Chapter 8, Problem 44SP

A 6000-kg truck traveling north at 5.0 m/s collides with a 4000-kg truck moving west at 15 m/s. If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?

To determine

The final speed and direction of locked trucks after collision, given that the truck of mass $6000\text{ kg}$ travelling north at the speed of $5.0\text{ m/s}$ collides with the truck of mass $4000\text{ kg}$ travelling west at the speed of $15\text{ m/s}$.

Answer to Problem 44SP

Solution: The speed of the locked trucks is $6.7\text{ m/s}$ and it is directed at an angle of $26.56°$ north of west.

Explanation of Solution

Given data:

The mass of the truck travelling north is $6000\text{ kg}$.

The speed of the truck travelling north is $5.0\text{ m/s}$.

The mass of the truck travelling west is $4000\text{ kg}$.

The speed of the truck travelling west is $15\text{ m/s}$.

Formula used:

The momentum of a body is expressed as,

$p=mv$

Here, $v$ is the velocity of the body, $m$ is the mass of the body, and $p$ is the momentum of the body.

Understand that during a collision between bodies 1 and 2, when the bodies after the collision combine to form a single body, the momentum in a particular direction before and after the collision is expressed as,

$p_{i1}+p_{i2}=p_f$

Here, $p_{i1}$ is the momentum of body 1 in a particular direction before collision, $p_{i2}$ is the momentum of body 2 in that direction before collision, and $p_{f1}$ is the momentum of the combined system in that direction after collision.

The resultant of the horizontal and vertical components of velocity is expressed as,

$v_R=\sqrt{v_H^2+v_V^2}$

Here, $v_R$ is the resultant velocity, $v_H$ is the velocity in horizontal direction and $v_V$ is the velocity in vertical direction.

The direction of inclination of the resultant velocity with the horizontal is expressed as,

$\tan \beta =\frac{v_V}{v_H}$

Here, $\beta$ is the inclination of the resultant velocity with the horizontal measured anticlockwise from the horizontal.

Explanation:

Consider the truck moving north as body A and the truck moving west as body B. Also, assuming positive x-direction as East direction and positive y-direction as North direction.

Now, consider the expression for initial momentum of body A in x-direction before collision

$p_{1ix}=m_1{v}_{1ix}$

Here, $v_{1ix}$ is the component of velocity of the body A in x-direction before collision, $p_{1ix}$ is the momentum of the body A in x-direction before collision, and $m_1$ is the mass of body A.

Understand that body A or the truck of mass $6000\text{ kg}$ is travelling north that is in y-direction. Therefore, body A has no initial velocity in x-direction. Therefore, substitute $6000\text{ kg}$ for $m_1$ and 0 for $v_{1ix}$.

$\begin{array}{c}{p}_{1ix}=\left(6000\text{ kg}\right)0\\ =0\text{ kg}\cdot \text{m/s}\end{array}$

Consider the expression for initial momentum of body B in x-direction before collision

$p_{2ix}=m_2{v}_{2ix}$

Here, $v_{2ix}$ is the component of velocity of body B in x direction before collision, $p_{2ix}$ is the momentum of the body B in x direction before collision, and $m_2$ is the mass of the body B.

Understand that the body B or the truck of mass $4000\text{ kg}$ is moving in the west direction, that is along the negative x-axis. Therefore, substitute $-15\text{ m/s}$ for $v_{2ix}$ and $4000\text{ kg}$ for $m_2$.

$\begin{array}{c}{p}_{2ix}=\left(4000\text{ kg}\right)\left(-15\text{ m/s}\right)\\ =-60000\text{ kg}\cdot \text{m/s}\end{array}$

Understand that according to the problem, after the collision, both the bodies form a single system and move as a single body having mass equal to sum of masses of body A and body B.

Consider the mass of the final combined body after the collision

$m_3=m_2+m_1$

Here, $m_3$ is the mass of the combined system formed by sticking of bodies A and B after collision.

Substitute $6000\text{ kg}$ for $m_1$ and $4000\text{ kg}$ for $m_2$

$\begin{array}{c}{m}_3=6000\text{ kg}+4000\text{ kg}\\ =10000\text{kg}\end{array}$

Consider the expression for the momentum of the combined system after collision

$p_{fx}=m_3{v}_x$

Here, $p_{fx}$ is the final momentum of the system in x-direction after collision and $v_x$ is the horizontal component of velocity of the combined system.

Substitute $\text{10}\text{000 kg}$ for $m_3$

$\begin{array}{c}{p}_{fx}=\left(\text{10}\text{000 kg}\right)v_x\\ =\text{10}\text{000}{v}_x\end{array}$

Consider the expression for conservation of momentum for the collision

$p_{1ix}+p_{2ix}=p_{fx}$

Substitute $\text{10000}{v}_x$ for $p_{fx}$, $0\text{ kg}\cdot \text{m/s}$ for $p_{1ix}$, and $-60000\text{ kg}\cdot \text{m/s}$ for $p_{2ix}$

$\begin{array}{c}\left(0\text{ kg}\cdot \text{m/s}\right)+\left(-60000\text{ kg}\cdot \text{m/s}\right)=10000v_x\\ 10000v_x=-60000\\ v_x=\frac{-60000}{10000}\\ v_x=-6\text{ m/s}\end{array}$

The negative sign indicates that the horizontal component of velocity is in the negative x-direction that is towards west. Therefore, the velocity of the combined system in x-direction is $6\text{ m/s}$ towards west.

Consider the expression for initial momentum of body A in y-direction before collision

$p_{1iy}=m_1{v}_{1iy}$

Here, $v_{1iy}$ is the component of velocity of the body A in y-direction before collision and $p_{1iy}$ is the momentum of the body A in y-direction before collision.

Understand that body A or the truck of mass $6000\text{ kg}$ is travelling north that is in the positive y-direction. Therefore, substitute $6000\text{ kg}$ for $m_1$ and $5.0\text{ m/s}$ for $p_{1iy}$

$\begin{array}{c}{p}_{1iy}=\left(6000\text{ kg}\right)\left(5.0\text{ m/s}\right)\\ =30000\text{ kg}\cdot \text{m/s}\end{array}$

Consider the expression for initial momentum of body B in y-direction before collision

$p_{2iy}=m_2{v}_{2iy}$

Here, $v_{2iy}$ is the component of velocity of body B in y-direction before collision and $p_{2iy}$ is the momentum of body B in y-direction before collision.

Understand that body B or the truck of mass $4000\text{ kg}$ is moving in the west direction that is along the negative x-axis. Therefore, body B has no motion in y-direction. Therefore, substitute $0\text{ m/s}$ for $v_{2iy}$ and $4000\text{ kg}$ for $m_2$.

$\begin{array}{c}{p}_{2iy}=\left(4000\text{ kg}\right)\left(0\text{ m/s}\right)\\ =0\text{ kg}\cdot \text{m/s}\end{array}$

Consider the expression for the momentum of the combined system after collision

$p_{fy}=m_3{v}_y$

Here, $p_{fy}$ is the final momentum of the system in y-direction after collision and $v_y$ is the vertical component of velocity of the combined system.

Substitute $\text{10}\text{000 kg}$ for $m_3$

$\begin{array}{c}{p}_{fy}=\left(\text{10}\text{000 kg}\right)v_y\\ =\text{10}\text{000}{v}_y\end{array}$

Consider the expression for conservation of momentum for the collision.

$p_{1iy}+p_{2iy}=p_{fy}$

Substitute $\text{10000}{v}_y$ for $p_{fy}$, $0\text{ kg}\cdot \text{m/s}$ for $p_{2iy}$, and $30000\text{ kg}\cdot \text{m/s}$ for $p_{1iy}$.

$\begin{array}{c}\left(30000\text{ kg}\cdot \text{m/s}\right)+\left(0\text{ kg}\cdot \text{m/s}\right)=10000v_y\\ 10000v_y=30000\\ v_y=\frac{30000}{10000}\\ v_y=3\text{ m/s}\end{array}$

The positive sign indicates that the vertical component of velocity is in the positive y-direction that is toward north. Therefore, the component of velocity of the combined system in y-direction is $3\text{ m/s}$ toward north.

Write the expression for resultant final velocity of the system.

$v=\sqrt{v_x^2+v_y^2}$

Substitute $3\text{ m/s}$ for $v_y$ and $-6\text{ m/s}$ for $v_x$.

$\begin{array}{c}v=\sqrt{{\left(-6\text{ m/s}\right)}^2+{\left(3\text{ m/s}\right)}^2}\\ =\sqrt{36+9}\\ =\sqrt{45}\\ =6.7\text{ m/s}\end{array}$

The velocity of locked trucks is $6.7\text{ m/s}$.

Consider the expression for angle made by the resultant velocity with the horizontal.

$\tan \theta =\frac{v_y}{v_x}$

Here, $\theta$ is the angle of resultant velocity of the locked trucks with the horizontal.

Substitute $3\text{ m/s}$ for $v_y$ and $-6\text{ m/s}$ for $v_x$.

$\begin{array}{l}\tan \theta =\frac{3\text{ m/s}}{-6\text{ m/s}}\\ \tan \theta =-\frac{1}{2}\\ \theta ={\tan }^{-1}\left(-\frac{1}{2}\right)\\ \theta =-26.56°\end{array}$

The negative sign indicates that the angle is measured in clockwise direction from the x-axis. Therefore, the resultant velocity is inclined at an angle $26.56°$ with the negative x-axis measured clockwise, that is, the resultant velocity is inclined at an angle of $26.56°$ north of west.

Conclusion:

The speed of the locked trucks is $6.7\text{ m/s}$ and it is directed at an angle of $26.56°$ north of west.