# Newton's 3rd law

1. Nov 16, 2006

### sunbunny

I'm having troubles with this problem:

An 80.0 kg spacewalking astronaut pushes off a 620kg satellite, exerting a 100N force for the 0.590s it takes him to straighten his arms.How far apart are the astronaut and the satellite after 1.20min ?

I know that i need to somehow set it up to be a kinematics problem and that i need to find the acceleration and how fast the astronaut and the satellite are moving.

I tried to use F=ma where a=100N/80kg and then the same for the orbital

for the astronaut i got the acceleration to equal 1.25m/s^2
for the satellite a=0.1613 m/s^2
and the change in time used was 72-0.59s=71.41s

I then put these into xf=xi +vi(t)+.5at^2 to find the distances

I then added their distances together to get the total distance between them however this way that I did it was wrong.

I got 3187.1m (astronaut) +411(orbit)

any feedbaclk would be great

2. Nov 16, 2006

### PhanthomJay

The astronaut and satellite are accelerating only when the force is applied over the 0.59 second period. After that, once contact is gone, they both must move at constant velocity, per newton's first law.

3. Nov 16, 2006

### sunbunny

Okay, thank you.

So if they are only accelerating during the 0.59s how would i go about finding the acceleration for them during this time interval?

I was using F=ma but this formula doesn't have time in it. How would you suggest that I go about this?

4. Nov 16, 2006

### PhanthomJay

Use F(delta t) =m(delta v)=m(v_f -v_i). That's the same as F=ma.

5. Nov 16, 2006

### sunbunny

Thank you so much!!!from the equation you gave me, i found the velocity of the astronaut to be 0.7373m/s and the satellite to be 0.095167m/s. from there, i put these velocities into:

xf=vi(delta t) and then I added the two distances and got the answer
59.5m. Thanks a alot I really appreciated it!

6. May 13, 2009

### brunettegurl

hi i have a similar question to this and when i tried to recreate your workings it wasnt working for me can u please explain to me what you did..thanks

7. May 13, 2009

### LowlyPion

Consider the force applied and the duration.

Applying the force to each mass results in a change in momentum - which gives you the speed of each.

The speeds are in opposite directions so simply add the speeds and determine the distance given the time.

8. May 15, 2009

### brunettegurl

so wld i be using the Fdeltat = mdeltav for each of the objects...using this velocity and the second time given in the question figure out the distance and add them together??

9. May 15, 2009

### LowlyPion

Basically yes.

Though I think of it more as determining their relative velocity first from the F*Δt on each and then applying the duration of their drift. (The magnitudes of the velocities add since they are in opposite directions,)

10. May 16, 2009

### brunettegurl

we assume for each object that the vintial is zero right?? so then wld the distance we get be a negative or a positive value??