Physics 30 question about conservation of momentum

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Homework Help Overview

The problem involves a scenario where a space person attempts to return to a spacecraft after throwing a phaser. The context centers around the conservation of momentum and the calculations related to the time and distance involved in the return journey.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum, questioning the values used for mass and velocity in the calculations. There are inquiries about specific calculations and the reasoning behind combining speeds.

Discussion Status

Some participants have provided guidance on unit conversions and suggested approaches to the problem. There is a mix of interpretations regarding the calculations, with differing answers being presented. The original poster expresses confusion about their results, while others share their own calculations and results.

Contextual Notes

Participants note the importance of unit consistency and question the assumptions made in the calculations. There is mention of the astronaut's mass including equipment, and the potential confusion arising from previous questions that may have influenced reasoning.

RRmy0440
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Homework Statement


A space person is motionless a distance of 500m away from the safety of the spacecraft . The person has exactly 11.32min of air left and the person's mass is 103.2kg, including equipment. The person throws a phaser at a velocity of 50.2km/h away from the spacecraft in order to get back. I f the mass of the phaser is 5.3 kg, how many minutes of extra air does the person have when they get back to the spacecraft ?(Ans: 0.28min)

I attempted to solve it , but get the answer of 0.25min.

please help!

T

Homework Equations


Conservation and momentum.
P=mv
Ft=mv
Impulse momentum theory

The Attempt at a Solution


① m1v1+m2v2=m1v1'+m2v2'
v1'=42.97m/min

②t=500/(836.9m/min+42.97m/min) =0.57min

③d=v1't=42.97m/min *0.57min=24.42m
④d(remaining)=500m-24m=475.6m
⑤t=d/v=475.6m/ (42.97m/min)=11.07min
⑥t(remaining)=11.32-11.07=0.25min

I try to solve the question, but got the wrong answer. Could anyone help me out? Thank you so much.
 
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RRmy0440 said:
m1v1+m2v2=m1v1'+m2v2'
What values are you using for m1 and m2 here?
RRmy0440 said:
②t=500/(836.9m/min+42.97m/min) =0.57min
Please explain this calculation.
 
I would leave it in km/h till the latest possible moment (unless you've been taught differently).
 
haruspex said:
What values are you using for m1 and m2 here?

Please explain this calculation.

d=500m
t1=11.32min
m1=103.2kg
m2=5.3kg
v2'=50.2km/h=836.7m/min
 
verty said:
I would leave it in km/h till the latest possible moment (unless you've been taught differently).
If you would like to solve it with the unit km/h , please feel free to do that.
 
So do you think I am on the right track of solving this?
 
It's up to you what way you want to solve it. By the way, I got a different answer: 0.31. I don't think 0.28 is correct.

PS. Your way of doing it is confusing to me so I won't comment anymore.
 
RRmy0440 said:
m1=103.2kg
That is the mass of the astronaut including all equipment.
I still do not see any explanation for your equations.

Also, why do you add the two speeds?
 
Last edited:
verty said:
I got a different answer: 0.31. I don't think 0.28 is correct.
I get 0.28. Check you did not get -0.31. If you did, see the bold text in my post #8.
 
  • #10
haruspex said:
I get 0.28. Check you did not get -0.31. If you did, see the bold text in my post #8.
Can you please explain how did you get the right answer?
 
  • #11
haruspex said:
That is the mass of the astronaut including all equipment.
I still do not see any explanation for your equations.

Also, why do you add the two speeds?

Does it mean that the phaser is also included to have the total mass of 103.2kg?

For the reason why I added the two speeds together...I think I had been misled by another question I just done
 
  • #12
Thank you guys, i get the right answer.
 

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