Question

A small space probe, of mass\(240\;{\rm{kg}}\)is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will land. At a time\(20.7\;{\rm{s}}\)after it is launched, the probe is at the location\(\left\langle {4.30 \times {{10}^3},8.70 \times {{10}^2},0} \right\rangle {\rm{m}}\)and at this same time its momentum is\(\)\(\left\langle {4.40 \times {{10}^4}, - 7.60 \times {{10}^3},0} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\)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\(\left\langle { - 7 \times {{10}^3}, - 9.2 \times {{10}^2},0} \right\rangle \;{\rm{N}}\)Assuming that the net force on the probe is approximately constant over this time interval, what are the momentum and position of the probe\(20.9\;{\rm{s}}\)after it is launched? Divide the interval into two-time steps, and use the approximation\({\vec v_{{\rm{avg}}}} \approx {\vec p_f}/m\)

Short answer

The required final momentum is\(\left\langle {4.26 \times {{10}^4}, - 7.784 \times {{10}^3},0} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\)andthe final position is\(\left\langle {4.34 \times {{10}^3},8.64 \times {{10}^2},0} \right\rangle \;{\rm{m}}\)

Step-by-step solution

Given

Mass \(240\;{\rm{kg}}\)

At given time location and momentum are given by \(20.7\;{\rm{s}}\)

\(\left\langle {4.30 \times {{10}^3},8.70 \times {{10}^2},0} \right\rangle {\rm{m}}\)

\(\left\langle {4.40 \times {{10}^4}, - 7.60 \times {{10}^3},0} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\)

Resistance acting on the probe is \(\left\langle { - 7 \times {{10}^3}, - 9.2 \times {{10}^2},0} \right\rangle \;{\rm{N}}\)

The concept of momentum and impulse

The product of an object's mass and its speed is defined as the momentum expression.

\(p = mv\)

The mass of the item is\(m\), and the speed of the object is\(v\).

The product of force\(F\)and time is defined as the impulse, which is stated as follows:

\(I = {F_{net}}\Delta t\)

Where,\(\Delta t\)is the time taken.

Find the value of final momentum by applying change in momentum formula

The change in momentum can alternatively be defined as an impulse:\(I = {p_f} - {p_i}\)

The ultimate momentum is\({p_f}\), while the beginning momentum is.

The time interval is now as follows to determine the final momentum:\(\begin{aligned}{c}\Delta t = (20.9\;{\rm{s}}) - (20.7\;{\rm{s}})\\ = 0.2\;{\rm{s}}\end{aligned}\)

The required time period can be divided into two halves, one for each stage.\(0.1\;\;{\rm{s}}\).

Rearrange the equation\(I = {p_f} - {p_i}\)for\({p_f}\)and substitute\({F_{{\rm{net }}}}\Delta t\)for\(I\)

\(\begin{aligned}{c}I &= {p_f} - {p_i}\\{p_f} &= {p_i} + I\\ &= {p_i} + \left( {{F_{{\rm{net }}}}\Delta t} \right)\end{aligned}\)

Substitute\( < - 7 \times {10^3}, - 9.2 \times {10^2},0 > {\rm{N}}\)for\({F_{{\rm{net }}}}\),\( < 4.40 \times {10^4}, - 7.60 \times {10^3},0 > {\rm{kg}} \cdot {\rm{m/s}}\)for\({p_i}\),and\(0.1\;\;{\rm{s}}\)for\(\Delta t\).

\(\begin{aligned}{c}{p_f} &= {p_t} + \left( {{F_{{\rm{net }}}}\Delta t} \right)n\\ &= \left( { < 4.40 \times {{10}^4}, - 7.60 \times {{10}^3},0 > {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}} \right) + \left( { < - 7 \times {{10}^3}, - 9.2 \times {{10}^2},0 > {\rm{N}}} \right)(0.1\;{\rm{s}})n\\ &= \left( { < 4.40 \times {{10}^4}, - 7.60 \times {{10}^3},0 > {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}} \right) + \left( { < - 0.7 \times {{10}^3}, - 0.92 \times {{10}^2},0 > {\rm{N}} \cdot {\rm{s}}} \right)\\ &= < 4.33 \times {10^4}, - 7.692 \times {10^3},0 > {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}\end{aligned}\)

Find the final position.

The approximation for given condition

\({v_{{\rm{avg }}}} = \frac{{{p_f}}}{m}\),

The expression for the position is,

\({r_f} = {r_i} + {v_{{\rm{avg }}}}\Delta t\)

Substitute\(\frac{{{p_f}}}{m}\)for\({v_{{\rm{avg}}}}\)

\({r_f} = {r_i} + \left( {\frac{{{p_f}}}{m}} \right)\Delta t\)

Substitute\(\left\langle {4.30 \times {{10}^3},8.70 \times {{10}^2},0} \right\rangle {\rm{m}}\)for\({r_i}\)\(\left\langle {4.33 \times {{10}^4}, - 7.692 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}\)for\({p_f}\),\(240\;{\rm{kg}}\)for\(m\), and\(0.1\;{\rm{s}}\)for\(\Delta t\).

\(\begin{aligned}{c}{r_f} &= {r_i} + \left( {\frac{{{p_f}}}{m}} \right)\Delta t\\ &= \left( {\left\langle {4.30 \times {{10}^3},8.70 \times {{10}^2}} \right\rangle ,0{\rm{m}}} \right) + \left( {\frac{{\left\langle {4.33 \times {{10}^4}, - 7.692 \times {{10}^3},0} \right\rangle \;{\rm{kg}} \cdot {\rm{m}}/{\rm{s}}}}{{(240\;{\rm{kg}})}}} \right)(0.1\;{\rm{s}})\\ &= \left( {\left\langle {4.30 \times {{10}^3},8.70 \times {{10}^2}} \right\rangle ,0{\rm{m}}} \right) + \left( {\left\langle {0.018 \times {{10}^3}, - 0.032 \times {{10}^2},0} \right\rangle {\rm{m}}} \right)\\ = \left\langle {4.318 \times {{10}^3},8.668 \times {{10}^2}} \right\rangle ,0{\rm{m}}\end{aligned}\)

The final position and momentum of the previous step is now the initial position and momentum for the second step of\(0.1\;{\rm{s}}\),Then it follows:

Substitute\(\left\langle { - 7 \times {{10}^3}, - 9.2 \times {{10}^2},0} \right\rangle {\rm{N}}\)for\({F_{{\rm{net }}}}\),\(\left\langle {4.33 \times {{10}^4}, - 7.692 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}\)for\({p_t}\),and\(0.1\;{\rm{s}}\), for\(\Delta t\).

\(\begin{aligned}{c}{p_f} &= {p_f} + \left( {{F_{net}}\Delta t} \right)n\\ &= \left( {\left\langle {4.33 \times {{10}^4}, - 7.692 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m/s}}} \right) + \left( {\left\langle { - 7 \times {{10}^3}, - 9.2 \times {{10}^2},0} \right\rangle {\rm{N}}} \right)(0.1\;{\rm{s}})\\ &= \left( {\left\langle {4.33 \times {{10}^4}, - 7.692 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m/s}}} \right) + \left( {\left\langle { - 0.7 \times {{10}^3}, - 0.92 \times {{10}^2},0} \right\rangle {\rm{N}} \cdot {\rm{s}}} \right)\\ &= \left\langle {4.26 \times {{10}^4}, - 7.784 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m/s}}\end{aligned}\)

Thus, the required final momentum is \(\left\langle {4.26 \times {{10}^4}, - 7.784 \times {{10}^3},0} \right\rangle \;{\rm{kg}} \cdot {\rm{m/s}}\)

Find the value of final position

Let Substitute\(\left\langle {4.318 \times {{10}^3},8.668 \times {{10}^2},0} \right\rangle {\rm{m}}\)for\({r_i}\),\(\left\langle {4.26 \times {{10}^4}, - 7.784 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m/s}}\)for\({p_f}\)\(240\;{\rm{kg}}\)for\(m\),and\(0.1\;{\rm{s}}\)for\(\Delta t\),\(\left\langle { - 1.37,0,0.96} \right\rangle {\rm{m}}\)for\({r_f}\),in\({r_f} = {r_i} + \left( {\frac{{{p_f}}}{m}} \right)\Delta t\)

\(\begin{aligned}{c}{r_f} &= {r_i} + \left( {\frac{{{p_f}}}{m}} \right)\Delta t\\ &= \left( {\left\langle {4.318 \times {{10}^3},8.668 \times {{10}^2},0} \right\rangle {\rm{m}}} \right) + \left( {\frac{{\left\langle {4.26 \times {{10}^4}, - 7.784 \times {{10}^3},0} \right\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}}}{{(240\;{\rm{kg}})}}} \right)(0.1\;{\rm{s}})\\ &= \left( {\left\langle {4.318 \times {{10}^3},8.668 \times {{10}^2},0} \right\rangle {\rm{m}}} \right) + \left( {\left\langle {0.0187 \times {{10}^3}, - 0.0324 \times {{10}^2},0} \right\rangle {\rm{m}}} \right)\\ &= \left\langle {4.34 \times {{10}^3},8.64 \times {{10}^2},0} \right\rangle \;{\rm{m}}\end{aligned}\)Thus, the required final position is \(\left\langle {4.34 \times {{10}^3},8.64 \times {{10}^2},0} \right\rangle \;{\rm{m}}\)