Chapter 11, Problem 32PQ

To determine

The velocity of the system just after the collision using Newton’s second law and the comparison of the answer with the result of Example 11.4.

Answer to Problem 32PQ

The velocity of the system just after the collision using Newton’s second law is $\left(9\le {\overrightarrow{\text{v}}}_{1i}\le 21\right)\widehat{\text{i}}\text{ m/s}$ and it is equal to the result of Example 11.4.

Explanation of Solution

The free-body diagram of the system is shown in figure 1.

The two-train system only moves in $x$ direction.

Write the expression for the Newton’s second law in $x$ direction.

  $\Sigma F_x=M{a}_x$                                                                                                               (I)

Here, $\Sigma F_x$ is the magnitude of the net force in $x$ direction, $M$ is the mass of the two-train system and $a_x$ is the acceleration in $x$ direction.

Refer to figure 1. The only force acting in $x$ direction is the force of kinetic friction and it acts in $-x$ direction.

Write the equation for $\Sigma F_x$ .

  $\Sigma F_x=-F_k$

Here, $F_k$ is the kinetic friction.

Put the above equation in equation (I).

  $-F_k=M{a}_x$                                                                                                              (II)

The kinetic friction is proportional to the normal force and the normal force is in turn equal to the weight of the train.

Write the expression for $F_k$ .

  $F_k={\mu }_k{F}_g$                                                                                                               (III)

Here, ${\mu }_k$ is the coefficient of kinetic friction and $F_g$ is the weight of the train.

Write the equation for $F_g$ .

  $F_g=Mg$

Here, $g$ is the acceleration due to gravity.

Put the above equation in equation (III).

  $F_k={\mu }_{k}Mg$

Put the above equation in equation (II) and rewrite it for $a_x$ .

  $\begin{array}{c}-{\mu }_{k}Mg=M{a}_x\\ a_x=-{\mu }_{k}g\end{array}$                                                                                                     (IV)

Initially only the freight train has momentum. Assume that the two trains move together with velocity $v_c$ after the collision. Apply the conservation of momentum for the two-train system before and after the collision.

  $m_1{v}_{1i}\widehat{\text{i}}=M{v}_c\widehat{\text{i}}$

Here, $m_1$ is the mass of the freight train, $v_{1i}$ is the initial speed of the freight train, $m_2$ is the mass of the passenger train and $v_c$ is the speed with the two trains move after collision.

Rewrite the above equation for $v_c$ .

  $v_c=\frac{m_1}{M}{v}_{1i}$

Now consider the motion of the system just after the collision to the moment it comes to rest.

Replace $v_c$ in the above equation by $v_{x,0}$ where $v_{x,0}$ denotes the speed of the system just after the collision.

  $v_{x,0}=\frac{m_1}{M}{v}_{1i}$                                                                                                              (V)

The speed of the two-train system as it comes to rest is zero.

Write the expression for the final speed of the system as it comes to rest.

  $v_x=0$                                                                                                                     (VI)

Here, $v_x$ is the final speed of the system as it comes to rest.

Write the constant-acceleration equation of motion.

  $v_x^2=v_{x,0}^2+2a_x\Delta x$

Here, $\Delta x$ is the distance travelled by the two-train system before it comes to rest.

Put equations (IV) to (VI) in the above equation and rewrite it for $v_{1i}$ .

  $\begin{array}{c}0={\left(\frac{m_1}{M}{v}_{1i}\right)}^2+2\left(-{\mu }_{k}g\right)\Delta x\\ \frac{m_1}{M}{v}_{1i}=\sqrt{2{\mu }_{k}g\Delta x}\\ v_{1i}=\frac{M}{m_1}\sqrt{2{\mu }_{k}g\Delta x}\end{array}$                                                                          (VII)

To find the range of possible initial freight train velocities, the extreme values of the coefficient of kinetic friction must be used.

Write the equation for $M$ .

  $M=m_1+m_2$                                                                                                        (VIII)

Here, $m_2$ is the mass of the passenger train.

Conclusion:

Given that the value of $m_1$ is $5.221\times {10}^6\text{ kg}$ , the value of $m_2$ is $4.1\times {10}^5\text{ kg}$ , the minimum value of the coefficient of kinetic friction is $0.3$ and its maximum value is $0.4$ and the distance travelled by the system before getting stopped is $370\text{ ft}$ . The value of $g$ is $9.81{\text{ m/s}}^2$ .

Substitute $5.221\times {10}^6\text{ kg}$ for $m_1$ and $4.1\times {10}^5\text{ kg}$ for $m_2$ in equation (VIII) to find the value of $M$ .

  $\begin{array}{c}M=5.221\times {10}^6\text{ kg}+4.1\times {10}^5\text{ kg}\\ =5.63\times {10}^6\text{ kg}\end{array}$

Substitute $5.63\times {10}^6\text{ kg}$ for $M$ , $5.221\times {10}^6\text{ kg}$ for $m_1$, $0.3$ for ${\mu }_k$ , $9.81{\text{ m/s}}^2$ for $g$ and $370\text{ ft}$ for $\Delta x$ in equation (I) to find the minimum value of the speed of the two-train system.

  $\begin{array}{c}{v}_{1i}\left(\text{min}\right)=\frac{5.63\times {10}^6\text{ kg}}{5.221\times {10}^6\text{ kg}}\sqrt{2\left(0.3\right)\left(9.81{\text{ m/s}}^2\right)\left(370\text{ ft}\frac{0.3048\text{ m}}{1\text{ ft}}\right)}\\ =\frac{5.63\times {10}^6\text{ kg}}{5.221\times {10}^6\text{ kg}}\sqrt{2\left(0.3\right)\left(9.81{\text{ m/s}}^2\right)\left(1.1\times {10}^2\text{ m}\right)}\\ =9\text{ m/s}\end{array}$

Here, $v_{1i}\left(\min \right)$ is the minimum speed of the system.

The system moves in $x$ direction. Write the expression for the minimum velocity of the system.

  ${\overrightarrow{\text{v}}}_{1i}\left(\min \right)=v_{1i}\left(\min \right)\widehat{\text{i}}$

Here, ${\overrightarrow{\text{v}}}_{1i}\left(\min \right)$ is the minimum velocity of the system.

Substitute $9\text{ m/s}$ for $v_{1i}\left(\min \right)$ in the above equation to find ${\overrightarrow{\text{v}}}_{1i}\left(\min \right)$ .

  ${\overrightarrow{\text{v}}}_{1i}\left(\min \right)=9\widehat{\text{i}}\text{ m/s}$

Substitute $5.63\times {10}^6\text{ kg}$ for $M$ , $5.221\times {10}^6\text{ kg}$ for $m_1$, $0.4$ for ${\mu }_k$ , $9.81{\text{ m/s}}^2$ for $g$ and $370\text{ ft}$ for $\Delta x$ in equation (I) to find the maximum value of the speed of the two-train system.

  $\begin{array}{c}{v}_{1i}\left(\text{max}\right)=\frac{5.63\times {10}^6\text{ kg}}{5.221\times {10}^6\text{ kg}}\sqrt{2\left(0.4\right)\left(9.81{\text{ m/s}}^2\right)\left(370\text{ ft}\frac{0.3048\text{ m}}{1\text{ ft}}\right)}\\ =\frac{5.63\times {10}^6\text{ kg}}{5.221\times {10}^6\text{ kg}}\sqrt{2\left(0.4\right)\left(9.81{\text{ m/s}}^2\right)\left(1.1\times {10}^2\text{ m}\right)}\\ =30\text{ m/s}\end{array}$

Here, $v_{1i}\left(\max \right)$ is the maximum speed of the system.

It is given that the train crosses the red signal at a speed of $48\text{ mph}$ . Convert this speed into units of m/s.

  $\begin{array}{c}48\text{ mph}=48\text{ mph}\frac{1609.344\text{ m}}{1\text{ mile}}\frac{1\text{ h}}{\left(60\times 60\right)\text{ s}}\\ =21\text{ m/s}\end{array}$

Since the train passed the red signal with the speed of $21\text{ m/s}$ , the maximum speed cannot be greater than this and the upper limit will be replaced by this.

Write the expression for the maximum velocity of the system.

  ${\overrightarrow{\text{v}}}_{1i}\left(\max \right)=v_{1i}\left(\max \right)\widehat{\text{i}}$

Here, ${\overrightarrow{\text{v}}}_{1i}\left(\max \right)$ is the maximum velocity of the system.

Substitute $21\text{ m/s}$ for $v_{1i}\left(\max \right)$ in the above equation to find ${\overrightarrow{\text{v}}}_{1i}\left(\max \right)$ .

  ${\overrightarrow{\text{v}}}_{1i}\left(\max \right)=21\widehat{\text{i}}\text{ m/s}$

The range of the speed of the two-train system is $\left(9\le {\overrightarrow{\text{v}}}_{1i}\le 21\right)\widehat{\text{i}}\text{ m/s}$ and the result obtained in Example 11.4 is the same.

Therefore, the velocity of the system just after the collision using Newton’s second law is $\left(9\le {\overrightarrow{\text{v}}}_{1i}\le 21\right)\widehat{\text{i}}\text{ m/s}$ and it is equal to the result of Example 11.4.