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A space probe of mass 400 kg drifts past location 0,3×104,-6×104m with momentum6×103,0,-3.6×103kg.m/s . Assuming the momentum of the probe does not change, what will be its position 2 minutes later?

Short Answer

Expert verified

The location of the space probe after 2 s is1.8×103,3×104,-6.108×104m.

Step by step solution

01

Identification of given data

The given data can be listed below,

  • The mass of the space probe is m=400kg
  • The initial location of the space probe is,r,⇶Ä=0,3×104,-6×104m
  • The momentum of the space probe is,p⇶Ä=6×103,0,-3.6×103m
02

Concept/Significance of linear velocity

The velocity relies on both the magnitude and direction. So, velocity is a vector quality. Speed, on the other hand, is a scalar number since its magnitude is all that matters.

03

Determination of position of space probe 2 minutes later

The velocity of the space probe is given by,

v⇶Ä=p⇶Äm

Here,pâ‡¶Ä is the momentum of the space probe and m is the mass of the space probe.

Substitute all the values in the above,

v⇶Ä=6×103,0,-3.6×103m400kg=15,0,-9m/s

The velocity of the space probe can also be expressed as,

v⇶Ä=rf⇶Ä-ri⇶Ä∆t

Here,râ‡¶Ä is the final location ofthe space probe,râ‡¶Ä is the initial location of thespace probeand∆t is the time elapsed.

Rearrange above expression for final location.

rf⇶Ä=ri⇶Ä+v⇶Ä∆t

Substitute all the values in the above expression.

rf⇶Ä=0,3×104,-6×104m+15,0,-9m/s2s=0,3×104,-6×104m+1.8×103,1.8×103m=1.8×103,3×104,6.108×104m

Thus, the location of the space probe after 2 s is1.8×103,3×104,-6.108×104m .

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