Collision with an unknown mass and velocity after

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A 2 kg block slides on an icy surface at 3 m/s towards a static block of mass M, colliding and reversing direction to -1 m/s. The total kinetic energy remains constant during the collision. The calculations yield a mass M of 4 kg and a final velocity of 2 m/s for mass M after the collision. The solution approach involved applying conservation of momentum and kinetic energy equations. The results were confirmed as correct, addressing concerns about potential calculation errors.
fantisism
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Homework Statement


You slide a 2 kg block on a flat icy surface towards a static block of mass M. Before the collision, the 2 kg mass is moving at 3 m/s. The total kinetic energy is not changed by the collision. After the collision, the 2 kg mass is moving at −1 m/s i.e. opposite to its original direction. Calculate the mass M in kg and the velocity of the mass M after the collision.

Homework Equations


Eq 1.) 0.5mv1i2+0.5Mv2i2=0.5mv1f2+0.5Mv2f2
Eq 2.) mv1i+Mv2i=mv1f+Mv2f

The Attempt at a Solution


I got v2f=2 m/s and M=4 kg. I solved for M for the second equation and substituted that in into the first equation.
 
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fantisism said:

Homework Statement


You slide a 2 kg block on a flat icy surface towards a static block of mass M. Before the collision, the 2 kg mass is moving at 3 m/s. The total kinetic energy is not changed by the collision. After the collision, the 2 kg mass is moving at −1 m/s i.e. opposite to its original direction. Calculate the mass M in kg and the velocity of the mass M after the collision.

Homework Equations


Eq 1.) 0.5mv1i2+0.5Mv2i2=0.5mv1f2+0.5Mv2f2
Eq 2.) mv1i+Mv2i=mv1f+Mv2f

The Attempt at a Solution


I got v2f=2 m/s and M=4 kg. I solved for M for the second equation and substituted that in into the first equation.
What is your question then? That result is correct.
 
ehild said:
What is your question then? That result is correct.
Oh sorry. I just wanted to make sure those were right because I tend to make small simple math errors. Thank you!
 
You are welcome:oldsmile:
 
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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