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A small space probe of mass 220 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.5 seconds after it is launched, the probe is at location <4600, 7700, 0> m, and at this same instant its momentum is <47000, -7000, 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 <-4100, -780, 0> N.
1. 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 22.2 seconds after the probe is launched to 22.6 seconds after the launch?
2. What is the the momentum of the probe at time 22.6 seconds after launch?

Respuesta :

akande212

Answer:

(1) The change in momentum of the space probe between the time t = 22.2 to 22.6 is (-1640, -312, 0) kgm/s

(2)The momentum at t = 22.6s is (46590, 6922, 0)

Explanation:

From Newtons second law the change in momentum us equal to impulse with is the product of force and time.

All the given quantities were grouped into the (x, y, z) axis and each of them treated separately.

Change in momentum = net force × change in time.

(2) the momentum at t=22.6s is equal to the sum of the momentum at 22.5s plus the change in momentum from t = 22.5s to t = 22.6s.

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ajeigbeibraheem

The change of the momentum of the probe in the time interval is [tex]\mathbf{ \Big \langle -1640,312,0 \Big \rangle \ kg.m/s}[/tex]

The momentum of the probe  [tex]\mathbf{ \Big \langle 45360,-7312,0 \Big \rangle \ kg.m/s}[/tex]

What is the change of momentum?

The change of momentum is the product mass and velocity of a body. it also depends on the force that acts in relation to the length of its time.

(1)

The change in the momentum is

[tex]\mathbf{\Delta p ^{\to} = F^{\to } \Delta t}[/tex]

The force during the interval is:

[tex]\mathbf{F^{\to} \Big \langle -4100, -780, 0 \Big \rangle N}[/tex]

  • The time interval:
  • Δt = 22.6 - 22.2
  • Δt = 0.4 s

[tex]\mathbf{ \Delta p^{ \to} = \Big \langle -4100,780 \Big \rangle(0.4 s ) }[/tex]

[tex]\mathbf{ \Delta p^{ \to} = \Big \langle -1640,-312,0 \Big \rangle \ kg.m/s}[/tex]

(2)

The change in the momentum can be computed by using the expression:

[tex]\mathbf{\Delta p^{\to} = \Delta p_{f}^{\to} -\Delta p_i^{\to}}[/tex]

Thus, the final momentum of the probe is:

[tex]\mathbf{\Delta p_{f}^{\to}= \Delta p^{\to} +\Delta p_i^{\to}}[/tex]

The initial momentum is:

[tex]\mathbf{p_i^{\to} = \Big \langle 47000, -7000,0 \Big \rangle \ kg.m/s}[/tex]

The change in the momentum is:

[tex]\mathbf{ \Delta p^{ \to} = \Big \langle -1640,-312,0 \Big \rangle \ kg.m/s}[/tex]

The final momentum is:

[tex]\mathbf{ p^{ \to} _f = \Big \langle -1640,-312,0 \Big \rangle + \Big \langle47000, -7000, 0 \Big \rangle \ kg.m/s}[/tex]

[tex]\mathbf{ p^{ \to} _f = \Big \langle 45360,-7312,0 \Big \rangle \ kg.m/s}[/tex]

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