91影视

The given data is listed below as,

• The mass of the rock 1 is,$m_1=5\mathrm{kg}$
• The initial velocity of the rock 1 is$v_{1i}=(3300,-3100,3400)\mathrm{m}/\mathrm{s}$
• The initial velocity of the rock 2 is,$v_{2i}=0$
• The final velocity of the first rock is,$v_{1i}=(2800,-2400,3700)\mathrm{m}/\mathrm{s}$

The law of conservation of momentum states that the momentum of a system before and after collision remains constant if no external force acts.

The law of conservation of energy states that the total initial kinetic energy of a system equals the addition of the total final kinetic energy and the system鈥檚 total thermal energy.

The expression for the initial momentum of the system can be expressed as,

$P_{initial}=m_1{v}_{1i}+m_2{v}_{2i}$

Here,$P_{initial}$ is the initial momentum of the system,$m_1$ is the mass of the rock 1,$m_2$ is the mass of the rock 2,$v_{1i}$ is the initial velocity of the rock 1 and$v_{2i}$ is the initial velocity of the rock 2.

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As the mass of the rock 2 is not given, so, it is assumed that the mass of the rock 2 is for${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{kg}$ simplifying the calculations.

For$m_1=5kg,v_{1i}=(3300,-3100,3400)m/s$

$$\begin{gathered} P_{initial}=\left(5kg\right)脳\left((3300,-3100,3400)m/s\right)+\left(8kg\right)脳(0m/s) \\ =\left((16500,-15500,17000)kg.m/s\right)\_+0 \\ =\left((16500,-15500,17000)kg.m/s\right) \end{gathered}$$

According to the law of conservation of momentum, the initial and the final rate of the system are the same. So, the final velocity of the system is expressed as,

$$\begin{gathered} P_{final}=\left(m_1{v}_{1i}\right)+\left(m_2{v}_{2i}\right) \\ {P}_{initial}=\left(m_1{v}_{1i}\right)+\left(m_2{v}_{2i}\right) \end{gathered}$$

Here,$v_{2i}$ is the final velocity of rock 2.

For$P_{initial}=\left((16500,-15500,17000)kg.m/s\right),m_1=5kg$ and$v_{1i}=(2800,-2400,3700)m/s$ .

$$\begin{gathered} \left((16500,-15500,17000)kg.m/s\right)=\left(5kg\right)脳\left((2800,-2400,3700)m/s\right)+\left(8kg\right)脳\left(v\right) \\ =\left((14000,-12000,18500)kg.m/s\right)+(8vkg) \\ \left((2500,-3500,-1500)kg.m/s\right)=(8vkg) \\ v=\left(312.5,-437.5,187.5\right)m/s \end{gathered}$$

Thus, the final momentum of rock 2 is$\left((2500,-3500,-1500)kg.m/s\right)$ .

The expression for the magnitude of the initial velocity of the rock 1 is expressed as,

$v_{1i}=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2+{\left(v_z\right)}^2}$

Here,role="math" localid="1657949364982" $v_x,v_y$ and$v_z$ are the velocities of the rock 1 at the x, y, and z coordinate respectively.

For $v_x=3300m/s,v_y=-3100m/s$and$v_z=3400m/s$ .

$$\begin{gathered} v_{1i}=\sqrt{{\left(3300\right)}^2+{\left(-3100\right)}^2+{\left(3400\right)}^2}m/s \\ =5662.155m/s \end{gathered}$$

The expression of the kinetic energy of the rock 1 before the collision can be expressed as-

$K.E_{.1i}=\frac{1}{2}{m}_1{v_{1i}}^2$

Here, is the mass of rock 1 and$v_{1i}$ is the initial velocity of rock 1.

For$m_1=5kg$ and$v_{1i}=5662.155m/s$ .

role="math" localid="1657950103667" $$\begin{gathered} \mathrm{K}.{\mathrm{E}}_{.1\mathrm{i}}=\frac{1}{2}脳\left(5\mathrm{kg}\right)脳{(5662.155\mathrm{m}/\mathrm{s})}^2 \\ =8.01脳{10}^7\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2脳\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =8.01脳{10}^7\mathrm{J} \end{gathered}$$

Thus, the kinetic energy of rock 1 before the collision is$8.01脳{10}^7\mathrm{J}$ .

As rock 2 was at rest before the collision, then the kinetic energy of rock 2 before the collision is 0.

Thus, the kinetic energy of rock 2 before the collision is$\mathbf{0}\mathbf{J}$.

The expression for the magnitude of the final velocity of the rock 1 is as follows,

$v_{1i}=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2+{\left(v_z\right)}^2}$

Here, $v_x,v_y$and$v_z$ are the velocity of the rock 1 at the x, y and z coordinate respectively.

For$v_x=2800m/s,v_y=-2400m/s$ and$v_z=3700m/s$ .

$$\begin{gathered} v_{1i}=\sqrt{{\left(2800\right)}^2+{\left(-2400\right)}^2+{\left(3700\right)}^2}m/s \\ =5223.98m/s \end{gathered}$$

The expression for the kinetic energy of the rock 1 after the collision can be expressed as,

$K.E_{.1f}=\frac{1}{2}{m}_1{v_{1f}}^2$

Here,$m_1$ is the mass of the rock and$v_{1f}$ is the final velocity of rock 1.

For$m_1=5kg$ and$v_{1f}=5223.98m/s$ .

role="math" localid="1657951558267" $$\begin{gathered} \mathrm{K}.{\mathrm{E}}_{.1\mathrm{f}}=\frac{1}{2}脳\left(5\mathrm{kg}\right)脳{(5223.98\mathrm{m}/\mathrm{s})}^2 \\ =6.82脳{10}^7\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2脳\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =6.82脳{10}^7\mathrm{J} \end{gathered}$$

Thus, the kinetic energy of rock 1 after the collision is$6.82脳{10}^7\mathrm{J}$ .

If the collision is elastic, then according to the energy balance, the kinetic energy of rock 2 will be the difference between the initial and the final kinetic energy of rock 1 as the kinetic energy remains constant.

The expression of the final kinetic energy of the rock 2 can be expressed as,

$K{E}_{2f}=K{E}_{1i}-K{E}_{1f}$

Here,$K{E}_{1i}$ and$K{E}_{1f}$ are the initial and the final energy of the rock 1 respectively.

For$K{E}_{1i}=8.01脳{10}^{7}J$ and$K{E}_{1f}=6.82脳{10}^{7}J$ .

$$\begin{gathered} K{E}_{2f}=8.01脳{10}^{7}J-6.82脳{10}^{7}J \\ =1.2脳{10}^{7}J \end{gathered}$$

Thus, the kinetic energy of the rock 2 is$1.2脳{10}^{7}J$ .

From the law of conservation of energy, the equation of the total final and the initial kinetic energy of the system can be expressed as,

$K.E_{.1i}+K.E_{.2i}=K.E_{.1f}+K.E_{.2f}+鈭�E_{thermal,1}+鈭�E_{thermal,2}$

Here,$K.E_{.1i}$ is the initial kinetic energy of the rock 1,$K.E_{.2i}$ is the initial kinetic energy of the rock 2,$K.E_{.1f}$ is the final kinetic energy of rock 1,$K.E_{.2f}$ is the final kinetic energy of rock 2 and$鈭�E_{thermal,1}+鈭�E_{thermal,1}$ is the change in the thermal energy of the rocks.

For $K{E}_{1i}=8.01脳{10}^{7}J,K{E}_{1f}=8.01脳{10}^{7}J,K{E}_{2i}=0$and$鈭�E_{thermal,1}+鈭�E_{thermal,2}=7.61脳{10}^{6}J$

$$\begin{gathered} 8.01脳{10}^{7}J+0=6.82脳{10}^{7}J+K.E_{.2f}+7.16脳{10}^{6}J \\ K.E_{.2f}=8.01脳{10}^{7}J-68.2脳{10}^{6}J-7.16脳{10}^{6}J \\ =4.8脳{10}^{6}J \end{gathered}$$

Thus, the final kinetic energy of rock 2 is$4.8脳{10}^{6}J$ .

Assuming this system is an isolated system, the transfer of energy Q will be 0 when the kinetic energy is converted to thermal energy.

Thus, the energy transfer from the surroundings into the two-rock system during the collision is 0.