Chapter 11, Problem 38AP

(a)

To determine

The velocity of Jacob and Ethan immediate thereafter.

Answer to Problem 38AP

The velocity of Jacob and Ethan immediate thereafter is $\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$.

Explanation of Solution

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The momentum is conserved in the isolated system of two boys then,

$\begin{array}{c}{\overrightarrow{\text{p}}}_{\text{i}}={\overrightarrow{\text{p}}}_{\text{f}}\\ m_1{v}_1\widehat{\text{i}}-m_2{v}_2\widehat{\text{i}}=\left(m_1+m_2\right){\overrightarrow{\text{v}}}_{\text{f}}\end{array}$

• $m_1$ is the mass of Jacob.
• $m_2$ is the mass of Ethan.
• $v_1$ is the speed of Jacob.
• $v_2$ is the speed of Ethan.
• ${\overrightarrow{\text{v}}}_{\text{f}}$ is the final velocity vector.

Substitute $45.0\text{ kg}$ for $m_1$, $31.0\text{ kg}$ for $m_2$, $8.0\text{m}/\text{s}$ for $v_1$ and $11.0\text{m}/\text{s}$ for $v_{\text{2}}$ in above equation to find ${\overrightarrow{\text{v}}}_{\text{f}}$.

$\begin{array}{c}\left(45.0\text{ kg}\right)\left(8.0\text{m}/\text{s}\right)\widehat{\text{i}}-\left(31.0\text{ kg}\right)\left(\text{11}.0\text{m}/\text{s}\right)\widehat{\text{i}}=\left(45.0\text{ kg}+31.0\text{ kg}\right){\overrightarrow{\text{v}}}_{\text{f}}\\ \left(\text{19}.0\text{kg}\cdot \text{m}/\text{s}\right)\widehat{\text{i}}=\left(\text{76}.0\text{ kg}\right){\overrightarrow{\text{v}}}_{\text{f}}\\ {\overrightarrow{\text{v}}}_{\text{f}}=\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}\end{array}$

Conclusion:

Therefore, the velocity of Jacob and Ethan immediate thereafter is $\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$.

(b)

To determine

The fraction of their original kinetic energy is still mechanical energy after their collision.

Answer to Problem 38AP

The fraction of their original kinetic energy is still mechanical energy after their collision is $0.00071$.

Explanation of Solution

Section 1:

To determine: The initial kinetic energy of the system.

Answer: The initial kinetic energy of the system is $3315.5\text{ J}$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The formula to calculate initial kinetic energy of the system is,

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

Substitute $45.0\text{ kg}$ for $m_1$, $31.0\text{ kg}$ for $m_2$, $8.0\text{m}/\text{s}$ for $v_1$ and $11.0\text{m}/\text{s}$ for $v_{\text{2}}$ in above equation to find $K{E}_{\text{i}}$.

$\begin{array}{c}K{E}_{\text{i}}=\frac{1}{2}\left(45.0\text{ kg}\right){\left(8.0\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(31.0\text{ kg}\right){\left(\text{11}.0\text{m}/\text{s}\right)}^2\\ =3315.5\text{ J}\end{array}$

Section 2:

To determine: The final kinetic energy of the system.

Answer: The final kinetic energy of the system is $2.375\text{ J}$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The formula to calculate final kinetic energy of the system is,

$K{E}_{\text{f}}=\frac{1}{2}\left(m_{\text{1}}+m_{\text{2}}\right)v_{\text{f}}^2$

Substitute $45.0\text{ kg}$ for $m_1$, $31.0\text{ kg}$ for $m_2$ and $0.25\text{m}/\text{s}$ for $v_{\text{f}}$ in above equation to find $K{E}_{\text{f}}$.

$\begin{array}{c}K{E}_{\text{f}}=\left(\left(45.0\text{ kg}\right)+\left(31.0\text{ kg}\right)\right){\left(\text{0}\text{.25 }\text{m}/\text{s}\right)}^2\\ =2.375\text{ J}\end{array}$

Section 3:

To determine: The fraction of kinetic energy of the system.

Answer: The fraction of energy of the system is $0.00071$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The formula to calculate fraction of kinetic energy is,

$\begin{array}{c}\frac{K{E}_{\text{f}}}{K{E}_{\text{i}}}=\frac{2.375\text{ J}}{3315.5\text{ J}}\\ =0.00071\end{array}$

Conclusion:

Therefore, the fraction of their original kinetic energy is still mechanical energy after their collision is $0.00071$.

(c)

To determine

The velocity of the centre of mass of Jacob and Ethan.

Answer to Problem 38AP

The velocity of the centre of mass of Jacob and Ethan is $\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$.

Explanation of Solution

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The velocity of the centre of mass of Jacob and Ethan is still remains same as calculated in part (a) because the conservation of momentum calculations will be same as part (a).

Then, the velocity of the centre of mass of Jacob and Ethan is,

${\overrightarrow{\text{v}}}_{\text{f}}=\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$

Conclusion:

Therefore, the velocity of the centre of mass of Jacob and Ethan is $\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$.

(d)

To determine

The angular speed of Jacob and Ethan.

Answer to Problem 38AP

The angular speed of Jacob and Ethan is $15.8\text{rad}/\text{s}$.

Explanation of Solution

Section 1:

To determine: The position of the centre of mass of the boys.

Answer: The position of the centre of mass of the boys is $0.489\text{ m}$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The position of the centre of mass of the boys is,

$y_{\text{CM}}=\frac{m_1{y}_1+m_{2}L}{\left(m_1+m_2\right)}$

Substitute $45.0\text{ kg}$ for $m_1$, $31.0\text{ kg}$ for $m_2$, $0$ for $y_1$ and $1.20\text{ m}$ for $L$ in above equation.

$\begin{array}{c}{y}_{\text{CM}}=\frac{\left(45.0\text{ kg}\right)\left(0\right)+\left(31.0\text{ kg}\right)\left(1.20\text{ m}\right)}{\left(45.0\text{ kg}+31.0\text{ kg}\right)}\\ =0.489\text{ m}\end{array}$

Section 2:

To determine: The angular speed of Jacob and Ethan.

Answer: The angular speed of Jacob and Ethan is $15.8\text{rad}/\text{s}$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The Jacob is $y_{\text{CM}}=0.489\text{ m}$ from the centre of mass and Ethan is $\left(L-y_{\text{CM}}\right)=0.711\text{ m}$ from the centre of mass.

The angular momentum is,

$\begin{array}{c}{m}_1{v}_1{y}_{\text{CM}}+m_2{v}_2\left(L-y_{\text{CM}}\right)=\left(m_1{y}_{\text{CM}}^2+m_2{\left(L-y_{\text{CM}}\right)}^2\right)\omega \\ \omega =\frac{m_1{v}_1{y}_{\text{CM}}+m_2{v}_2\left(L-y_{\text{CM}}\right)}{\left(m_1{y}_{\text{CM}}^2+m_2{\left(L-y_{\text{CM}}\right)}^2\right)}\end{array}$

Substitute $45.0\text{ kg}$ for $m_1$, $31.0\text{ kg}$ for $m_2$, $8.0\text{m}/\text{s}$ for $v_1$, $11.0\text{m}/\text{s}$ for $v_{\text{2}}$, $0.711\text{ m}$ for $\left(L-y_{\text{CM}}\right)$ and $0.489\text{ m}$ for $y_{\text{CM}}$ in above equation.

$\begin{array}{c}\omega =\frac{\left(45.0\text{ kg}\right)\left(8.0\text{m}/\text{s}\right)\left(0.489\text{ m}\right)+\left(31.0\text{ kg}\right)\left(\text{11}.0\text{m}/\text{s}\right)\left(0.711\text{ m}\right)}{\left(\left(45.0\text{ kg}\right){\left(0.489\text{ m}\right)}^2+\left(31.0\text{ kg}\right){\left(0.711\text{ m}\right)}^2\right)}\\ =\frac{418\text{kg}\cdot {\text{m}}^2/\text{s}}{26.4\text{ kg}\cdot {\text{m}}^2}\\ =15.8\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular speed of Jacob and Ethan is $15.8\text{rad}/\text{s}$.

(e)

To determine

The fraction of their original kinetic energy that is still mechanical energy after they link arms.

Answer to Problem 38AP

The fraction of their original kinetic energy that is still mechanical energy after they link arms is $1$.

Explanation of Solution

Section 1:

To determine: The initial and final kinetic energy of the system when they link arms.

Answer: The initial and final kinetic energy of the system when they link arms is $3315.5\text{ J}$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

Refer to the section 1 of part (b), the initial kinetic energy is,

$K{E}_{\text{i}}=3315.5\text{ J}$

The formula to calculate final kinetic energy of the system is,

$K{E}_{\text{f}}=\frac{1}{2}\left(m_{\text{1}}+m_{\text{2}}\right)v_{\text{CM}}^2+\frac{1}{2}\left(m_1{y}_{\text{CM}}^2+m_2{\left(L-y_{\text{CM}}\right)}^2\right){\omega }^2$

Substitute $45.0\text{ kg}$ for $m_1$, $31.0\text{ kg}$ for $m_2$, $0.25\text{m}/\text{s}$ for $v_{\text{CM}}$, $0.711\text{ m}$ for $\left(L-y_{\text{CM}}\right)$, $0.489\text{ m}$ for $y_{\text{CM}}$ and $15.8\text{rad}/\text{s}$ for $\omega$ in above equation to find $K{E}_{\text{f}}$.

$\begin{array}{c}K{E}_{\text{f}}=\left[\begin{array}{l}\frac{1}{2}\left(\left(45.0\text{ kg}\right)+\left(31.0\text{ kg}\right)\right){\left(0.25\text{m}/\text{s}\right)}^2\\ +\frac{1}{2}\left(\left(45.0\text{ kg}\right){\left(0.489\text{ m}\right)}^2+\left(31.0\text{ kg}\right){\left(0.711\text{ m}\right)}^2\right){\left(15.8\text{rad}/\text{s}\right)}^2\end{array}\right]\\ =\frac{1}{2}\left(76.0\text{ kg}\right){\left(0.25\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(26.4\text{ kg}\cdot {\text{m}}^2\right){\left(15.8\text{rad}/\text{s}\right)}^2\\ =3315.5\text{ J}\end{array}$

Section 3:

To determine: The fraction of kinetic energy of the system.

Answer: The fraction of energy of the system is $1$.

Given information:

The mass of Jacob is $45.0\text{ kg}$, the mass of Ethan is $31.0\text{ kg}$, the speed of Jacob is $8.0\text{m}/\text{s}$ and the speed of Ethan is $11.0\text{m}/\text{s}$.

The fraction of kinetic energy is,

$\begin{array}{c}\frac{K{E}_{\text{f}}}{K{E}_{\text{i}}}=\frac{3315.5\text{ J}}{3315.5\text{ J}}\\ =1\end{array}$

Conclusion:

Therefore, the fraction of their original kinetic energy is still mechanical energy after they link arms is $1$.

(f)

To determine

To explain: The reason for the answer of part (b) and part (e) is so different.

Answer to Problem 38AP

the answer of part (b) and part (e) is so different because the head on collision between similarly sized objects are grossly inefficient.

Explanation of Solution

The deformation is a process in which one form of energy changed into the other form. The head on collision between similarly sized objects are grossly inefficient. So the answers are so different. If the two kids were the same size and had the same velocity; conservation of the momentum states that they lose all their kinetic energy. On the other hand, glancing blows like this one allow a lot of energy to be transferred into the rotational kinetic energy, causing much less energy to be lost on impact.

Conclusion:

Therefore, the answer of part (b) and part (e) is so different because the head on collision between similarly sized objects are grossly inefficient.