91影视

A property of matter by which it remains at rest or in uniform motion in the same straight line unless acted upon by some external force.

Angular momentum is a property of objects which are changing the angle of their position vector with respect to a reference point.

Expression for the momentum is defined as the product of the mass of an object and its speed.\(p = mv\)

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

There moment of inertia is defined as the products of the mass\(m\)of each particle with the square of its distance\(r\)and it is expressed as follows:

\(I = \frac{1}{2}m{r^2}\)

Conservation of momentum states that when there is no external force applied to the system then momentum of the system is conserved. That is the total momentum of the objects before the collision is equal to the total momentum of the objects after the collision.

\({p_i} = {p_f}\)

Here, \({p_f}\)is the final momentum and \({p_i}\)is the initial momentum.

\({p_i} = {p_f}\)

Now, for the given system the above equation can be reduced as follows:

\(\begin{aligned}{l}{p_i} = {p_f}\\{m_1}{v_1} + {m_2}{v_2} = {m_1}{v_f} + {m_2}{v_3}\end{aligned}\)

Here,\({m_1}\)is the mass of the satellite,\({m_2}\)is the mass of the piece of the space junk,\({v_1}\)is the original velocity of the satellite,\({v_2}\)is the velocity of the piece of the space junk, \({v_f}\)is the final velocity of the satellite after the collision, and\({v_3}\)is the new velocity of the piece of the space junk where it hit the satellite.

Substitute\(205kg\)for\({m_1},4.1kg\)for\({m_2},(2600m/s)i\)for\({v_1},( - 2200m/s)i\)for\({v_2},\)and\(( - 1300m/s)i + (480m/s)j\)for\({v_3}.\)

\(\begin{aligned}{l}{m_1}{v_1} + {m_2}{v_2} = {m_1}{v_f} + {m_2}{v_3}\\(205kg)(2600m/s)i + (4.1kg)( - 2200m/s)i = \\(205kg){v_f} + (4.1kg)( - 1300m/s)i + (480m/s)j\\{v_f} = \frac{{(529310kg.m/s)i - (1968kg.m/s)j}}{{(205kg)}}\\ = (2582m/s)i - (9.6m/s)j\end{aligned}\)

Then the above expression for the final velocity can be written in terms of the\(x\)and\(y\)components as follows:

\({v_x} = (2582m/s)\)

\({v_y} = - (9.6m/s)\)

Therefore, the required components of the velocity of the satellite after the collision is

\(\left\langle {2582, - 9.6,0} \right\rangle m/s\)

Now, calculate the magnitude of the velocity of the satellite after collision by using the expression as follows:

\(V = \sqrt {{v_x}^2 + {v_x}^2} \)

Substitute\(2582m/s\)for\({v_x}\)and\( - 9.6m/s\)for\({v_y}.\)

\(\begin{aligned}{l}V = \sqrt {{v_x}^2 + {v_x}^2} \\ = \sqrt {{{(2582m/s)}^2} + {{( - 9.6m/s)}^2}} \\ = 2582.02m/s\end{aligned}\)

Now, calculate the moment of inertia\(I\)of the satellite by using the expression for the moment of inertia of the satellite about its axis of rotation is as follows:

\(I = \frac{2}{5}m{r^2}\)

Substitute\(205kg\)for\(m\)and\(4.8m\)for\(r.\)

\(\begin{aligned}{l}I = \frac{2}{5}m{r^2}\\ = \frac{2}{5}(205kg){(4.8m)^2}\\ = 1889.3kg.{m^2}\end{aligned}\)

Now, calculate the final rotational speed of the satellite after collision by applying the conservation of angular momentum for the given system as follows:

\(I{\omega _i} + {m_2}{v_2}r = I{\omega _f} + {m_2}{v_3}r\)

Here,\({\omega _i}\)is the rational speed of the satellite before the collision and\({\omega _f}\)is the rational speed of the satellite after the collision.

Substitute\( - 1889.3kg.{m^2}\)for\(I,2rad/s\)for\({\omega _i},4.2kg\)for\({m_2}, - 2200m/s\)for\({v_2},1300m/s\)for\({v_3},\)and\(4.8m\)for\(r.\)

\(I{\omega _i} + {m_2}{v_2}r = I{\omega _f} + {m_2}{v_3}r\)

\((1189.3kg.{m^2})(2rad/s) + (4.1kg)( - 2200m/s)(4.8m) = (1889.3kg.{m^2}){\omega _f}\)

\( + (4.1kg)( - 1300m/s)(4.8m)\)

\({\omega _f} = \frac{{( - 13933.44kg.{m^2}/s)}}{{(1889.3kg.{m^2})}}\)

\( = - 7.375rad/s\)

Here, the negative sign indicates that the direction of the rotational speed after the collision is out of the page.

Therefore, the required rotational speed is\(7.375rad/s\)and is out of the page.

\(\Delta E = {E_i} - {E_f}\)

Here,\({E_i}\)is the energy before the collision and\({E_f}\)is the energy after the collision.

Calculate the energy before the collision\({E_i}\)for the given system by using the expression as:

\({E_f} = I{\omega _f}^2 + \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2\)

Substitute\(1889.3kg.{m^2}\)for\(I,2rad/s\)for\({\omega _i},205kg\)for\({m_1},4.1kg\)for\({m_2}, - 2200m/s\)for\({v_2},\)and\(2600m/s\)for\({v_1}.\)

\(\begin{aligned}{l}{E_i} = I{\omega _i}^2 + \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2\\ = (1889.3kg.{m^2}){(2rad/s)^2} + \frac{1}{2}(205kg){(2600m/s)^2}\\ + \frac{1}{2}(4.1kg){( - 2200m/s)^2}\\ = 702829557.2J\end{aligned}\)

Similarly, calculate the energy after the collision\({E_f}\)for the given system by using the expression as:

\({E_f} = I{\omega _f}^2 + \frac{1}{2}{m_1}{V^2} + \frac{1}{2}{m_2}{v_3}^2\)

Substitute\(1889.3kg.{m^2}\)for\(I,\)\(7.375rad/s\)for\({\omega _f},\)\(205kg\)for\({m_1},\)\(4.1kg\)for\({m_2},\)\(( - 1300m/s + 480m/s)\)for\({v_3},\)and\(2582.02m/s\)for\(V.\)

\(\begin{aligned}{l}{E_f} = I{\omega _f}^2 + \frac{1}{2}{m_1}{V^2} + \frac{1}{2}{m_2}{v_3}^2\\ = (1889.3kg.{m^2}){(7.375rad/s)^2} + \frac{1}{2}(205kg){(2582.02m/s)^2}\\ + \frac{1}{2}(4.1kg){( - 1300m/s + 480m/s)^2}\\ = 6844830976.45J\\7.375rad/sI,1889.3k\end{aligned}\)

Therefore, the change in internal energy becomes:

\(\Delta E = {E_i} - {E_f}\)

Substitute\(684830976.45J\)for\({E_f}\)and\(702829557.2J\)for\({E_i}.\)

\(\begin{aligned}{l}\Delta E = {E_i} - {E_f}\\ = (702829557.2J) - (684830976.45J)\\ = 1.779 \times {10^7}J\end{aligned}\)

Hence, the required change in internal energy for the given system is\(1.79 \times {10^7}J\)