Question

In outer space, rock 1 whose mass is$\mathbf{5}\mathbf{}\mathbf{Kg}$ and whose velocity was$\mathbf{(}\mathbf{3300}\mathbf{,}\mathbf{-}\mathbf{3100}\mathbf{,}\mathbf{3400}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$ struck rock 2, which was at rest. After the collision, rock 1’s velocity is$\mathbf{(}\mathbf{2800}\mathbf{,}\mathbf{-}\mathbf{2400}\mathbf{,}\mathbf{3700}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$ . (a) What is the final momentum of rock 2? (b) Before the collision, what was the kinetic energy of rock 1? (c) Before the collision, what was the kinetic energy of rock 2? (d) After the collision, what is the kinetic energy of rock 1? (e) Suppose that the collision was elastic (that is, there was no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2? (f) On the other hand, suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, where$\mathbf{∆}{\mathbf{E}}_{\mathbf{thermal}\mathbf{,}\mathbf{1}}\mathbf{+}\mathbf{∆}{\mathbf{E}}_{\mathbf{thermal}\mathbf{,}\mathbf{2}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{16}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{J}$ . What is the final kinetic energy of rock 2? (g) In this case (some of the kinetic energy being converted to thermal energy), what was the transfer of energy Q (microscopic work) from the surroundings into the two-rock system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)

Short answer

$$\begin{gathered} \mathrm{a})\left((2500,-3500,-1500)\mathrm{kg}.\mathrm{m}/\mathrm{s}\right) \\ \mathrm{b})8.01\times {10}^7\mathrm{J} \\ \mathrm{c})\mathbf{}\mathbf{0}\mathbf{}\mathbf{J} \\ \mathrm{d})6.82\times {10}^7\mathrm{J} \\ \mathrm{e})1.2\times {10}^7\mathrm{J} \\ \mathrm{f})4.8\times {10}^6\mathrm{J} \\ \mathrm{g})0 \end{gathered}$$

Step-by-step solution

Identification of the given data

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,$v_{1i}=(2800,-2400,3700)\mathrm{m}/\mathrm{s}$

Significance of the law of conservation of momentum and energy

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’s total thermal energy.

Determination of the final momentum of the rock 2

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

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

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$v_{1i}=(2800,-2400,3700)m/s$ .

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

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

Determination of the kinetic energies of the rock 1

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

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

Determination of the kinetic energy of the rock 2 before the collision

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$\mathbf{0}\mathbf{J}$.

Determination of the kinetic energy of the rock 1 after the collision

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

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

Determination of the kinetic energy of the rock 2 after collision

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

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

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

Determination of the final kinetic energy of the rock 2

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,

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

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$∆E_{thermal,1}+∆E_{thermal,1}$ is the change in the thermal energy of the rocks.

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

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

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

Determination of the transfer of energy

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.