Chapter 11, Problem 67A

(a)

To determine

The momentum of first railroad car before collision.

Answer to Problem 67A

  ${\text{p}}_{i1}=4.0\times {10}^6\text{ kg}\text{.m/s}$

Explanation of Solution

Given:

Mass of each railroad car $1$ is $m_1=5.0\times {10}^5\text{ kg}$ .

Initially the railroad car $1$ is moving and the railroad car $2$ is at stationary.

Initial velocity of the railroad car $1$ is $v_{i1}=8.0\text{ m/s}$ .

Formula used:

Momentum $\left(p\right)$ is the product of mass $\left(m\right)$ and velocity $\left(v\right)$ of a moving body. That is,

  $p=mv$

The momentum of first railroad car before collision can be calculated using the equation,

  ${\text{p}}_{i1}=m_1{v}_{i1}$  $\left(1\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  ${\text{p}}_{i1}=\left(5.0\times {10}^5\text{ kg}\right)\left(8.0\text{ m/s}\right)$

  $=4.0\times {10}^6\text{ kg}\text{.m/s}$

Conclusion:

The momentum of first railroad car before collision is $4.0\times {10}^6\text{ kg}\text{.m/s}$ .

(b)

To determine

Total momentum of the two cars after collision.

Answer to Problem 67A

  ${\text{p}}_f=4.0\times {10}^6\text{ kg}\text{.m/s}$

Explanation of Solution

Given:

Mass of each railroad car $1$ is $m_1=5.0\times {10}^5\text{ kg}$

Mass of each railroad car $2$ is $m_2=5.0\times {10}^5\text{ kg}$

Final velocity of the cars is $v_f=4.0\text{ m/s}$

Formula used:

Total momentum of the two cars after collision can be calculated using the equation,

  ${\text{p}}_f=m{v}_f$  $\left(2\right)$

Calculation:

After collision, the two cars lock together. Hence, the total mass of the cars is $m=\left(5.0\times {10}^5\text{ kg}\right)+\left(5.0\times {10}^5\text{ kg}\right)\text{ = 10}\times {10}^5\text{ kg}$

Substituting the numerical values in equation $\left(2\right)$ ,

  ${\text{p}}_f=\left(\text{10}\times {10}^5\text{ kg}\right)\left(4.0\text{ m/s}\right)$

  $=4.0\times {10}^6\text{ kg}\text{.m/s}$

Conclusion:

Total momentum of the two cars after collision is $4.0\times {10}^6\text{ kg}\text{.m/s}$ .

(c)

To determine

The kinetic energies of the two cars before and after the collision.

Answer to Problem 67A

  ${\text{KE}}_{i1}=1.60\times {10}^7\text{ J}$

  ${\text{KE}}_{i2}=0.0\text{ J}$

  $K{E}_f=8.0\times {10}^6\text{ J}$

Explanation of Solution

Given:

Mass of each railroad car $1$ is $m_1=5.0\times {10}^5\text{ kg}$

Mass of each railroad car $2$ is $m_2=5.0\times {10}^5\text{ kg}$

Initial velocity of the railroad car $1$ is $v_{i1}=8.0\text{ m/s}$

Initial velocity of the railroad car $2$ is $v_{i2}=0.0\text{ m/s}$

Final velocity of the cars is $v_f=4.0\text{ m/s}$

Formula used:

Kinetic energy $\left(KE\right)$ of a body of mass $\left(m\right)$ moving with velocity $\left(v\right)$ is given by,

  $KE=\frac{1}{2}m{v}^2$

Kinetic energy of the railroad car $1$ before the collision can be calculated using the equation,

  ${\text{KE}}_{i1}=\frac{1}{2}{m}_1{v}_{i1}^2$  $\left(3\right)$

Kinetic energy of the railroad car $2$ before the collision can be calculated using the equation,

  ${\text{KE}}_{i2}=\frac{1}{2}{m}_2{v}_{i2}^2$  $\left(4\right)$

Kinetic energy of the railroad cars after the collision can be calculated using the equation,

  $K{E}_f=\frac{1}{2}m{v}_f^2$  $\left(5\right)$

Calculation:

After the collision the two cars lock together, hence the total mass of the cars is $m=\left(5.0\times {10}^5\text{ kg}\right)+\left(5.0\times {10}^5\text{ kg}\right)\text{ = 10}\times {10}^5\text{ kg}$

Kinetic energy of the railroad car $1$ before the collision can be calculated by substituting the numerical values in equation $\left(3\right)$ ,

  ${\text{KE}}_{i1}=\frac{1}{2}\left(5.0\times {10}^5\text{ kg}\right){\left(8.0\text{ m/s}\right)}^2$

  $=\frac{1}{2}\left(5.0\times {10}^5\text{ kg}\right)\left(64.0{\text{ m}}^2{\text{/s}}^2\right)$

  $=1.60\times {10}^7\text{ kg}{\text{.m}}^2{\text{/s}}^2$

  $=1.60\times {10}^7\text{ J}$

Kinetic energy of the railroad car $2$ before the collision can be calculated by substituting the numerical values in equation $\left(4\right)$ ,

  ${\text{KE}}_{i2}=\frac{1}{2}\left(5.0\times {10}^5\text{ kg}\right){\left(0.0\text{ m/s}\right)}^2$

  $=0.0\text{ kg}{\text{.m}}^2{\text{/s}}^2$

  $=0.0\text{ J}$

Kinetic energy of the railroad cars after the collision can be calculated by substituting the numerical values in equation $\left(5\right)$ ,

  $K{E}_f=\frac{1}{2}\left(\text{10}\times {10}^5\text{ kg}\right){\left(4.0\text{ m/s}\right)}^2$

  $=\frac{1}{2}\left(10.0\times {10}^5\text{ kg}\right)\left(16.0{\text{ m}}^2{\text{/s}}^2\right)$

  $=8.0\times {10}^6\text{ kg}{\text{.m}}^2{\text{/s}}^2$

  $=8.0\times {10}^6\text{ J}$

Conclusion:

Kinetic energy of the railroad car $1$ before the collision is $1.60\times {10}^7\text{ J}$ .

Kinetic energy of the railroad car $2$ before the collision is $0.0\text{ J}$ .

Kinetic energy of the railroad cars after the collision is $8.0\times {10}^6\text{ J}$ .

(d)

To determine

To explain: The change of kinetic energy.

Answer to Problem 67A

The loss in kinetic energy during the collision is converted in to the thermal energy and sound energy.

Explanation of Solution

Given:

Mass of each railroad car $1$ is $m_1=5.0\times {10}^5\text{ kg}$

Mass of each railroad car $2$ is $m_2=5.0\times {10}^5\text{ kg}$

Initial velocity of the railroad car $1$ is $v_{i1}=8.0\text{ m/s}$

Initial velocity of the railroad car $2$ is $v_{i2}=0.0\text{ m/s}$

Final velocity of the cars is $v_f=4.0\text{ m/s}$

Formula used:

Kinetic energy $\left(KE\right)$ of a body of mass $\left(m\right)$ moving with velocity $\left(v\right)$ is given by,

  $KE=\frac{1}{2}m{v}^2$

The momentum of the body is conserved during the collision but the kinetic energy of the body is not conserved during the collision.

As calculated in the part (c), the kinetic energy of the railroad cars before collision is $1.60\times {10}^7\text{ J}$ and the kinetic energy of the railroad cars after collision is $8.0\times {10}^6\text{ J}$ .

Therefore, the loss in kinetic energy during the collision is,

  $K{E}_f=1.60\times {10}^7\text{ J}-8.0\times {10}^6\text{ J = }8.0\times {10}^6\text{ J}$

The loss in kinetic energy during the collision is converted in to the thermal energy and sound energy.