A small space probe of mass 260 kg is launched from a spacecraft near Mars. It t

Question

A small space probe of mass 260 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.6 seconds after it is launched, the probe is at location m, and at this same instant its momentum is kgmiddotmiddot m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is -4700, -780, 0 > N. 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 22.6 seconds after the probe is launched to 22.8 seconds after the launch? Delta p = kg middot m/s What is the momentum of the probe at time 22.8 seconds after launch? Delta r = kg middot m/s

Explanation / Answer

here the time interval is 22.8-22.6 = 0.2 sec

by newton's second law of motion

Fnet = dp / dt

Fx = dPx/dt

dPx = Fx*dt = -4700*0.2 = -940 kg m/sec

dpy = Fy*dt = -780*0.2 = -156 kg m/sec

delta p = < -940 , -156 ,0 >

dp = sqrt(940^2+156^2) = 953 kg m/sec

at 22.8 sec

dpx = pfx - pix

pfx = dpx + pix = -940+47000 = 46060 kg m/sec

pfy = dpy + piy = -156-7900 = -8056 kg m/sec

pf = < 46060 , -8056 ,o > kg m/sec

pf = sqrt(46060^2+8056^2) = 46759 kg m/sec

initial velocity along x-axis is vix = 47000/260 = 181 m/sec

initial velocity along y-axis is viy = -7900/260 = -30.4 m/sec

final velocity along X-axis is vfx = 46060/260 = 177.15 m/sec

final velocity along Y-axis is vfy = -8056/260 = -31 m/sec

final position along x-axis is x = xo + (vix*t)+(0.5*ax*t^2)

ax = Fx/m = -4700 / 260 = 18.1 m/s^2

then

x = 5500+(181*0.2)+(0.5*18.1*0.2^2)

x = 5536.56 m

final position along y-axis is

y = yo + (viy*t)+(0.5*ay*t^2)

ay = Fy/m = -780/260 = -3 m/s^2

then

y = 8800-(30.4*0.2)-(0.5*3*0.2^2)

y = 8794 m

delta r = < 5536-5500 , 8794-8800,0 > m

delta r = < 36 , -6,0 >

delta r = sqrt(36^2+6^2+0^2) = 36.5 m

rf = < 5536 , 8794 ,0 >

rf = sqrt(5536^2+8794^2) = 10391.5 m

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