1. Question Details SerCP 6.P.025 My Notes Ask Your An astronaut in her space su
ID: 2031605 • Letter: 1
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1. Question Details SerCP 6.P.025 My Notes Ask Your An astronaut in her space suit has a total mass of m1 = 71.6 kg, including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her oxygen tank of mass m2 12.0-kg away from her spacecraft with a speed v 7.70 m/s to propel herself back toward it (see figure). m, (a) Determine the maximum distance she can be from the craft and still return within 1.90 min (the amount of time the air in her helmet remains breathable) (b) Explain in terms of Newton's laws of motion why this strategy works. Show My Work (optional) Questian Details SerCPs &P022; My Nebes Ask Your A rifle with a weight of 40 N fires a 4.0 g bullet with a speed of 300 m/s (a) Find the recoil speed of the rifle. m/s (b) If a 650 N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle. m/sExplanation / Answer
momentum before throwing Pi = 0
momentum after throwing cylinder Pf = (m1-m2)*v1 + m2*v2
momentum is conserved
Pf = Pi
m1*v1 + m2*v2 = 0
(71.6-12)*v1 + 12*7.7 = 0
v1 = -1.55 m/s
speed of astronaut = 1.55 m/s
distance = v1*t = 1.55*1.9*60 = 176.7 m
(b)
from newton's third law of motion
to every action there is an equal and opposite reaction
here the external force Fext = 0
from newtons second law
Fext = m*a = 0
dP/dt = 0
dp = 0
change in momentum = 0
momentum remains constant
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momentum before firing Pi = 0
momentum after firing Pf = Mgun*Vgun + Mbullet*Vbullet = 0
from momentum conservation
total momentum is conserved
Pf = Pi
(40/9.8)*Vgun + 4*10^-3*300 = 0
Vgun = -0.294 m/s
speed of gun = 0.294 m/s
(b)
initial momentum of man+gun Pi = 0
final momentum of man+gun Pf = Mman*Vman + Mgun*Vgun
from momentum conservation
total momentum is conserved
Pf = Pi
Mman*Vman + Mgun*Vgun = 0
(650/9.8)*Vman - (40/9.8)*0.294 = 0
Vman = 0.02 m/s
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(a)
Impulse J = change in momentum
J = m*(vf - vi)
J = 82.7*(0 - 5.41) = -447.4 kg m/ s
(b)
force F = J/t = 596.5 N
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mommentum of base ball = momentum of bullet
Mball*Vball = Mbullet*Vbullet
0.15*Vball = 3.5*10^-3*1.5*10^3
speed of base ball = Vball = 35 m/s
(b)
kinetic energy K = P^2/(2m)
P = momentum
m = mass
here both have same momentum
smaller rhe mass greater the kinetic energy
bullet has greater kinetic energy
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