In outer space a rock whose mass is 3kg and whose velocity was$\mathbf{(}\mathbf{3900}\mathbf{,}\mathbf{-}\mathbf{2900}\mathbf{,}\mathbf{3000}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$struck a rock with mass 13kg and velocity$\mathbf{(}\mathbf{220}\mathbf{,}\mathbf{-}\mathbf{260}\mathbf{,}\mathbf{300}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$. After the collision, the 3kg rock’s velocity is$\mathbf{(}\mathbf{3500}\mathbf{,}\mathbf{-}\mathbf{2300}\mathbf{,}\mathbf{3500}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s}$. (a) What is the final velocity of the 13kg rock? (b) What is the change in the internal energy of the rocks? (c) Which of the following statements about Q (transfer of energy into the system because of a temperature difference between system and surroundings) are correct? (1)$\mathbf{Q}\mathbf{\approx }\mathbf{0}$ because the duration of the collision was very short. (2)$\mathbf{Q}\mathbf{=}{\mathbf{E}}_{\mathbf{thermal}}$ of the rocks. (3)$\mathbf{Q}\mathbf{\approx }\mathbf{0}$ because there are no significant objects in the surroundings. (4)$\mathbf{Q}\mathbf{=}\mathbf{∆}\mathbf{k}$ of the rocks.

The given data is listed below as-

• The mass of the first rock is,$m_1=3kg$ .
• The velocity of the first rock is,$V_{1i}=(3900,-2900,3000)m/s$
• The mass of the second rock is,$m_2=13kg$
• The velocity of the second rock is,$V_{2i}=(220,-260,300)m/s$
• After the collision, the velocity of the first rock is,$V_{1f}=(3500,-2300,3500)m/s$

The first law of thermodynamics states that the change in the internal energy of an object is equal to the difference between the energy transfer and the work done by the system.

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

The work-energy theorem states that the net work done by the system is equal to the change in the system’s kinetic energy.

The law of conservation of momentum gives the final velocity of the second rock, and the work-energy theorem. The first law of thermodynamics gives the change in the internal energy of the rocks.

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

$p_{initial}=\left(m_2{v}_{2i}\right)+\left(m_1{v}_{1i}\right)$

Here,$p_{initial}$is the initial momentum of the system$m_1$and$m_2$are the mass of the first and the second rock respectively and$v_{1i}$and$v_{2i}$are the initial velocities of the first and the second rock respectively.

For localid="1657877689140" $m_1=3kg,m_2=13kg,v_{1i}=(3900,-2900,3000)m/s$and$v_{2i}=(220,-260,300)m/s$.

$$\begin{gathered} P_{initial}=\left(3kg\right)\times \left((3900,-2900,3000)m/s\right)+(13kg)\times \left((220,-260,300)m/s\right) \\ =\left((11700,-8700,9000)kg,m/s\right)+\left((2860,-3380,3900)kg,m/s\right) \\ =\left((14560,-12080,12900)kg.m/s\right) \end{gathered}$$

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,

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

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

For $P_{initial}=\left(14560,-12080,2900\right)kg.m/s,m_1=3kg,m_2=13kg$and $v_{1,i}=(3500,-2300,3500)m/s$.

$$\begin{gathered} \left(\left(14560,-12080,12900\right)kg.m/s\right)=\left(3kg\right)\times \left(\left(3500,-2300,3500\right)m/s\right)+\left(13kg\right)\times \left(v_{2,i}\right) \\ =\left(\left(10500,-6900,10500\right)kg.m/s\right)+\left(13\left(v_{2,i}\right)kg\right) \\ \left(\left(4060,-5180,2400\right)kg.m/s\right)=\left(13\left(v_{2,i}\right)kg\right) \\ {v}_{2,i}\approx (312,-398,185)m/s \end{gathered}$$

Thus, the final velocity of the$13kg$ rock is$(312,-398,185)m/s$ .

The magnitude of the initial velocity of the first rock is expressed as,

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

Here, $v_{x,}{v}_y$and $v_z$are the initial velocities of the first rock at the $x,y$and $z$coordinate respectively.

For $V_x=3900m/s,V_y=-2900m/s$and $V_z=3000m/s$.

$$\begin{gathered} V_{1i}=\sqrt{{\left(3900\right)}^2+{\left(-2900\right)}^2+{\left(3000\right)}^{2}m/s} \\ =5711.39m/s \end{gathered}$$

The magnitude of the initial velocity of the second rock is expressed as,

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

Here,$v_x,v_y$and $v_z$are the initial velocities of the second rock at the $x,y$and $z$coordinate respectively.

For $v_x=220m/s,v_y=-260m/s$and $V_z=3000m/s$.

$$\begin{gathered} v_{2i}=\sqrt{{\left(220\right)}^2+{\left(-260\right)}^2+{\left(3000\right)}^2}m/s \\ =3019.27m/s \end{gathered}$$

The magnitude of the final velocity of the first rock is expressed as,

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

Here, $V_x,V_Y$and $V_z$are the final velocities of the first rock at the $x,y$and $z$coordinate respectively.

For $v_x=3500m/s,v_y=-2300m/s$and $v_z=3500m/s$.

$$\begin{gathered} v_{1i}=\sqrt{{\left(3500\right)}^2+{\left(-2300\right)}^2+{\left(3500\right)}^2}m/s \\ =5458.02m/s \end{gathered}$$

The magnitude of the final velocity of the second rock is expressed as,

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

Here,role="math" localid="1657883192673" $v_x,v_y$and $v_z$are the final velocities of the second rock at the $x,y$and$z$coordinate respectively.

For $v_x=312m/s,v_y=-398m/s$and$v_z=185m/s$.

$$\begin{gathered} v_{2i}=\sqrt{{\left(312\right)}^2+{\left(-398\right)}^2+{\left(185\right)}^2}m/s \\ =538.49m/s \end{gathered}$$

From the first law of thermodynamics, the equation of the change in the internal energy of the rocks can be expressed as,

$∆l=Q-W$

Here, Q and W are the transfer of energy and work done by the system respectively.

For the first rock,

According to the work-energy theorem, the net change in the work done changes the system’s kinetic energy. So, the value is approximately Q as the duration of the collision is very short.

Hence, the equation of the change in the internal energy can be expressed as,

role="math" localid="1657885101892" $$\begin{gathered} ∆l=0-(finalK.E,-initialK.E.) \\ =initialK.E.-finalK.E. \end{gathered}$$ …(1)

The equation of the total initial kinetic energy can be expressed as,

$K{E}_i=\frac{1}{2}{m}_1{v^2}_{1i}+\frac{1}{2}{m}_1{v^2}_{2i}$

For $m_1=3kg,m_2=13kg,v_{1i}=5711.39m/s$and$v_{2i}=3019.27m/s,$

$$\begin{gathered} K{E}_i=\left(\frac{1}{2}\left(3kg\right){(5711.39m/s)}^2+\frac{1}{2}\left(13kg\right){(3019.27m/s)}^2\right) \\ =(1.5kg)\times (3.2\times {10}^7)m^2/s^2+(6.5kg)\times (9.1\times {10}^6)m^2/s^2 \\ =48.92\times {10}^{6}kg.m^2/s^2+59.25\times {10}^{6}kg.m^2/s^2 \\ =108.17\times {10}^{6}kg.m^2/s^2\times \frac{1J}{1kg.m^2/s^2} \\ =108.17\times {10}^{6}J \end{gathered}$$

The equation of the total final kinetic energy can be expressed as,

$K{E}_i=\frac{1}{2}{m}_1{v^2}_{1i}+\frac{1}{2}{m}_2{v^2}_{2i}$

For$m_1=3kg,m_2=13kg,v_{1i}=5458.02m/s$ and$v_{2,i}=538.49m/s$ .

$$\begin{gathered} K{E}_i=\left(\frac{1}{2}\left(3kg\right){(5458.02m/s)}^2+\frac{1}{2}\left(13kg\right){(538.49m/s)}^2\right) \\ =(1.5kg)\times (2.9\times {10}^7)m^2/s^2+(6.5kg)\times \left(2.8\times {10}^5\right)m^2/s^2 \\ =44.68\times {10}^{6}kg.m^2/s^{2^2}+1.82\times {10}^{6}kg.m^2/s^2 \\ =46.68\times {10}^{6}kg.m^2/s^2\times \frac{1J}{1kg.m^2/s^2} \\ =46.68\times {10}^{6}J \end{gathered}$$

Substituting all the values in equation (1).

$$\begin{gathered} ∆l=108.17\times {10}^{6}J-46.48\times {10}^{6}J \\ =61.69\times {10}^{6}J \end{gathered}$$

Thus, the change in the internal energy of the rocks is$61.69\times {10}^{6}J$ .

As it is assumed that the collision is short, the value of the is approximately 0.

Statement 2 says$Q=E_{thermal}$ for the rocks. However, this statement is incorrect as, during a short collision, the value cannot be high as the$E_{thermal}$ .

Statement 3 says the value is approximately zero as there are no significant objects in the surroundings. However, this statement is incorrect as the value is zero as the collision is short, not because of the surroundings.

Statement 4 says$Q=∆k$ that for the rocks. This statement is incorrect because the value is zero due to the short collision.

Statement 1 states that$Q\approx 0$ as the collision is short. However, this statement is correct as the collision is fast with the rocks.

Thus, statement 1 is correct.

Thus, the duration of the collision was very short.