Chapter 6, Problem 63AP

A 2.0-g particle moving at 8.0 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

To determine

The speed of each particle after collision.

Answer to Problem 63AP

The speeds of the particles are $2.67\text{m}\cdot {\text{s}}^{-1}$ and $10.7\text{m}\cdot {\text{s}}^{-1}$

Explanation of Solution

Explanation

Given Info:

Mass of the moving particle is $2.0\text{g}$, speed of the moving particle is $8.0\text{m}\cdot {\text{s}}^{-1}$ , mass of the particle which is at rest is $1.0\text{g}$, the initial speed of particle at rest is zero.

Apply conservation of momentum for both particles before and after collision,

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

• $m_1$ is the mass of the moving particle
• $v_1$ is initial the speed of the particle
• $m_2$ is the mass of the particle which is at rest
• $v_2$ is the speed of particle at rest.
• $v_1{}^{\prime }$ is the speed of $m_1$ after collision
• $v_2{}^{\prime }$ is the final speed of $m_2$ after collision

Use $0\text{m}/\text{s}$ for $v_2$ in the above equation and rewrite in terms of ${v^{\prime }}_2$.

$v_2{}^{\prime }=\left(\frac{m_1}{m_2}\right)\left(v_1-v_1{}^{\prime }\right)$ (I)

Apply conservation of mechanical energy for both particles,

$\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2=\frac{1}{2}{m}_1{\left(v_1{}^{\prime }\right)}^2+\frac{1}{2}{m}_2{\left(v_2{}^{\prime }\right)}^2$

Use $0\text{m}/\text{s}$ for ${\left({v^{\prime }}_2\right)}^2$ in the above equation and rewrite in terms of ${v^{\prime }}_2$.

${\left(v_2{}^{\prime }\right)}^2=\left(\frac{m_1}{m_2}\right)\left(v_1{}^2-{\left(v_1{}^{\prime }\right)}^2\right)$ (II)

Use $\left(m_1/m_2\right)\left(v_1-v_1{}^{\prime }\right)$ for $v_2{}^{\prime }$ in the above equation and rewrite in terms of $v_1{}^{\prime }$.

$\begin{array}{c}{\left(\left(\frac{m_1}{m_2}\right)\left(v_1-v_1{}^{\prime }\right)\right)}^2=\left(\frac{m_1}{m_2}\right)\left(v_1{}^2-{\left(v_1{}^{\prime }\right)}^2\right)\\ \left(\frac{m_1}{m_2}\right)\left(v_1-v_1{}^{\prime }\right)\left(v_1-v_1{}^{\prime }\right)=\left(v_1-v_1{}^{\prime }\right)\left(v_1+v_1{}^{\prime }\right)\\ v_1\left(\frac{m_1-m_2}{m_2}\right)=v_1{}^{\prime }\left(\frac{m_1+m_2}{m_2}\right)\\ v_1{}^{\prime }=v_1\left(\frac{m_1-m_2}{m_1+m_2}\right)\end{array}$ (III)

Substitute $8.0\text{m}\cdot {\text{s}}^{-1}$ for $v_1$ , $2.0\text{g}$ for $m_1$ and $1.0\text{g}$ for $m_2$ to find $v_1{}^{\prime }$.

$\begin{array}{c}{v}_1{}^{\prime }=\left(8.0\text{m}\cdot {\text{s}}^{-1}\right)\left(\frac{2.0\text{g}-1.0\text{g}}{2.0\text{g+}1.0\text{g}}\right)\\ =\frac{8}{3}\text{m}\cdot {\text{s}}^{-1}\\ =2.67\text{m}\cdot {\text{s}}^{-1}\end{array}$

Substitute $2.67\text{m}\cdot {\text{s}}^{-1}$ for $v_1{}^{\prime }$, $8.0\text{m}\cdot {\text{s}}^{-1}$ for $v_1$ , $2.0\text{g}$ for $m_1$ and $1.0\text{g}$ for $m_2$ in (1) to calculate $v_2{}^{\prime }$,

$\begin{array}{c}{v}_2{}^{\prime }=\left(\frac{2.0\text{g}}{1.0\text{g}}\right)\left(8.0\text{m}\cdot {\text{s}}^{-1}-2.67\text{m}\cdot {\text{s}}^{-1}\right)\\ =10.66\text{m}\cdot {\text{s}}^{-1}\\ \approx 10.7\text{m}\cdot {\text{s}}^{-1}\end{array}$

Conclusion:

Therefore the speeds of the particles are $2.67\text{m}\cdot {\text{s}}^{-1}$ and $10.7\text{m}\cdot {\text{s}}^{-1}$

To determine

The speed of each particle after collision.

Answer to Problem 63AP

The speeds of the particles are $-5.33\text{m}\cdot {\text{s}}^{-1}$ and $2.67\text{m}\cdot {\text{s}}^{-1}$

Explanation of Solution

Explanation

Given Info:

Mass of the moving particle is $2.0\text{g}$, speed of the moving particle is $8.0\text{m}\cdot {\text{s}}^{-1}$ , mass of the particle which is at rest is $10.0\text{g}$, the initial speed of particle at rest is zero.

Formula to calculate speed of the moving particle after collision is,

$v_1{}^{\prime }=v_1\left(\frac{m_1-m_2}{m_1+m_2}\right)$ (IV)

Substitute $8.0\text{m}\cdot {\text{s}}^{-1}$ for $v_1$ , $2.0\text{g}$ for $m_1$ and $10.0\text{g}$ for $m_2$ in (IV) to find $v_1{}^{\prime }$.

$\begin{array}{c}{v}_1{}^{\prime }=\left(8.0\text{m}\cdot {\text{s}}^{-1}\right)\left(\frac{2.0\text{g}-10.0\text{g}}{2.0\text{g+}10.0\text{g}}\right)\\ =-\frac{64}{12}\text{m}\cdot {\text{s}}^{-1}\\ =-5.33\text{m}\cdot {\text{s}}^{-1}\end{array}$

Formula to calculate speed of the particle of mass $m_2$ at rest after collision is,

$v_2{}^{\prime }=\left(\frac{m_1}{m_2}\right)\left(v_1-v_1{}^{\prime }\right)$ (V)

Substitute $5.33\text{m}\cdot {\text{s}}^{-1}$ for $v_1{}^{\prime }$, $8.0\text{m}\cdot {\text{s}}^{-1}$ for $v_1$ , $2.0\text{g}$ for $m_1$ and $10.0\text{g}$ for $m_2$ in the above equation to calculate $v_2{}^{\prime }$,

$\begin{array}{c}{v}_2{}^{\prime }=\left(\frac{2.0\text{g}}{10.0\text{g}}\right)\left(8.0\text{m}\cdot {\text{s}}^{-1}+5.33\text{m}\cdot {\text{s}}^{-1}\right)\\ =2.67\text{m}\cdot {\text{s}}^{-1}\end{array}$

Conclusion:

Therefore the speeds of the particles are $-5.33\text{m}\cdot {\text{s}}^{-1}$ and $2.67\text{m}\cdot {\text{s}}^{-1}$

To determine

The magnitude of kinetic energy of the moving particle for given velocities.

Answer to Problem 63AP

The kinetic energies of the moving particle are $7.13\times {10}^{-3}\text{J}$ and $28.41\times {10}^{-3}\text{J}$ (a)

Explanation of Solution

Explanation

Given Info:

The speeds of the moving particle are. $2.67\text{m}\cdot {\text{s}}^{-1}$ and $-5.33\text{m}\cdot {\text{s}}^{-1}$ in part (a) and part (b) respectively,

Formula to calculate kinetic energy of the moving particle is,

$KE=\frac{1}{2}{m}_1{\left(v_1{}^{\prime }\right)}^2$

Substitute $2.0\text{g}$ for $m_1$ and $2.67\text{m}\cdot {\text{s}}^{-1}$ for $v_1{}^{\prime }$ from part (a) in the above equation to calculate KE.

$\begin{array}{c}KE=\frac{1}{2}\left(2.0\text{g}\right)\left(\frac{1\text{g}}{1000\text{kg}}\right){\left(2.67\text{m}\cdot {\text{s}}^{-1}\right)}^2\\ =7.13\times {10}^{-3}\text{J}\end{array}$

Substitute $2.0\text{g}$ for $m_1$ and $2.67\text{m}\cdot {\text{s}}^{-1}$ for $v_1{}^{\prime }$ from part (b) in the above kinetic energy equation to calculate KE.

$\begin{array}{c}KE=\frac{1}{2}\left(2.0\text{g}\right)\left(\frac{1\text{g}}{1000\text{kg}}\right){\left(-5.33\text{m}\cdot {\text{s}}^{-1}\right)}^2\\ =28.4\times {10}^{-3}\text{J}\end{array}$

Conclusion:

Therefore the kinetic energies of the moving particle are $7.13\times {10}^{-3}\text{J}$ and $28.4\times {10}^{-3}\text{J}$

The incident particle loose more energy in part (a)