Question

A small space probe, of mass 240 kg, is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will land. At a time 20.7 s after it is launched, the probe is at the location 4.30×10-, 8.70×100, 0 m, at at this same time its momentum is 4.40×102, −7.60×10-, 0 kg⋅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 −7×10-, −9.2×100, 0 N. Assuming that the net force on the probe is approximately constant over this time interval, what are the momentum and position of the probe 20.9 s after it is launched? Divide the time interval into two time steps, and use the approximation Vavg = Pf / m

Answer 1

Given that :

mass of small space probe, m = 240 kg

probe location, r = (4.3 x 10³, 8.7 x 10², 0) m

mometum, p = (4.4 x 10², -7.6 x 10³, 0) kg.m/s

net force on the probe, F_net = (-7 x 10³, -9.2 x 10², 0) N

initial time taken, t₁ = 20.7 sec

(a) Assuming that the net force on the probe is approximately constant over this time interval, the momentum and position of the probe after it is launched which given as :

final momentum of the probe, p_f = p_i + F_nett                                                           { eq.1 }

where, t = t₂ - t₁ = (20.9 - 20.7) sec = 0.2 sec

inserting the values in above eq.

p_f = [(4.4 x 10², -7.6 x 10³, 0) kg.m/s] + [(-7 x 10³, -9.2 x 10², 0) N] (0.2 sec)

p_f = [(4.4 x 10², -7.6 x 10³, 0) kg.m/s] + [(-1.4 x 10³, -1.84 x 10², 0) N.sec]

p_f = [(44000 - 1400), (-7600 - 184), (0 - 0] kg.m/s

p_f = (42600, -7784, 0) kg.m/s

p_f = (4.26 x 10⁴, -7.78 x 10³, 0) kg.m/s

And   Final location of the probe, r_f = r_i + v_avgt   

r_f = r_i + [(v_i + v_f) / 2] t   

where, p = m v

or   r_f = r_i + [p_i + p_f / 2m] t                                                                                      { eq.2 }

inserting the values in eq.2,

r_f = [(4.3 x 10³, 8.7 x 10², 0) m] + [(4.4 x 10², -7.6 x 10³, 0) kg.m/s + (4.26 x 10⁴, -7.78 x 10³, 0) kg.m/s / 2 (240 kg)] (0.2 sec)

r_f = [(4.3 x 10³, 8.7 x 10², 0) m] + [(43040, 15380, 0) / (480 kg)] (0.2 sec)

r_f = (4336, 864, 0) m

r_f = (4.33 x 10³, 8.64 x 10², 0) m