Chapter 12, Problem 6IP

Lineman A has a mass of 100 kg and is traveling with a velocity of 4 m/s when he collides head-on with lineman B, who has a mass of 90 kg and is traveling at 4.5 m/s. If both players remain on their feet, what will happen? (Answer: Lineman B will push lineman A backward with a velocity of 0.03 m/s.)

Summary Introduction

To determine: The kinetic energy of lineman A.

Answer to Problem 6IP

The kinetic energy of lineman A is $800\text{J}$.

Explanation of Solution

Calculation:

Write an expression to find the kinetic energy of lineman A.

${\left(KE\right)}_A=\frac{1}{2}{m}_A{v}_A^2$

Here, ${\left(KE\right)}_A$ is the kinetic energy of lineman A, $m_A$ is the mass of the lineman A, and $v_A$ is the velocity of the lineman A.

Substitute 100 kg for $m_A$ and $4\text{m}/\text{s}$ for $v_A$ to find the kinetic energy of lineman A.

$\begin{array}{c}{\left(KE\right)}_A=\frac{1}{2}\left(100\text{kg}\right){\left(4\text{m}/\text{s}\right)}^2\\ =800\text{kg}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\left(\frac{\text{1}\text{J}}{\text{kg}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =800\text{J}\end{array}$

Therefore, the kinetic energy of lineman A is $800\text{J}$.

Conclusion

Therefore, the kinetic energy of lineman A is $800\text{J}$.

Summary Introduction

To determine: The kinetic energy of lineman B.

Answer to Problem 6IP

The kinetic energy of lineman B is $911\text{J}$.

Explanation of Solution

Calculation:

Write an expression to find the kinetic energy of lineman B.

${\left(KE\right)}_B=\frac{1}{2}{m}_B{v}_B^2$

Here, ${\left(KE\right)}_B$ is the kinetic energy of lineman B, $m_B$ is the mass of the lineman B, and $v_B$ is the velocity of the lineman B.

Substitute 90 kg for $m_B$ and $4.5\text{m}/\text{s}$ for $v_B$ to find the kinetic energy of lineman B.

$\begin{array}{c}{\left(KE\right)}_B=\frac{1}{2}\left(90\text{kg}\right){\left(4.5\text{m}/\text{s}\right)}^2\\ =911.25\text{kg}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\left(\frac{\text{1}\text{J}}{\text{kg}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ \approx 911\text{J}\end{array}$

Therefore, the kinetic energy of lineman B is $911\text{J}$.

Conclusion

Therefore, the kinetic energy of lineman B is $911\text{J}$.

Summary Introduction

To determine: The final velocity of combined mass.

Answer to Problem 6IP

Therefore, the kinetic energy of lineman B is higher than lineman A, thus lineman B will push lineman A backward with a magnitude of velocity $0.03\text{m}/\text{s}$.

Explanation of Solution

Calculation:

Write an expression of conservation of linear momentum to find the final velocity of combined mass.

$m_A{u}_A-m_B{u}_B=\left(m_A+m_B\right)v$

Here, $u_A$ is the velocity of lineman A, $u_B$ is the velocity of the lineman B, and $v$ is the velocity of the combined mass.

Substitute 100 kg for $m_A$, $4\text{m}/\text{s}$ for $u_A$, 90 kg for $m_B$, and $4.5\text{m}/\text{s}$ for $u_B$ to find the velocity of combined mass.

$\left(100\text{kg}\times 4\text{m}/\text{s}\right)-\left(90\text{kg}\times 4.5\text{m}/\text{s}\right)=\left(100\text{kg}+90\text{kg}\right)v$

$\begin{array}{c}400\text{kg}\text{m}/\text{s}-405\text{kg}\text{m}/\text{s}=\left(\text{190}\text{kg}\right)v\\ v=\frac{-5}{190}\text{m}/\text{s}\\ =-0.026\text{m}/\text{s}\\ \approx -0.03\text{m}/\text{s}\end{array}$

Therefore, the velocity of the combined mass is $-0.03\text{m}/\text{s}$.

Conclusion

Therefore, the kinetic energy of lineman B is higher than lineman A, thus, lineman B will push lineman A backward with a magnitude of velocity $0.03\text{m}/\text{s}$.