An astronaut (mass 62.0kg) loses connection to the international space station,

Question

An astronaut (mass 62.0kg) loses connection to the international space station, and is moving at a velocity of 0.85 m/s away from the station. The astronaut holds a small drill, mass m = 5.0 kg. The astronaut throws the drill at a velocity of 10.0 m/s in the direction she is moving. (Important: This is the velocity with respect to the space station, NOT with respect to the astronaut.) What is the velocity of the astronaut after she throws the drill? (With respect to the station.) What is the net work done by the astronaut? What is the velocity of the drill, with respect to the astronaut? If she instead wanted to just stop moving with respect to the station, at what velocity should she have thrown the drill?

Explanation / Answer

here,

mass of asternaut , ma = 62 kg

mass of drill , md = 5 kg

velocity of drill , vd = 10 m/s

velocity of asternaut , va = 0.85 m/s

(a)

let the velocity of asternaut be v'

change in momentum = 0

ma * ( v' - (-va)) - md * vd = 0

62 * ( v' + 0.85) - 5 * 10 = 0

v' = - 0.0435 m/s

the speed of asternaut is 0.0435 m/s

(b)

the net work done by asternaut , W = change in kinetic energy

W = 0.5 * ( md + ma) * va^2 - 0.5 * ma * v'^2 - 0.5 * md * vd^2

W = 0.5 * ( 5 + 62) * 0.85^2 - 0.5 * 62 * 0.0435^2 - 0.5 * 5 * 10^2

W = -225.85 J

(c)

the velocity of drill w.r.t asternaut , v = vd - va

v = 10 - 0.85 = 9.15 m/s

(d)

if it has to just stop moving

let the speed of dill be vd'

using conservation of momentum

md * vd = ma* va

5 * vd = 62 * 0.85

vd = 10.54 m/s

the velocity of drill to stop her is 10.54 m/s

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