Chapter 8, Problem 48TP

To determine

(a)

Whether the collision is elastic or inelastic.

Answer to Problem 48TP

The initial kinetic energy is not equal to the final kinetic energy, so the collision is inelastic.

Explanation of Solution

Given:

Let the mass of A be $m_1$.

The mass $m_2$ of B is $m_2=2m_1$.

The initial velocity of mass B is $v_2=0\text{m}/\text{s}$.

The final velocity of both the masses after collisions is $v^{\prime }=6\text{m}/\text{s}$.

Let the initial velocity of mass A be $v_1$.

Formula used:

The formula from law of conservation of momentum is

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

For the collision to be conserved, the final momentum of the system should be equal to the initial momentum.

As the collision is totally inelastic then the above equation becomes,

  $m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v^{\prime }$

Here, $v^{\prime }$ is the final velocity of both the masses.

The initial kinetic energy ${\text{KE}}_{\text{i}}$ of the system is given by

  ${\text{KE}}_{\text{i}}=\frac{1}{2}\left(m_1{v}_1^2+m_2{v}_2^2\right)$

The final kinetic energy ${\text{KE}}_{\text{f}}$ of the system is given by

  ${\text{KE}}_{\text{f}}=\frac{1}{2}\left(m_1+m_2\right){\left(v^{\prime }\right)}^2$

Calculation:

The initial velocity of mass A is calculated as

  $\begin{array}{c}{m}_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v^{\prime }\\ \left(m_1\right)v_1+\left(2m_1\right)\left(0\text{m}/\text{s}\right)=\left(m_1+2m_1\right)\left(6\text{m}/\text{s}\right)\\ m_1{v}_1=18m_1\\ v_1=18\text{m}/\text{s}\end{array}$

The initial kinetic energy ${\text{KE}}_{\text{i}}$ of the system is calculated as

  $\begin{array}{c}{\text{KE}}_{\text{i}}=\frac{1}{2}\left(m_1{v}_1^2+m_2{v}_2^2\right)\\ =\frac{1}{2}\left[\left(m_1\right){\left(18\text{m}/\text{s}\right)}^2+\left(2m_1\right){\left(0\text{m}/\text{s}\right)}^2\right]\\ =162m_1\text{J}\end{array}$

The final kinetic energy ${\text{KE}}_{\text{f}}$ of the system is calculated as

  $\begin{array}{c}{\text{KE}}_{\text{f}}=\frac{1}{2}\left(m_1+m_2\right){\left(v^{\prime }\right)}^2\\ =\frac{1}{2}\left(m_1+2m_1\right){\left(6\text{m}/\text{s}\right)}^2\\ =\left(3m_1\right)\frac{36{\text{m}}^2/{\text{s}}^2}{2}\\ =54m_1\text{J}\end{array}$

Compare the initial and the final kinetic energy.

  $162m_1\text{J}\ne 54m_1\text{J}$

Conclusion:

Therefore, the initial kinetic energy is not equal to the final kinetic energy, so the collision is inelastic.

To determine

(b)

The initial velocity of the mass A.

Answer to Problem 48TP

The initial velocity of mass A is $18\text{m}/\text{s}$.

Explanation of Solution

Calculation:

The initial velocity of mass A is calculated as

  $\begin{array}{c}{m}_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v^{\prime }\\ \left(m_1\right)v_1+\left(2m_1\right)\left(0\text{m}/\text{s}\right)=\left(m_1+2m_1\right)\left(6\text{m}/\text{s}\right)\\ m_1{v}_1=18m_1\\ v_1=18\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the initial velocity of mass A is $18\text{m}/\text{s}$.