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Problem 2.22 (Multistep) A small space probe of mass 160 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.7 seconds after it is launched, the probe is at location <4400, 7200, 0m, and at this same instant its momentum is <46000, -7600, 0> kg m/s. At this instant, thanat o on due to the gravitational pull of Mars plus the air resistance acting on the probe is -4200, -760, 0> N. Collapse question part Part 1 (a) Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.7 seconds after the probe is launched to 23.0 seconds after the launch? kg-mis By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Attempts: 0 of 9 used SAVE FOR LATER SUBMIT ANSWE Part 2 (b) What is the momentum of the probe at time 23.0 seconds after launch? kg m/s By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Attempts: 0 of 9 used SAVE TOR LATER SUBMIT ANSWE Part 3 (c) Assuming that the force is nearly constant during the time interval from 22.7 seconds after the probe is launched to 23.0 seconds after the launch, so that the velocity is changing at a constant rate, what is the change of the position of Phm. the probe during this time interval? Divide the time interval into two equal parts and use the approximation Vavg By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Attempts: 0 of 9 used SAVE FOR LATER SUBMIT ANSWE Part 4 (d) What is the location of the probe 23.0 seconds afte aunch? By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor

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The solution of the above problems as explained below..

and the formulas which are used in this problem as shown below.Given thob Ans t2- 23.0 kc . Diferuce Jn time Inteaval 23.0 s-24.1 s. de o.3 usn Meston s Second lou We noco hat net Given, fy -ヒ) 2 Py--16ON X 0.3 sec : Chame ofe nomentuor of the prok is, The momentum. of the p be at tineマ3,0kcís And we have 民ー-1260 m1600 am ke C) Caleulaken of Choge of the posihion ot the probe dungiven time Jntoel First e calcalate nitial and finol Velocihe X-and Y-akis. Tnihial Velociy alon x-ais is Pix may od probe6 ibal velocih al axi may a e Poe ,60 F veloiy x-axis, Oc mow of the P»be. 44, 740 g/se 160 may of the Pro be16 2 - 48,925 jc Finol Posin alongxa is, OSi hion may of te Probe. And jven 4400 m+ C237.fm/kXO. 38%) Co-5 x -2 6.25 m с 4400 m + 86.25m-3.9375m |X-4432.3mFinal 컹 l 6o 건. 15 m s 1200 m +((-41.5mb) xa3wj ec ør- 4432. 3-4400 7185-72000>mre the loakon o te pmoke 21.0 atter lhunch is, こ$468.48 m

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