Chapter 11, Problem 11.46AP

Review. Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45.0 kg, is gliding to the right at 8.00 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 11.0 m/s along the same line. When they meet, they grab each other and hang on. (a) What is their velocity immediately thereafter? (b) What fraction of their original kinetic energy is still mechanical energy after their collision? That was so much fun that the boys repeat the collision with the same original velocities, this time moving along parallel lines 1.20 m apart. At closest approach, they lock arms and start rotating about their common center of mass. Model the boys as particles and their arms as a cord that does not stretch. (c) Find the velocity of their center of mass. (d) Find their angular speed. (e) What fraction of their original kinetic energy is still mechanical energy after they link arms? (f) Why are the answers to parts (b) and (e) so different?

To determine

The velocity of Jacob and Ethan immediate thereafter.

Answer to Problem 11.46AP

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

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,

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

Here, $m_1$ 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 and ${\overrightarrow{\text{v}}}_{\text{f}}$ 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 $\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$.

To determine

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

Answer to Problem 11.46AP

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

Explanation of Solution

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 $K{E}_{\text{i}}$.

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

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 $K{E}_{\text{f}}$.

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

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 $0.00071$.

To determine

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

Answer to Problem 11.46AP

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

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 $\left(\text{0}\text{.25 }\text{m}/\text{s}\right)\widehat{\text{i}}$.

To determine

The angular speed of Jacob and Ethan.

Answer to Problem 11.46AP

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

Explanation of Solution

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.

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

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 $15.8\text{rad}/\text{s}$.

To determine

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

Answer to Problem 11.46AP

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

Explanation of Solution

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 $K{E}_{\text{f}}$.

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

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 $1$.

To determine

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

Answer to Problem 11.46AP

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.