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

AI Thread Summary
The discussion focuses on calculating the kinetic energy and momentum of two spaceships, a Vulcan and a Romulan, with given masses and engine force. The Vulcan spaceship has a mass of 65,000 kg, while the Romulan is twice as massive, and both generate a force of 9.5 x 10^6 N for 100 seconds. The initial calculations for the Vulcan's kinetic energy and momentum are presented, but a participant suggests a more efficient method using impulse to reduce error and save time. The conversation concludes with an acknowledgment that the alternative approach is more straightforward and will be useful for future calculations. Understanding these principles is essential for accurately determining kinetic energy and momentum in physics problems.
yellowgators
Messages
15
Reaction score
0

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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top