Chapter 9, Problem 9.91AP

A 2.00-g particle moving at 8.00 m/s makes a perfectly elastic head-on collision with a resting 1.00-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 10.0 g. (c) Find the final kinetic energy of the incident 2.00-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 velocity of each particle after collision.

Answer to Problem 9.91AP

The velocity of incident particle after collision is $2.67\text{m}/\text{s}$ and velocity of target particle after collision is $10.7\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of incident particle is $2.00\text{g}$, the initial velocity of the particle $8.00\text{m}/\text{s}$ and mass of target particle is $1.00\text{g}$.

Write the condition for velocity of incident particle after collision.

$v_{1\text{f}}=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_{1\text{i}}+\left(\frac{2m_2}{m_1+m_2}\right)v_{2\text{i}}$

Here,

$m_1$ is the mass of the incident particle.

$m_2$ is the mass of the target particle.

$v_{1\text{f}}$ is the final velocity of the incident particle.

$v_{1\text{i}}$ is the initial velocity of incident particle.

$v_{2\text{f}}$ is the final velocity of target particle.

$v_{2\text{i}}$ is the initial velocity of target particle.

The initial velocity of the target particle is $0$.

Substitute $0$ for $v_{2\text{i}}$, $2.00\text{g}$ for $m_1$, $1.00\text{g}$ for $m_2$ and $8.00\text{m}/\text{s}$ for $v_{1\text{i}}$ in above equation.

$\begin{array}{c}{v}_{1\text{f}}=\left(\frac{2.00\text{g}-1.00\text{g}}{2.00\text{g}+1.00\text{g}}\right)\left(8.00\text{m}/\text{s}\right)+\left(\frac{2\left(1.00\text{g}\right)}{2.00\text{g}+1.00\text{g}}\right)\left(0\right)\\ =2.666\text{m}/\text{s}\\ \approx 2.67\text{m}/\text{s}\end{array}$

Thus, the value of $v_{\text{1f}}$ is $2.67\text{m}/\text{s}$.

Write the condition for velocity of target particle after collision.

$v_{2\text{f}}=\left(\frac{2m_1}{m_1+m_2}\right)v_{\text{1i}}$

Substitute $2.00\text{g}$ for $m_1$, $1.00\text{g}$ for $m_2$ and $8.00\text{m}/\text{s}$ for $v_{1\text{i}}$ in above equation.

$\begin{array}{c}{v}_{2\text{f}}=\left(\frac{2\times 2.00\text{g}}{\left(2.00\text{g}\right)+\left(1.00\text{g}\right)}\right)\left(8.00\text{m}/\text{s}\right)\\ =10.666\text{m}/\text{s}\\ \approx 10.7\text{m}/\text{s}\end{array}$

Thus, the value of $v_{2\text{f}}$ is $10.7\text{m}/\text{s}$.

Conclusion:

Therefore, the velocity of incident particle after collision is $2.67\text{m}/\text{s}$ and velocity of target particle after collision is $10.7\text{m}/\text{s}$.

To determine

The velocity of each particle after collision.

Answer to Problem 9.91AP

The velocity of incident particle after collision is $-5.33\text{m}/\text{s}$ and velocity of target particle after collision is $2.7\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of incident particle is $2.00\text{g}$, the initial velocity of the particle $8.00\text{m}/\text{s}$ and mass of target particle is $10.00\text{g}$.

Write the condition for velocity of incident particle after collision.

$v_{1\text{f}}=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_{1\text{i}}+\left(\frac{2m_2}{m_1+m_2}\right)v_{2\text{i}}$

Here,

$m_1$ is the mass of the incident particle.

$m_2$ is the mass of the target particle.

$v_{1\text{f}}$ is the final velocity of the incident particle.

$v_{1\text{i}}$ is the initial velocity of incident particle.

$v_{2\text{f}}$ is the final velocity of target particle.

$v_{2\text{i}}$ is the initial velocity of target particle.

The initial velocity of the target particle is $0$.

Substitute $0$ for $v_{2\text{i}}$, $2.00\text{g}$ for $m_1$, $10.00\text{g}$ for $m_2$ and $8.00\text{m}/\text{s}$ for $v_{1\text{i}}$ in above equation.

$\begin{array}{c}{v}_{1\text{f}}=\left(\frac{2.00\text{g}-10.00\text{g}}{2.00\text{g}+10.00\text{g}}\right)\left(8.00\text{m}/\text{s}\right)+\left(\frac{2\left(1.00\text{g}\right)}{2.00\text{g}+1.00\text{g}}\right)\left(0\right)\\ =-5.33\text{m}/\text{s}\end{array}$

Thus, the value of $v_{\text{1f}}$ is $-5.33\text{m}/\text{s}$.

Write the condition for velocity of target particle after collision.

$v_{2\text{f}}=\left(\frac{2m_1}{m_1+m_2}\right)v_{\text{1i}}$

Substitute $2.00\text{g}$ for $m_1$, $10.00\text{g}$ for $m_2$ and $8.00\text{m}/\text{s}$ for $v_{1\text{i}}$ in above equation.

$\begin{array}{c}{v}_{2\text{f}}=\left(\frac{2\times 2.00\text{g}}{\left(2.00\text{g}\right)+\left(10.00\text{g}\right)}\right)\left(8.00\text{m}/\text{s}\right)\\ =2.666\text{m}/\text{s}\\ \approx 2.7\text{m}/\text{s}\end{array}$

Thus, the value of $v_{2\text{f}}$ is $2.7\text{m}/\text{s}$.

Conclusion:

Therefore, the velocity of incident particle after collision is $-5.33\text{m}/\text{s}$ and velocity of target particle after collision is $2.7\text{m}/\text{s}$.

To determine

The kinetic energy of the incident particle in the situation described in part (a) and (b) and the case in which more kinetic energy is lost.

Answer to Problem 9.91AP

The kinetic energy of the incident particle in the situation described in part (a) is $7.11\times {10}^{-3}\text{J}$ and in part (b) is $2.84\times {10}^{-2}\text{J}$ and the incident particle loses more energy in case (a).

Explanation of Solution

Given info: The mass of incident particle is $2.00\text{g}$, the initial velocity of the particle $8.00\text{m}/\text{s}$ and mass of target particle is $10.00\text{g}$.

Case (a);

From part (a), the velocity of incident particle after collision is $2.67\text{m}/\text{s}$.

Write the expression for final kinetic energy of incident particle for case (a).

$K{E}_{1\text{f}}=\frac{1}{2}{m}_1{v}_{1\text{f}}{}^2$

Here,

$K{E}_{\text{1f}}$ is the final kinetic energy of incident particle for case (a).

Substitute  $2.00\text{g}$ for $m_1$ and $2.67\text{m}/\text{s}$ for $v_{1\text{f}}$ in above equation.

$\begin{array}{c}K{E}_{1\text{f}}=\frac{1}{2}\times 2.00\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\times {\left(2.67\text{m}/\text{s}\right)}^2\\ \approx 7.11\times {10}^{-3}\text{J}\end{array}$

Thus, the value of $K{E}_{\text{1f}}$ is $7.11\times {10}^{-3}\text{J}$.

Case (b);

From part (b), the velocity of incident particle after collision is $-5.33\text{m}/\text{s}$.

Write the expression for final kinetic energy of incident particle case (b).

$K{E}_{1\text{f}}=\frac{1}{2}{m}_1{v}_{1\text{f}}{}^2$

Here,

$K{E}_{\text{1f}}$ is the final kinetic energy of incident particle for case (a).

Substitute  $2.00\text{g}$ for $m_1$ and $-5.33\text{m}/\text{s}$ for $v_{1\text{f}}$ in above equation.

$\begin{array}{c}K{E}_{1\text{f}}=\frac{1}{2}\times 2.00\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\times {\left(-5.33\text{m}/\text{s}\right)}^2\\ =2.84\times {10}^{-2}\text{J}\end{array}$

Thus, the value of $K{E}_{\text{1f}}$ is $2.84\times {10}^{-2}\text{J}$.

Since, the incident kinetic energy is almost same in both cases.

The incident particle loses more kinetic energy in case (a) where the mass of the incident particle is $1.00\text{g}$.

Conclusion:

Therefore, the kinetic energy of the incident particle in the situation described in part (a) is $7.11\times {10}^{-3}\text{J}$ and in part (b) is $2.84\times {10}^{-2}\text{J}$ and the incident particle loses more energy in case (a).