Did i do this Kinetic Energy lost Collision Problem Right?

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SUMMARY

The forum discussion centers on solving a kinetic energy loss collision problem involving two objects with masses of 0.08 kg and 0.06 kg, both initially moving at 50 m/s. The user initially miscalculated the total mass as 140 kg instead of 0.14 kg, leading to an incorrect final velocity (vf). After correcting the mass, the final velocity was recalculated to be 35.7 m/s, resulting in a lost kinetic energy of 85.7 J. The discussion emphasizes the importance of using consistent units and the conservation of momentum in collision problems.

PREREQUISITES
  • Understanding of kinetic energy formulas, specifically KE = 0.5 * m * v^2
  • Knowledge of momentum conservation principles in collisions
  • Familiarity with vector components in physics
  • Ability to perform calculations in the M.K.S. (Meter-Kilogram-Second) system
NEXT STEPS
  • Review the conservation of momentum in elastic and inelastic collisions
  • Learn how to decompose vectors into their x and y components
  • Study the effects of mass and velocity on kinetic energy changes
  • Practice solving collision problems with varying angles and masses
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision problems, as well as educators looking for examples of kinetic energy and momentum conservation in real-world scenarios.

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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