Did i do this Kinetic Energy lost Collision Problem Right?

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The discussion focuses on a kinetic energy loss collision problem involving two objects with initial momentums at right angles. The user initially miscalculated the total mass, using 140 kg instead of 0.14 kg, which affected the final velocity and kinetic energy calculations. After correcting the mass, the user recalculated the final velocity to be 35.7 m/s and determined the kinetic energy lost to be 85.7 J. The importance of using proper units and vector components in momentum calculations was emphasized for accuracy. Overall, the corrections led to a more accurate understanding of the collision dynamics.
Lori

Homework Statement



upload_2017-12-16_17-20-11.png

Homework Equations


Total Ke initial - Total Ke final = lost KE
P= m1v1 + m2v2

The Attempt at a Solution


Initial P : sqrt((0.08*50)^2 + (0.06+50)^2)
Final P : (m1+m2)vf = 140vf
Initial P = Final P and solve for vf , vf = 0.0357

KE initially : .5(0.080)(50^2) + .5(0.060)(50^2)
Ke final: .5(140)(0.0357)^2

Lost KE = KE initial - KE final = ~174964 KJ = 175 J
 

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(Editing: This first paragraph you can skip)..This one is somewhat complicated because their initial momentums are each at right angles to each other. You need to first compute the complete vector (having an ## x ## and ## y ## component), for the combined initial momentum, and also compute the amplitude of that vector. You then need to set the initial momentum equal to the final momentum,(because of conservation of momentum), from which you can determine a final velocity vector for the objects that are stuck together. After that, you can work on finding any changes in kinetic energy. ## \\ ## For starters, can you write out the initial total momentum vector ## \vec{p} ##? ## \\ ## Editing: I see you did some of that already=it was a little hard to read your solution=let me have a second look at it... ## \\ ## I spotted one error: Your final ## m_{total}=.14 ## kg, (not 140 because you need to stay in M.K.S.) ## v_f ## will thereby be much larger. (Instinctively, ## v_f ## should be in the 50 m/s range=maybe 35 m/sec, but somewhere in that ballpark). In writing out the final momentum, and also the final kinetic energy, you incorrectly used ## m=140 ## in both cases. ## \\ ## Additional item: Please show your computed results for ## K.E._{initial} ## and ## K.E._{final} ##. It will help in checking the arithmetic.
 
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Charles Link said:
(Editing: This first paragraph you can skip)..This one is somewhat complicated because their initial momentums are each at right angles to each other. You need to first compute the complete vector (having an ## x ## and ## y ## component), for the combined initial momentum, and also compute the amplitude of that vector. You then need to set the initial momentum equal to the final momentum,(because of conservation of momentum), from which you can determine a final velocity vector for the objects that are stuck together. After that, you can work on finding any changes in kinetic energy. ## \\ ## For starters, can you write out the initial total momentum vector ## \vec{p} ##? ## \\ ## Editing: I see you did some of that already=it was a little hard to read your solution=let me have a second look at it... ## \\ ## I spotted one error: Your final ## m_{total}=.14 ## kg, (not 140 because you need to stay in M.K.S.) ## v_f ## will thereby be much larger. (Instinctively, ## v_f ## should be in the 50 m/s range=maybe 35 m/sec, but somewhere in that ballpark). In writing out the final momentum, and also the final kinetic energy, you incorrectly used ## m=140 ## in both cases. ## \\ ## Additional item: Please show your computed results for ## K.E._{initial} ## and ## K.E._{final} ##. It will help in checking the arithmetic.
Thank you. I always forget that they give us the mass in grams. I did the problem again and calculated a new vf. So the lost in KE is 85.7. I used .140 grams instead of 140 grams and got vf= 35.7
 
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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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