Can the Crew's Plan to Fragment an Asteroid Succeed?

  • Thread starter Thread starter phalanx123
  • Start date Start date
  • Tags Tags
    Asteroid
AI Thread Summary
The crew's plan to fragment the asteroid involves a rocket colliding with it to generate sufficient energy. The rocket has a kinetic energy of 1.25 million joules, which initially seems adequate to break the rock. However, the rocket's speed must be calculated relative to the asteroid, not the space station, resulting in a relative speed of 1100 m/s. This adjustment is crucial for accurately determining the kinetic energy available for fragmentation. The discussion highlights the importance of considering relative velocities in collision scenarios.
phalanx123
Messages
29
Reaction score
0
Hi I got this question, it looks like its simple. But I don't know if I've done it right or not? Please help me check it.

A 2kg rock is heading straight for a space station at a speed of 100 m/s. The crew plan to break the rock up into small fragments by hitting it with a rocket of 3kg traveling at 1000 m/s. If it takes 1.0 *10^6 J to fragment the rock, show wether or not their plan can succeed. (Speed are given in the station' s frame.



I simply said that the rocket will have a kinetic energy of 1/2mv^2=1/2*3*1000^2=1.25*10^6 J. SO this is higher than the minimum energy required to break the rock, so the plan will work.

But it is so simple I think I've missed some information heere. Could somebody give me a little help please?:confused: Thank you
 
Physics news on Phys.org
Although I imagine the answer would still be "yes", not that the 1000 m/s you use for the rocket's velocity is relative to the space station not the asteroid. Of course, it is the rocket's velocity relative to the asteroid that will be converted into energy to break it apart. If the rocket is going 1000 m/s directly away from the space station, and the asteroid is going 100 m/s directly toward the space station, what is the rocket's speed relative to the asteroid?
 
Oh I see, so I have to use 1100 m/s instead of 1000 m/s. Ah that makes more sense. Thank you very much. I am also wondering if it's correct to do it this way: calculate the rocket's kinetic energy relative to the station first then take away the kinetic energy of the rock relative to the station from the rocket's kinetic energy? Anybody got any ideas? Thanks.
 
Last edited:
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