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

Given that

The initial velocity of the astronut with respect to the stationis (Vias) =0.85m/s

Mass of the astronut (M) =62kg

A small drill of mass(m) =5kg

Astronut throws the drill with a speed of (vds) =10m/s

Now from the conservation of momentum we get

MVias =Mfvas +mvds

62*0.85=62*fvas+5*10

Then fvas =0.044m/s

b)

The work done by the astronut =KEi -KEf

=(1/2)*62*[(0.85)2 -(0.044)2] =22.34J

c)

Velocity of the dril with respect to the astronut is

Vda =vds-fvas =10-0.044 =9.956m/s

d)

Now from the conservation of linear momentum

62*0.85 =5*v

Then v =62*0.85/5 =10.54m/s moving in the same direction

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