# Homework Help: Astronaut floats outside spacecraft

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1. Mar 7, 2015

### HJ^2

1. The problem statement, all variables and given/known data
A 91-kg astronaut floats outside a 3,131 kg spacecraft. She's initially stationary with respect to the spacecraft. Then she pushes against the spacecraft, and moves away at 0.195 m/s to the left. Find the SPEED of the recoiling spacecraft.

2. Relevant equations
Force = Mass*Acc
Kinematics; Xf=Xi+Vit+(at^2/2)
Vf=Vi+at
Vf^2=Vi^2+2a(Xf-Xi)

3. The attempt at a solution
Used f=ma to calculate force of astronaut (-17.745). Used that number with Newtons 3rd law as force for space ship (17.745) and with f=ma found acceleration to be 0.00566

That is as far as I got!

2. Mar 7, 2015

### Staff: Mentor

Can you show how you used f=ma to find the "force of astronaut"?

3. Mar 7, 2015

### HJ^2

Mass of astronaut; 91kg
Acceleration; 0.195
f=ma
Solve for f
f=(91)*(0.195) = -17.745

4. Mar 7, 2015

### Staff: Mentor

Re-read the problem statement. Is 0.195 an acceleration? What units were given?

You'll need another approach. What conservation law might be applicable?

5. Mar 7, 2015

### HJ^2

Oh wait! Initially stationary. Starting with 0 m/s then final acceleration is 0.195 meters per second.

Would the conservation of momentum apply here? I haven't learned it yet - 2 chapters ahead from my current work.

6. Mar 7, 2015

### Staff: Mentor

No, the 0.195 figure is associated with units of velocity, m/s. Accelerations have the units m/s/s (that is, m/s2). You weren't given any time period or distance over which this change in speed took place, so you can't determine the actual acceleration that happened.
Conservation of momentum would definitely apply here and make things very simple. If you haven't learned about it yet then you'll have to "sneak up on it" somehow.

If you assume that whatever force was applied was constant for some unknown time interval Δt, and that Newton's 3rd law applies as you previously said, then you should be able to use Newton's 2nd (f = ma) and simple kinematics to find the speed of the spacecraft; You'll find that the Δt cancels out along the way.

7. Mar 7, 2015

### HJ^2

Ok, so I know initial and final velocities of both objects. I dont understand how I would be able to solve this problem with kinematics.

8. Mar 7, 2015

### Staff: Mentor

You know the final velocity of one of the objects. You want to find the final velocity of the other.

Imagine that you were given the acceleration a1 for the first object, and that the acceleration took place over a time interval Δt. What expression would you write to find the final velocity v1 for that object?

9. Mar 7, 2015

### HJ^2

Vf = Vi + at and solve for Vf, correct?

10. Mar 7, 2015

### Staff: Mentor

Well, that is the basic kinematic equation you would use. Good. How would you write it for the given problem in particular (use the variables that have been defined: v1, a1, Δt; and the given initial conditions).

11. Mar 7, 2015

### HJ^2

Would it be... Vf = Vi + a(Tf - Ti) ?

Mass wouldn't fit in anywhere and I can't find the speed and the time isn't indicated.

I'm really sorry I've looked at the same problem for an hour and most of this isn't getting to me at all.

12. Mar 7, 2015

### Staff: Mentor

We're going step by step here, so don't worry if you don't see all the given quantities at once.

The idea is to take the basic kinematic formula and use its variables as placeholders for the quantities that pertain to your particular problem.

Right now we're looking at the motion of the astronaut. What is Vi for the astronaut? If it's zero, drop it from the formula -- no need to carry along variables that are zero valued.

To distinguish variables that pertain to the two different objects, give them slightly different names. So I suggest designating the astronaut the first object and the spacecraft the second object, making the final velocity of the astronaut "v1" and its acceleration "a1". (Tf - Ti) is just Δt. No need to introduce more time variables like Ti and Tf.

Write the expression for the velocity v1 in terms of a1 and Δt.

13. Mar 7, 2015

### HJ^2

Vi for the astronaut is 0, got that.

The final velocity is 0.195 as stated in the problem? Wouldn't acceleration also be zero because theres no change of direction?

14. Mar 7, 2015

### Staff: Mentor

No numbers will be used until the end. Stick to variables for now. So use "v1" in place of 0.195 m/s.

The acceleration a1 cannot be zero because the astronaut goes from rest to the velocity v1. Acceleration does not always imply a change in direction (although a change in direction can sometimes be involved). Acceleration is any change in velocity.

15. Mar 7, 2015

### HJ^2

V1 = a1*Δt ? Since Vi is zero and is disregarded?

16. Mar 7, 2015

### Staff: Mentor

Yes! Excellent.

Now, Newton's 2nd law relates force to mass and acceleration. For the astronaut with mass m1 and acceleration a1 you have f = m1*a1, right?

Replace a1 in your V1 expression with a1 from Newton's 2nd. You should end up with an expression for V1 that involves force f, mass m1, and time Δt.

17. Mar 7, 2015

### HJ^2

as... in... v1=(f/m)*Δt ?

18. Mar 7, 2015

### Staff: Mentor

Yes. Remember to distinguish the different masses! Thus v1=(f/m1)*Δt .

While we're at it, write the similar expression for the spacecraft's velocity. The only difference is the mass, right? Newton's 3rd makes the force the same, and the time is the same for both, too.

19. Mar 7, 2015

### HJ^2

v2=(f/m2)*Δt ? But when solving.. there isn't any time so wouldn't it just be the force divided by the mass - respectively for each system?

20. Mar 7, 2015

### Staff: Mentor

Yes.
You mean there isn't a known time. If there were no time in the expression then you would not end up with a velocity -- the units would not be correct. Just hang on! All will be revealed.

Now. Return to the velocity expression for the astronaut. Isolate f from there to replace f in the spacecraft velocity expression. You should find that both "f" and "Δt" will disappear leaving only quantities that you have values for.

21. Mar 7, 2015

### HJ^2

SO... ok so if you isolate f1 in the first equation (v1 = f) and use it in place of f2 in the second... would it be V2=(m1/m2)? Is that right?

22. Mar 7, 2015

### Staff: Mentor

No, velocity is not a force. And mass divided by mass has no units (they cancel) so is not a velocity. Pay attention to units, they will tell you when things are going off track.

You want to isolate f from v1 = (f/m1)Δt. Don't drop any of variables! They can only disappear if they cancel somewhere, and that doesn't happen until after you complete the substitution into the v2 expression.

23. Mar 7, 2015

### HJ^2

v1=(f/m)*Δt --> f=(m/v1)*Δt?

24. Mar 8, 2015

### Staff: Mentor

Nope. Check your algebra. And keep the mass as "m1".

25. Mar 8, 2015

### HJ^2

f = (v/Δt) * m1 ?