Principles of conservation of energey

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SUMMARY

The discussion centers on the application of the principles of conservation of momentum and kinetic energy in a collision scenario involving two trains. A train with a mass of 50 tonnes traveling at 60 m/s collides with a stationary train of 30 tonnes, resulting in a post-collision velocity of 37.5 m/s. The kinetic energy before the collision is calculated to be 90 MJ, while the kinetic energy after the collision is 56.25 MJ, leading to a loss of kinetic energy of 33.75 MJ. The calculations confirm the conservation of momentum while highlighting the non-conservation of kinetic energy in inelastic collisions.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Knowledge of kinetic energy calculations
  • Familiarity with basic physics equations, specifically M1V1 + M2V2 = MfVf
  • Ability to perform unit conversions (tonnes to kilograms)
NEXT STEPS
  • Study inelastic collisions and their characteristics
  • Learn about energy loss in collisions and how to calculate it
  • Explore advanced topics in momentum conservation in multi-body collisions
  • Investigate real-world applications of momentum and energy conservation in transportation systems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of collisions and energy transformations in mechanical systems.

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Homework Statement



A train of mass 50 onnes traveling at 6o m/s collides head on into a stationary train which has a mass of 30 tonnes. If the two vehicles lock together after impact, calculate using the principles of conservation of momentum, what will be

a) their velocity after impact?
b) the loss of kenitec energy

i have had a attempt at A but i do no no where to start with B or what formula to use. this is my attempt at A)

Homework Equations



M1V1+M2V2=MfVf
M1= 50000kg
M2= 30000kg
V1 = 60m/s
V2 = 0 m/s

The Attempt at a Solution



MfVf = (50000*60)+(30000*0) = 3000000
3000000/ (50000+30000) = 37.5m/s

Could some one please help with Part b and help me with what formula to use
 
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Although you titled this "conservation of energy" I assume you realize that there is NO conservation of energy here. There is, however, conservation of momentum and that is what you used, correctly, to find the speed after the collision.

The kinetic energy before the collision is (1/2)mv^2= (1/2)(50)(60)^2. The kinetic energy after the collision is (1/2)(50+ 30)(37.5)^2= (1/2)(80)(37.5)^2. Subtract the energy after the collision from the kinetic energy after the collision to find the loss of kinetic energy.
 
hi hallsofivy thank you for the very prompt reply.

sorry about the title, that was a typo.

ive manged to work out before the collision as 90 MJ and after as 56.2 5MJ so the answer is 33.75 MJ?
 

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