How Do You Calculate Kinetic Energy and Momentum for Spaceships in Motion?

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SUMMARY

The discussion focuses on calculating the kinetic energy and momentum of two spaceships: a Vulcan spaceship with a mass of 65,000 kg and a Romulan spaceship that is twice as massive. Both spaceships generate a force of 9.5 x 10^6 N for 100 seconds. The Vulcan spaceship's kinetic energy is calculated to be 6.93 x 10^12 J and its momentum is 9.49 x 10^8 kg m/s. An alternative method using impulse is suggested, yielding a slightly different final velocity and energy calculation, emphasizing the importance of accuracy in calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Familiarity with the equations for kinetic energy (K=1/2mv^2) and momentum (p=mv)
  • Basic knowledge of impulse and its relation to momentum
  • Concept of conservation of energy in physics
NEXT STEPS
  • Study the concept of impulse and its applications in physics
  • Learn about the conservation of energy and its implications in mechanical systems
  • Explore advanced calculations involving relativistic effects on momentum and energy
  • Investigate the differences between classical mechanics and relativistic mechanics in space travel
USEFUL FOR

Students in physics, aerospace engineers, and anyone interested in the dynamics of spacecraft motion and energy calculations.

yellowgators
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Homework Statement


A Vulcan spaceship has a mass of 65,000 kg and a Romulan spaceship is twice as massive. Both have engines that generate the same total force of 9.5 x 10^6 N. Neglecting any changes in mass due to whatever is expelled by the engines, calculate the kinetic energy and momentum of each spaceship if each spaceship fires its engines for 100s, starting from rest.


Homework Equations


K=1/2mv^2
p (momentum)=mv


The Attempt at a Solution


First I dealt with the Vulcan:
F=ma--> F/m=a 9.5 x 10^6 N/ 6.5 x 10^4 kg= 146 m/s^2
Vf-Vi=a(delta-t) Vf=146m/s^2(100s)=14600 m/s^2
K=1/2mv^2=1/2(6.5 x 10^4kg)(14600 m/s)^2= 6.93 x 10^12 J

p=mv=(6.5 x 10^4kg)(14600m/s)=9.49x10^8 kg m/s

I just need to know... Am I on the right track? Are there factors I'm not taking into account?
 
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yellowgators said:

Homework Statement


A Vulcan spaceship has a mass of 65,000 kg and a Romulan spaceship is twice as massive. Both have engines that generate the same total force of 9.5 x 10^6 N. Neglecting any changes in mass due to whatever is expelled by the engines, calculate the kinetic energy and momentum of each spaceship if each spaceship fires its engines for 100s, starting from rest.


Homework Equations


K=1/2mv^2
p (momentum)=mv


The Attempt at a Solution


First I dealt with the Vulcan:
F=ma--> F/m=a 9.5 x 10^6 N/ 6.5 x 10^4 kg= 146 m/s^2
Vf-Vi=a(delta-t) Vf=146m/s^2(100s)=14600 m/s^2
K=1/2mv^2=1/2(6.5 x 10^4kg)(14600 m/s)^2= 6.93 x 10^12 J

p=mv=(6.5 x 10^4kg)(14600m/s)=9.49x10^8 kg m/s

I just need to know... Am I on the right track? Are there factors I'm not taking into account?
I'm not checking your arithmetic, but your method seems good to me :approve:.
 
Thanks!
 
touche yellowgators!
I only have 1 reproche; the method that you used makes the percent of error higher than it schoul really be. In addition your method is time consuming. Try using this method next time:
1)P=F*t=(m*v1)-(m*v2) ****v1 is initial velocity and v2 is final velocity*******
The impulse P = 9.5 x 10^6 N * 100s
P=9.5 x 10^8 N*s

2) v2 is found by:
P=(m*v2)-(m*v1) *******v1 equals to 0 *******
P=m*v2
v2=9.5 x 10^8 N*s/6.5 x 10^4 kg = 14615.38m/s

3)Potential energy(PE) = F*d
PE=9.5 x 10^6 N* (14615.38m/s/2) *100s ****d=average velocity*time********
PE=6.94*10^12J

A difference of 0.01*10^12 might not seem much but look at it again :10000000000. Wow, that is a big number!. that number schould really be larger because i aslo rounded.
well, here you go
 
Last edited:
tiale11 said:
the method that you used makes the percent of error higher than it schoul really be. In addition your method is time consuming. Try using this method next time
I looked over that method. Thanks! Using the law of conservation of energy is more straightforward than how I approached the problem. That will be helpful for future problems.
 

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