocity = 20.

M1U1 + M2U2 = M1V1 + M2V2
(80x0) + (4x0) = (80V) + (4x20)
0=80V + 80
80V = -80
V = -80/80
V= -1m/s.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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