Chapter 20, Problem 62AP

(a)

To determine

The appropriate models for the system of two bullets for the time interval before to after the interval.

Answer to Problem 62AP

It is an isolated system. The kinetic energy is converted into internal energy. Here, the momentum is conserved and the collision is a perfectly inelastic collision.

Explanation of Solution

Given info: The mass of one bullet is $12.0\text{g}$, mass of second bullet is $8.00\text{g}$, speed of one bullet which moves to the right direction is $300\text{m}/\text{s}$, speed of second bullet which moves to the left direction is $400\text{m}/\text{s}$, the temperature at collision is $30°\text{C}$.

It is an isolated system. The energy of the system is conserved but the kinetic energy is converted into internal energy. No external force exists. Here, the momentum is conserved and the collision is a perfectly inelastic collision.

Conclusion:

Therefore, it is an isolated system. The kinetic energy is converted into internal energy. Here, the momentum is conserved and the collision is a perfectly inelastic collision.

(b)

To determine

The speed of the combined bullets after the collision.

Answer to Problem 62AP

The speed of the combined bullets after the collision is $20.0\text{m}/\text{s}$ which moves to the right.

Explanation of Solution

Given info: The mass of one bullet is $12.0\text{g}$, mass of second bullet is $8.00\text{g}$, speed of one bullet which moves to the right direction is $300\text{m}/\text{s}$, speed of second bullet which moves to the left direction is $400\text{m}/\text{s}$, the temperature at collision is $30°\text{C}$.

Write the equation for conservation of momentum.

$\begin{array}{c}{m}_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2=\left(m_1+m_2\right)\overrightarrow{v}\\ \overrightarrow{v}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}\end{array}$

Here,

$m_1$ is the mass of one bullet.

$m_2$ is the mass of one bullet.

${\overrightarrow{v}}_1$ is the speed of one bullet.

${\overrightarrow{v}}_2$ is the speed of second bullet.

$\overrightarrow{v}$ is the speed of combined bullets.

Consider positive sign when bullet moves to the right and negative sign when bullet moves to the left.

Substitute $12.0\text{g}$ for $m_1$, $300\text{m}/\text{s}$ for ${\overrightarrow{v}}_1$, $8.00\text{g}$ for $m_2$ and $-400\text{m}/\text{s}$ for ${\overrightarrow{v}}_2$ in the above equation to get the speed of the combined bullets after the collision. $\begin{array}{c}\overrightarrow{v}=\frac{\left(12.0\text{g}\right)\left(300\text{m}/\text{s}\right)+\left(8.00\text{g}\right)\left(-400\text{m}/\text{s}\right)}{12.0\text{g}+8.00\text{g}}\\ =20.0\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the combined bullets after the collision is $20.0\text{m}/\text{s}$ which moves to the right.

(c)

To determine

The amount of initial kinetic energy which is converted into internal energy of the system after the collision.

Answer to Problem 62AP

The amount of initial kinetic energy which is converted into internal energy of the system after the collision is $1176\text{J}$.

Explanation of Solution

Given info: The mass of one bullet is $12.0\text{g}$, mass of second bullet is $8.00\text{g}$, speed of one bullet which moves to the right direction is $300\text{m}/\text{s}$, speed of second bullet which moves to the left direction is $400\text{m}/\text{s}$, the temperature at collision is $30°\text{C}$.

Write the equation for change in kinetic energy to calculate the amount of kinetic energy which is converted into internal energy of the system after the collision. $\begin{array}{c}\Delta K.E.=\left|K.E_{\text{final}}-K.E_{\text{initial}}\right|\\ =\left|\frac{1}{2}\left(m_1+m_2\right){\overrightarrow{v}}^2-\frac{1}{2}\left(m_1{\overrightarrow{v}}_1^2+m_2{\overrightarrow{v}}_2^2\right)\right|\end{array}$

Substitute $12.0\text{g}$ for $m_1$, $300\text{m}/\text{s}$ for ${\overrightarrow{v}}_1$, $8.00\text{g}$ for $m_2$, $-400\text{m}/\text{s}$ for ${\overrightarrow{v}}_2$ and $20\text{m}/\text{s}$ for $\overrightarrow{v}$ in the above equation to get the amount of kinetic energy which is converted into internal energy.

$\begin{array}{c}\Delta K.E.=\left|\left[\begin{array}{l}\left\{\frac{1}{2}\left(\left(12.0\text{g}+8.00\text{g}\right)\left(\frac{1\text{kg}}{1000\text{g}}\right)\right){\left(20.0\text{m}/\text{s}\right)}^2\right\}\\ -\left\{\frac{1}{2}\left[\begin{array}{l}\left(12.0\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right){\left(300\text{m}/\text{s}\right)}^2\\ +\left(8.00\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right){\left(-400\text{m}/\text{s}\right)}^2\end{array}\right]\right\}\end{array}\right]\right|\\ =\left|4\text{J}-1180\text{J}\right|\\ \text{=}\left|-\text{1176}\text{J}\right|\\ =1176\text{J}\end{array}$

Conclusion:

Therefore, the amount of initial kinetic energy which is converted into internal energy of the system after the collision is $1176\text{J}$.

(d)

To determine

Whether all the lead is melt due to the collision or not.

Answer to Problem 62AP

No, all the lead is not melt due to the collision.

Explanation of Solution

Given info: The mass of one bullet is $12.0\text{g}$, mass of second bullet is $8.00\text{g}$, speed of one bullet which moves to the right direction is $300\text{m}/\text{s}$, speed of second bullet which moves to the left direction is $400\text{m}/\text{s}$, the temperature at collision is $30°\text{C}$.

Write the equation to calculate the energy Q required to change the temperature of mass m.

$\begin{array}{c}Q=mc\Delta T\\ =mc\left(T_{\text{f}}-T_{\text{i}}\right)\end{array}$

Here,

$c$ is the specific heat of the aluminum.

$T_{\text{f}}$ is the melting temperature of lead.

$T_{\text{i}}$ is the temperature at collision.

The specific heat of aluminum is $128\text{J}/\text{kg}\cdot °\text{C}$.

The melting point of lead is $327.5°\text{C}$.

Substitute $20.0\text{g}$ for m, $128\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $327.5°\text{C}$ for $T_{\text{f}}$ and $30°\text{C}$ for $T_{\text{i}}$ in the above equation to get the energy required to change the temperature of mass $m$.

$\begin{array}{c}Q=\left(20.0\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(128\text{J}/\text{kg}\cdot °\text{C}\right)\left(\left(327.5°\text{C}\right)-\left(30°\text{C}\right)\right)\\ =761.6\text{J}\end{array}$

Initial energy generated by the collision is $1176\text{J}$, so the available heat for the melting process is calculated by,

$\begin{array}{c}{Q}_{\text{available}}=E-Q\\ =1176\text{J}-761.6\text{J}\\ =414.5\text{J}\end{array}$

Write the equation to calculate the amount of heat required to melt all of the lead.

$Q=mL$

Here,

L is the latent heat of the fusion for lead.

The latent heat of fusion for lead is equal to $2.32\times {10}^4\text{J}/\text{kg}$.

Substitute $20.0\text{g}$ for $m$ and $2.32\times {10}^4\text{J}/\text{kg}$ for $L$ in the above equation. $\begin{array}{c}Q=\left(20.0\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(2.32\times {10}^4\text{J}/\text{kg}\right)\\ =464\text{J}\end{array}$

Here, only $414.5\text{J}$ heat is available, so the all the lead does not melt.

Conclusion:

Therefore, all the lead is not melt due to the collision.

(e)

To determine

The temperature of the combined bullets after the collision.

Answer to Problem 62AP

The temperature of the combined bullets after the collision is $327.5°\text{C}$.

Explanation of Solution

Given info: The mass of one bullet is $12.0\text{g}$, mass of second bullet is $8.00\text{g}$, speed of one bullet which moves to the right direction is $300\text{m}/\text{s}$, speed of second bullet which moves to the left direction is $400\text{m}/\text{s}$, the temperature at collision is $30°\text{C}$.

The available heat to melt the bullets is $414.5\text{J}$ and heat required to melt the combined bullets is $464\text{J}$.

The temperature of the combined bullets is the melting temperature of the bullets which is equal to $327.5°\text{C}$.

Conclusion:

Therefore, the temperature of the combined bullets after the collision is $327.5°\text{C}$.

(f)

To determine

The phase of the combined bullets after the collision.

Answer to Problem 62AP

The phase of the combined bullets after the collision is such that $2.2\text{g}$ of solid lead and $17.8\text{g}$ of liquid lead.

Explanation of Solution

Given info: The mass of one bullet is $12.0\text{g}$, mass of second bullet is $8.00\text{g}$, speed of one bullet which moves to the right direction is $300\text{m}/\text{s}$, speed of second bullet which moves to the left direction is $400\text{m}/\text{s}$, the temperature at collision is $30°\text{C}$.

Write the equation to calculate the total melted mass of lead.

$\begin{array}{c}Q=mL\\ m=\frac{Q}{L}\end{array}$

Substitute $414.5\text{J}$ for $Q$ and $2.32\times {10}^4\text{J}/\text{kg}$ for $L$ in the above equation to get the mass of liquid lead.

$\begin{array}{c}m=\frac{414.5\text{J}}{2.32\times {10}^4\text{J}/\text{kg}}\\ =0.01786\text{kg}\times \frac{{10}^3\text{g}}{1\text{kg}}\\ \text{=17}\text{.8}\text{kg}\end{array}$

Calculate the mass of solid lead.

$m_{solid}=m_1-m$

Substitute $20.0\text{g}$ for $m_1$ and $17.8\text{g}$ for $m$ in the above equation to get the mass of solid lead.

$\begin{array}{c}{m}_{solid}=20.0\text{g}-17.8\text{g}\\ \text{=2}\text{.2}\text{g}\end{array}$

Conclusion:

Therefore, the phase of the combined bullets after the collision is such that $2.2\text{g}$ of solid lead and $17.8\text{g}$ of liquid lead.