Chapter 9, Problem 47AP

A 0.500-kg sphere moving with a velocity expressed as $(2.00\hat{i}-3.00\hat{j}+1.00\hat{k})\text{m}/\text{s}$ strikes a second, lighter sphere of mass 1.50 kg moving with an initial velocity of $(-1.00\hat{i}+2.00\hat{j}-3.00\hat{k})\text{m}/\text{s}$. (a) The velocity of the 0.500-kg sphere after the collision is $(-1.00\hat{i}+3.00\hat{j}-8.00\hat{k})\text{m}/\text{s}$. Find the final velocity of the 1.50-kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic). (b) Now assume the velocity of the 0.500-kg sphere after the collision is $(-0.250\hat{i}+0.750\hat{j}-2.00\hat{k})\text{m}/\text{s}$. Find the final velocity of the 1.50-kg sphere and identify the kind of collision. (c) What If? Take the velocity of the 0.500-kg sphere after the collision as $(-1.00\hat{i}+3.00\hat{j}-a\hat{k})\text{m}/\text{s}$. Find the value of a and the velocity of the 1.50-kg sphere after an elastic collision.

To determine

The final velocity of the $1.5\text{kg}$ sphere after the collision and the type of collision.

Answer to Problem 47AP

The final velocity of the $1.5\text{kg}$ sphere is 0 and this collision is inelastic.

Explanation of Solution

According to the law of conservation of momentum,

    $m_1{\overrightarrow{\text{v}}}_1+m_2{\overrightarrow{\text{v}}}_2=m_1{\overrightarrow{{\text{v}}^{\prime }}}_1+m_2{\overrightarrow{{\text{v}}^{\prime }}}_2$        (1)

Here, $m_1$ is the mass of lighter sphere, $m_2$ is the mass of heavier sphere, ${\overrightarrow{\text{v}}}_1$ is the initial velocity of the lighter sphere, ${\overrightarrow{\text{v}}}_2$ is the initial velocity of the heavier sphere, ${\overrightarrow{{\text{v}}^{\prime }}}_1$ is the final velocity of the lighter sphere and ${\overrightarrow{{\text{v}}^{\prime }}}_2$ is the final velocity of the heavier sphere.

Substitute $0.5\text{kg}$ for $m_1$, $1.5\text{kg}$ for $m_2$, $\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_1$, $\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}$ for ${\overrightarrow{\text{v}}}_2$ and $\left(-1\widehat{\text{i}}+3\widehat{\text{j}}-8\widehat{\text{k}}\right)\text{m}/\text{s}$ for ${\overrightarrow{{\text{v}}^{\prime }}}_1$ in above equation.

    $\begin{array}{c}\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}+\\ \left(1.5\text{kg}\right)\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}\end{array}\right]=\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(-1\widehat{\text{i}}+3\widehat{\text{j}}-8\widehat{\text{k}}\right)\text{m}/\text{s}\\ +1.5\text{kg}\times {\overrightarrow{{\text{v}}^{\prime }}}_2\end{array}\right]\\ \left(-0.5\widehat{\text{i}}+1.5\widehat{\text{j}}-1\text{4}\widehat{\text{k}}\right)\text{kg}\cdot \text{m}/\text{s}=\left(-0.5\widehat{\text{i}}+1.5\widehat{\text{j}}-1\text{4}\right)\text{kg}\cdot \text{m}/\text{s}+1.5\text{kg}\times {\overrightarrow{{\text{v}}^{\prime }}}_2\\ 1.5\text{kg}\times {\overrightarrow{{\text{v}}^{\prime }}}_2=0\\ {\overrightarrow{{\text{v}}^{\prime }}}_2=0\end{array}$

The final velocity of the heavier sphere is zero due to which the kinetic energy of the sphere is also zero. This is due to the loss of energy. Thus, the collision is inelastic.

Conclusion:

Therefore, the final velocity of the $1.5\text{kg}$ sphere is $0$  and this collision is inelastic.

To determine

The final velocity of the $1.5\text{kg}$ sphere.

Answer to Problem 47AP

The final velocity of the $1.5\text{kg}$ sphere is $\left(-0.25\widehat{\text{i}}+0.75\widehat{\text{j}}-2\widehat{\text{k}}\right)\text{m}/\text{s}$ and this collision is perfectly inelastic.

Explanation of Solution

From equation (1), the law of conservation of momentum is,

    $m_1{\overrightarrow{\text{v}}}_1+m_2{\overrightarrow{\text{v}}}_2=m_1{\overrightarrow{{\text{v}}^{\prime }}}_1+m_2{\overrightarrow{{\text{v}}^{\prime }}}_2$

Substitute $0.5\text{kg}$ for $m_1$, $1.5\text{kg}$ for $m_2$, $\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_1$, $\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_2$ and $\left(-0.25\widehat{\text{i}}+0.75\widehat{\text{j}}-2\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_1^{\text{'}}$ in above equation.

    $\begin{array}{c}\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}+\\ \left(1.5\text{kg}\right)\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}\end{array}\right]=\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(-0.25\widehat{\text{i}}+0.75\widehat{\text{j}}-2\widehat{\text{k}}\right)\text{m}/\text{s}\\ +1.5\text{kg}\times v_2^{\text{'}}\end{array}\right]\\ \left(-0.5\widehat{\text{i}}+1.5\widehat{\text{j}}-1\text{4}\widehat{\text{k}}\right)\text{kg}\cdot \text{m}/\text{s}=\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(-0.25\widehat{\text{i}}+0.75\widehat{\text{j}}-2\widehat{\text{k}}\right)\text{m}/\text{s}\\ +1.5\text{kg}\times v_2^{\text{'}}\end{array}\right]\\ v_2^{\text{'}}=\left(-0.25\widehat{\text{i}}+0.75\widehat{\text{j}}-2\widehat{\text{k}}\right)\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the final velocity of the $1.5\text{kg}$ sphere is $\left(-0.25\widehat{\text{i}}+0.75\widehat{\text{j}}-2\widehat{\text{k}}\right)\text{m}/\text{s}$ and this collision is perfectly inelastic.

To determine

The value of $a$ and the velocity of the $1.5\text{kg}$ sphere after an elastic collision.

Answer to Problem 47AP

The value of $a$ and the velocity of the $1.5\text{kg}$ sphere after an elastic collision is $-6.74$ and $-0.419\widehat{\text{k}}\text{m}/\text{s}$ respectively.

Explanation of Solution

From equation (1), the law of conservation of momentum is,

    $m_1{\overrightarrow{\text{v}}}_1+m_2{\overrightarrow{\text{v}}}_2=m_1{\overrightarrow{{\text{v}}^{\prime }}}_1+m_2{\overrightarrow{{\text{v}}^{\prime }}}_2$

Substitute $0.5\text{kg}$ for $m_1$, $1.5\text{kg}$ for $m_2$, $\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_1$, $\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_2$ and $\left(-1\widehat{\text{i}}+3\widehat{\text{j}}+a\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_1^{\text{'}}$ in above equation.

    $\begin{array}{c}\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}+\\ \left(1.5\text{kg}\right)\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}\end{array}\right]=\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(-1\widehat{\text{i}}+3\widehat{\text{j}}+a\widehat{\text{k}}\right)\text{m}/\text{s}\\ +1.5\text{kg}\times v_2^{\text{'}}\end{array}\right]\\ \left(-0.5\widehat{\text{i}}+1.5\widehat{\text{j}}-\text{4}\widehat{\text{k}}\right)\text{kg}\cdot \text{m}/\text{s}=\left[\begin{array}{l}\left(0.5\text{kg}\right)\left(-1\widehat{\text{i}}+3\widehat{\text{j}}+a\widehat{\text{k}}\right)\text{m}/\text{s}\\ +1.5\text{kg}\times v_2^{\text{'}}\end{array}\right]\\ -4\text{kg}\cdot \text{m}/\text{s}=\left(0.5\text{kg}\right)\left(a\widehat{\text{k}}\right)\text{m}/\text{s}+1.5\text{kg}\times v_2^{\text{'}}\\ v_2^{\text{'}}=\left(-2.67-0.333a\right)\widehat{\text{k}}\text{m}/\text{s}\end{array}$        (2)

According to the law of conservation of energy,

    $\begin{array}{c}K{E}_{\text{initial}}=K{E}_{\text{final}}\\ \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}{m}_1{\left(v_1^{\text{'}}\right)}^2+\frac{1}{2}{m}_2{\left(v_2^{\text{'}}\right)}^2\\ m_1{v}_1^2+m_2{v}_2^2=m_1{\left(v_1^{\text{'}}\right)}^2+m_2{\left(v_2^{\text{'}}\right)}^2\end{array}$

Substitute $0.5\text{kg}$ for $m_1$, $1.5\text{kg}$ for $m_2$, $\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_1$, $\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_2$ and $\left(-1\widehat{\text{i}}+3\widehat{\text{j}}+a\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_1^{\text{'}}$ and $-2.67\text{kg}\cdot \text{m}/\text{s}-0.333\left(a\widehat{\text{k}}\right)\text{m}/\text{s}$ for $v_2{}^{\prime }$ in above equation.

    $\begin{array}{l}\left[\begin{array}{l}0.5\text{kg}\times {\left(\left(2\widehat{\text{i}}-3\widehat{\text{j}}+1\widehat{\text{k}}\right)\text{m}/\text{s}\right)}^2\\ +1.5\text{kg}\times {\left(\left(-1\widehat{\text{i}}+2\widehat{\text{j}}-3\widehat{\text{k}}\right)\text{m}/\text{s}\right)}^2\end{array}\right]=\left[\begin{array}{l}0.5\text{kg}{\left(\left(-1\widehat{\text{i}}+3\widehat{\text{j}}+a\widehat{\text{k}}\right)\text{m}/\text{s}\right)}^2\\ +1.5\text{kg}{\left(-2.67\text{kg}\cdot \text{m}/\text{s}-0.333\left(a\widehat{\text{k}}\right)\text{m}/\text{s}\right)}^2\end{array}\right]\\ \left[\begin{array}{l}0.5\text{kg}\times \left(\left(2^2+1^2+3^2\right){{\text{m}}^2/\text{s}}^2\right)\\ +1.5\text{kg}\times {\left(\left(1^2+2^2+3^2\right){\text{m}}^2/{\text{s}}^2\right)}^2\end{array}\right]=\left[\begin{array}{l}0.5\text{kg}{\left(\left(1^2+3^2+a^2\right){\text{m}}^2/{\text{s}}^2\right)}^2\\ +1.5\text{kg}{\left(2.67\text{kg}\cdot \text{m}/\text{s}+0.333\left(a\widehat{\text{k}}\right)\text{m}/\text{s}\right)}^2\end{array}\right]\\ 14.0\text{ J}=\text{2}\text{.50 J}+0.250a^2+5.33\text{ J}+1.33a+0.0833a^2\end{array}$

Solve the above equation for $a$

    $a=-6.74\text{m}/{\text{s}}^2\text{ (or) 2}\text{.74 }\text{m}/{\text{s}}^2$

Substitute $-6.74\text{m}/{\text{s}}^2$ for $a$ in (2) to find $v_2^{\text{'}}$

  $\begin{array}{c}{v}_2^{\text{'}}=\left(-2.67-0.333\left(-6.74\text{m}/{\text{s}}^2\right)\right)\widehat{\text{k}}\text{m}/\text{s}\\ =-0.419\widehat{\text{k}}\text{m}/\text{s}\end{array}$

Substitute $2.74\text{m}/{\text{s}}^2$ for $a$ in (2) to find $v_2^{\text{'}}$

  $\begin{array}{c}{v}_2^{\text{'}}=\left(-2.67-0.333\left(2.74\text{m}/{\text{s}}^2\right)\right)\widehat{\text{k}}\text{m}/\text{s}\\ =-3.58\widehat{\text{k}}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the value of $a$ and the velocity of the $1.5\text{kg}$ sphere after an elastic collision are $-6.74{\text{ m/s}}^2$ and $-0.419\widehat{\text{k}}\text{m}/\text{s}$ or $2.74{\text{ m/s}}^2$ and $-3.58\widehat{\text{k}}\text{m}/\text{s}$.