A small space probe of mass 260 kg is launched from a spacecraft near Mars. It travels toward the...
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 22.7 seconds after it is launched, the probe is at location < 4500, 7200, 0 > m, and at this same instant its momentum is < 41000, -8400, 0 > kg{eq}\cdot {/eq}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 < -4050, -780, 0 > N.
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 23.1 seconds after the launch?
b. What is the momentum of the probe at time 23.1 seconds after launch?
c. Assuming that the force is nearly constant during the time interval from 22.7 seconds after the probe is launched to 23.1 seconds after the launch, so that the velocity is changing at a constant rate, what is the change of the position of the probe during this time interval?
d. What is the location of the probe 23.1 seconds after launch?
Second law of Newton
Newton's Second Law states that the sum of the external forces (net force) acting on a body is equal to the variation of its momentum over time:
{eq}\displaystyle \vec{F}_{net}=\sum_{i=1}^N\vec{F}_i=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t} {/eq}
In cases where the body mass is constant, the well-known expression is obtained:
{eq}\displaystyle \vec{F}_{net}=\sum_{i=1}^N\vec{F}_i=m\vec{a} {/eq}
Answer and Explanation: 1
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{eq}\displaystyle \vec{F}_{net}=\frac{\mathrm{d}\vec{p}}{\mathrm{d}t}\\ \displaystyle \vec{F}_{net}=constant\Rightarrow...
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