Conservation of momentum problem and kinematics ?

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SUMMARY

The discussion revolves around a conservation of momentum problem involving an 80 kg astronaut and a 2.0 kg piece of equipment thrown at 0.5 m/s. The astronaut calculates his recoil velocity using the formula Pf = Po, leading to a recoil speed of 0.0125 m/s. The astronaut incorrectly applies kinematic equations to find the time to reach his spaceship, resulting in an answer of 1600 seconds, while the correct time is 800 seconds due to the uniform motion post-throw.

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mychellbella
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Conservation of momentum problem and kinematics!?

Hi, I've been working on this problem for the last three hours, I hope someone can help. In fact, the solution I keep getting is double the correct answer (the answers are in the back of the book, but not the how).
An 80 kg astronaut is stranded 10 m from his spaceship in free space. In order to get back to his ship, he throws a 2.0 kg piece of equipment w/ a speed of 0.5 m/s directly away from the ship. How long will it take for him to reach the ship?

man object
m1=80kg m2= 2.0kg
vo1=0 vo2= 0
vf1= ? vf2= .5m/s

so, i started by treating this like a gun and bullet problem since he will be recoiling.

Pf(momentum final) = Po (momentum orignal) = 0

Pf=m1vf1+ m2vf2
0= m1vf1 +m2vf2
vf1= (-m2/m1)*vf2
vf1=(-2.0kg/80kg)*.5m/s
vf1= .0125 m/s

okay, now i find the acceleration, so i can then use a kinematic equation to find the time. I only use the info for the man and vo is .0125 m/s and vf=0. x=10m.

a=(vf^2-vo^2)/2x
a=(0-.0125m/s^2)/2(10m)
a=-7.81*10^-6 m/s^2

t=(vf-vo)/a
t=(0-.0125m/s)/-7.81*10^6 m/s^2
t =1600 seconds

The problem, is that the book is saying the anwer is 800 seconds. What am I doing wrong! Please help! thank you!
 
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Hi mychellbella, welcome to PF.
No external force is acting on the astronaut in the direction of his motion. So after throwing the packet, he will move with the uniform velocity.
 

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