Question

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -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 <-4000, -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 22.9 seconds after the launch?

Answer 1

The change in the space probe's momentum over the 0.3 seconds interval is calculated by multiplying the net force by the time interval, resulting in a change of <-1200, -234, 0> kg·m/s.

Step-by-step explanation:

To calculate the change in momentum of the space probe, we can use the relationship between net force, change in momentum, and the time interval. The net force on the object is equal to the rate of change of its momentum (Newton's second law). Given that the force is approximately constant during the time interval, we can use the formula:

Change in Momentum (Δp) = Net Force (F) × Time Interval (Δt)

Using the net force <-4000, -780, 0> N and the time interval from 22.6 to 22.9 seconds, which is 0.3 seconds, we can calculate the change in momentum as follows:

Δp = F × Δt = <-4000, -780, 0> N × 0.3 s = <-1200, -234, 0> kg·m/s

Therefore, the change in the probe's momentum over this time interval is <-1200, -234, 0> kg·m/s.

Answer 2

The change in momentum is
`\Delta p = <-1200 , -234 , 0> kg \cdot m/s`

Step-by-step explanation:

From the question we are told that

The mass of the probe is
`m = 170 kg`

The location of the prob at time t = 22.9 s is
`A = <5600, 7200,0>`

The momentum at time t = 22.9 s is
`p = < 51000, -7000, 0> kg m/s`

The net force on the probe is
`F = <-4000 , -780 , 0> N`

Generally the change in momentum is mathematically represented as

`\Delta p = F * \Delta t`

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

`\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)`

`\Delta p = <-4000 , -780 , 0> * (0.3)`

`\Delta p = <-1200 , -234 , 0> kg \cdot m/s`