Q102P - University Physics with Modern Physics

Question

For the Texas Department of Public Safety, you are investigating an accident that occurred early on a foggy morning in a remote section of the Texas Panhandle. A 2012 Prius traveling due north collided in a highway intersection with a 2013 Dodge Durango that was traveling due east. After the collision, the wreckage of the two vehicles was locked together and skidded across the level ground until it struck a tree. You measure that the tree is $\mathbf{35}\mathbf{\text{ ft}}$ from the point of impact. The line from the point of impact to the tree is in a direction $\mathbf{39}\mathbf{°}$ north of east. From experience, you estimate that the coefficient of kinetic friction between the ground and the wreckage is $\mathbf{0}\mathbf{.}\mathbf{45}$. Shortly before the collision, a highway patrolman with a radar gun measured the speed of the Prius to be $\mathbf{50}\mathbf{\text{ mph}}$ and, according to a witness, the Prius driver made no attempt to slow down. Four people with a total weight of $\mathbf{460}\mathbf{\text{ lb}}$  were in the Durango. The only person in the Prius was the $\mathbf{\text{150 lb}}$ driver. The Durango with its passengers had a weight of $\mathbf{6500}\mathbf{\text{ lb}}$ , and the Prius with its driver had a weight of $\mathbf{\text{3042 lb}}$. (a) What was Durango’s speed just before the collision? (b) How fast was the wreckage traveling just before it struck the tree?

Step-by-Step Solution

Answer

(a) The speed of Durango before collision is $28.85\text{ mph}$.

(b) The speed of wreckage before hitting the tree is $13\text{ mph}$. 

Step 1: A momentum:

Momentum, the product of a particle's mass and its velocity.

The law of conservation of momentum states that in an isolated system, the total momentum of two or more bodies acting on each other remains constant unless an external force acts.

Step 2: Given data and required formulas:

Consider the known data as below.

The Prius with its driver had a weight, $W_{\text{P}}=3042\text{ lb}$

Acceleration due to gravity, $g=32.2\frac{\text{ft}}{{\text{s}}^{\text{2}}}$

The y-component of momentum of Prius before the accident, $p_{P,iy}=500\text{ mph}$

Mass of the Durango, $W_{\text{D}}=6500\text{ lb}$

The line of impact, $\theta =39°$

The coefficient of the friction between block and surface,   

Distance, ${\mu }_{\text{k}}=0.45$

The coefficient of the friction between block and surface,   

Distance, $d=35\text{ ft}$

Step 3: Momentum of Prius:

Before Accident:

To find the x-component of momentum of Prius,

$p_{\text{P,ix}}=m_{\text{P}}{v}_{\text{P,ix}}$

Here, $v_{\text{P,ix}}$ is the x-component of the velocity of Prius before the accident, $m_{\text{P}}$ is the mass of Prius, and $p_{\text{P,ix}}$ is the x-component of momentum of Prius before the accident.

 

Assume $+x$ as eastward and $+y$ as a northward.

 

As Prius is heading north,

Therefore the momentum of Prius toward x-direction is,

 $p_{\text{P,ix}}=0$ 

The expression for the y-component of momentum of Prius before the accident is,

$p_{\text{P,iy}}=m_{\text{P}}{v}_{\text{P,iy}}$

Here, $v_{\text{P,iy}}$ is the y-component of the velocity of Prius before the accident, $m_{\text{P}}$ is the mass of Prius, and $p_{\text{P,iy}}$ is the y-component of momentum of Prius before the accident.

 

The expression to find the weight of Prius.

$$\begin{gathered} W_{\text{P}}=m_{\text{P}}g \\ {m}_{\text{P}}=\frac{W_{\text{P}}}{g} \end{gathered}$$

Substitute known numerical values in the above equation.

  $m_{\text{P}}=\frac{3042\text{ lb}}{32.2{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}$

 

Now putting the above value of $m_{\text{P}}$  and $50\text{ mph}$ for $p_{\text{P,iy}}$ in the expression of momentum.

$\begin{array}{rcl}{p}_{\text{P,iy}}& =& \frac{{\displaystyle 3042\text{ lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}\left(50\text{ mph}\right)\\ & =& \frac{{\displaystyle 3042\text{ lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}\left(50\text{ mph}\right)\left(\frac{1.47{\text{ fts}}^{-1}}{1\text{ mph}}\right)\\ & =& \frac{{\displaystyle 223587\text{ lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}\\ & =& 6943.69\text{ lb}·\text{s}\end{array}$

Step 4: Momentum of Durango:

The expression for the x-component of Durango is,

$p_{\text{D,ix}}=m_{\text{D}}{v}_{\text{D,ix}}$

Here, $v_{\text{D,ix}}$ is the y-component of the velocity of Durango before the accident, $m_{\text{D}}$ is the mass of Prius, and $p_{\text{D,ix}}$ is the x-component of momentum of Durango before the accident.

 

The expression to find weight is,

$m_{\text{D}}=\frac{W_{\text{D}}}{g}$

Substitute known values in the above equation.

$m_{\text{D}}=\frac{\text{6500 lb}}{32.2{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}$

Substitute the above value of $m_{\text{D}}$ in the expression of momentum for Durango.

$\begin{array}{rcl}{p}_{\text{D,ix}}& =& \frac{{\displaystyle \text{6500 lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}{v}_{\text{D,ix}}\\ & =& \left(201.86\times v_{D,ix}\right)\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2\end{array}$

As Durango is heading east, therefore the velocity of Durango toward the y-direction is,

$v_{\text{D,iy}}=0$

Here, $v_{\text{D,iy}}$ is the y-component of the velocity of Durango before the accident. 

The expression for the y-component of momentum 

$\begin{array}{rcl}{p}_{\text{D,iy}}& =& m{v}_{\text{D,iy}}\\ & =& 0\end{array}$

Step 5: Total momentum:

The expression for the total x-component of momentum before the accident is,

$\begin{array}{rcl}{p}_{\text{ix}}& =& p_{\text{P,ix}}+p_{\text{D,ix}}\\ & =& 0\text{ lb}+\left(201.86\times v_{\text{D,ix}}\right)\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2\\ & =& \left(201.86\times v_{\text{D,ix}}\right)\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2\end{array}$

The expression for the total of y-component of the momentum before the accident is,

$\begin{array}{rcl}{p}_{\text{iy}}& =& p_{\text{P,iy}}+p_{\text{D,iy}}\\ & =& 6943.69\text{ lb}·\text{s}+0\text{ lb}·\text{s}\\ & =& 6943.69\text{ lb}·\text{s}\end{array}$

Step 6: The momentum of wrecker:

Mass of wreckage is,

$\begin{array}{rcl}{m}_{\text{w}}& =& \frac{W_{\text{P}}+W_{\text{D}}}{g}\\ & =& \frac{{\displaystyle \left(3042\text{+650}\right)\text{0 lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}\\ & =& 296.33\text{ lb}·{\text{ft}}^{-2}·{\text{s}}^2\end{array}$

The expression for x-component of momentum of wrecker is,

$p_{\text{wx}}=m_{\text{w}}{v}_{\text{wi}}\cos \theta$

Here, $v_{\text{wi}}$ is the velocity of the wrecker at the time of accident, $m_{\text{w}}$ is the mass of the wrecker, $p_{\text{wx}}$ is the x-component of momentum of the wrecker after the accident, and $\theta$ is the line of impact.

Substitute known values in the above equation.

$\begin{array}{rcl}{p}_{\text{wx}}& =& \left(296.33{\text{ lbft}}^{-1}{\text{s}}^2\right)v_{\text{wi}}\cos 39°\\ & =& \left(296.33{\text{ lbft}}^{-1}{\text{s}}^2\right)\left(0.777\right)v_{\text{wi}}\\ & =& \left(230.25{\text{ lbft}}^{-1}{\text{s}}^2\right)v_{\text{wi}}\end{array}$

The expression for y-component of momentum of wrecker is,

$p_{\text{wy}}=m_{\text{w}}{v}_{\text{wi}}sin\theta$

Here, $p_{\text{wy}}$ is the y-component of the momentum of the wrecker after the accident.

Substitute known values in the above equation.

$\begin{array}{rcl}{p}_{\text{wy}}& =& \left(296.33{\text{ lbft}}^{-1}{\text{s}}^2\right)v_{\text{wi}}\sin 39°\\ & =& \left(296.33{\text{ lbft}}^{-1}{\text{s}}^2\right)\left(0.63\right)v_{\text{wi}}\\ & =& \left({\text{186.7 lbft}}^{-1}{\text{s}}^2\right)v_{\text{wi}}\end{array}$

Step 7: (a) Use the law of conservation of momentum:

Equate the momentum along y-axis:

$p_{\text{wy}}=p_{\text{iy}}$

Substitute the value of momentum which is along y-direction; after collision,

$\left(186.7\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2\right)v_{\text{wi}}=6943.69\text{ lb}·\text{s}$

$\begin{array}{rcl}{v}_{\text{wi}}& =& \frac{6943.69\text{ lb}·\text{s}}{186.7\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2}\\ & =& 37.19\text{ ft}·\text{s}{}^{-1}\end{array}$

Equate the momentum along the x-axis:

$p_{\text{wx}}=p_{\text{ix}}$

Substitute the value of momentum,

$$\begin{gathered} \left(\text{230.25 lb}·{\text{ft}}^{-1}·{\text{s}}^2\right)\left(37.19\text{ ft}·{\text{s}}^{-1}\right)=\left(\text{201.86}\times v_{\text{D,ix}}\right)\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2 \\ {v}_{\text{D,ix}}=\frac{\left(\text{230.25 lb}·{\text{ft}}^{-1}·{\text{s}}^2\right)\left(37.19\text{ ft}·{\text{s}}^{-1}\right)}{\left(\text{201.86}\right)\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2} \\ {v}_{\text{D,ix}}=42.42\text{ ft}·{\text{s}}^{-1} \end{gathered}$$

Further solve,

$\begin{array}{rcl}{v}_{\text{D,ix}}& =& 42.42\text{ ft}·{\text{s}}^{-1}\left(\frac{1\text{ mph}}{1.47\text{ ft}·{\text{s}}^{-1}}\right)\\ & =& 28.85\text{ mph}\end{array}$

Hence, the speed of Durango before the collision is $\begin{array}{rcl}& & 28.85\text{ mph}\end{array}$.

Step 8: (b) The speed of wreckage before hitting the tree:

After accident: 

The kinetic energy of wreckage at the time of the accident is,

$K_{\text{i}}=\frac{1}{2}{m}_{\text{w}}{\left(v_{\text{wi}}\right)}^2$

Here, $K_{\text{i}}$ is the kinetic energy of wreckage at the time of the accident and $v_{\text{wi}}$ is the velocity of wreckage at the time of the accident.

 

Substitute known values in the above equation.

$\begin{array}{rcl}{K}_{\text{i}}& =& \frac{1}{2}\left(296.33\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2\right){\left(37.19\text{ ft}·{\text{s}}^{-1}\right)}^2\\ & =& 204926.4\text{ lb}·\text{ft}\end{array}$

 The expression to find the work done by the friction is,

$W_{\text{f}}={\mu }_{\text{k}}\left(m_{\text{D}}+m_{\text{P}}\right)gd$

Here, $W_{\text{f}}$ is the work done by the friction, ${\mu }_{\text{k}}$ is the coefficient of the friction between block and surface, $g$ is the acceleration due to gravity, $m_{\text{P}}$ is the mass of Prius, and  $m_{\text{D}}$ is the mass of Durango.

Substitute known values in the above equation.

$\begin{array}{rcl}{W}_{\text{f}}& =& 0.45\left(\frac{{\displaystyle \text{6500 lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}+\frac{{\displaystyle \text{3042 lb}}}{32.2{{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}^2}\right)32.2{\raisebox{1ex}{${\displaystyle \text{ft}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.}^2\times 35\text{ ft}\\ & =& 1\text{50286.5 lb}·\text{ft}\end{array}$

The kinetic energy of wreckage while hitting the tree is,

$K_{\text{f}}=\frac{1}{2}{m}_{\text{w}}{\left(v_{\text{wf}}\right)}^2$

Here,   is the kinetic energy of wreckage while hitting the tree and $v_{\text{wf}}$ is the velocity of wreckage while hitting the tree.

 

Substitute known value in the above equation.

$\begin{array}{rcl}{K}_{\text{f}}& =& \frac{1}{2}\left(296.33\text{ lb}·{\text{ft}}^{-1}·{\text{s}}^2\right){\left(v_{\text{wf}}\right)}^2\\ & =& \left(\text{148.16 lb}·{\text{ft}}^{-1}·{\text{s}}^2\right){\left(v_{\text{wf}}\right)}^2\end{array}$

Step 9: Use conservation of momentum:

From the law of conservation of energy:

$$\begin{gathered} K_{\text{f}}=K_{\text{i}}-W_{\text{f}} \\ \left(\text{148.16 lb}·{\text{ft}}^{-1}·{\text{s}}^2\right){\left(v_{\text{wf}}\right)}^2=204926.4\text{ lb}·\text{ft}-1\text{50286.5 lb}·\text{ft} \end{gathered}$$

${\left(v_{\text{wf}}\right)}^2=\frac{\text{54639.9 lb}·\text{ft}}{\left(\text{148.16 lb}·{\text{ft}}^{-1}·{\text{s}}^2\right)}$

$\begin{array}{rcl}{v}_{\text{wf}}& =& \sqrt{368.8 \raisebox{1ex}{${\text{ft}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}\\ & =& 19.20 \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\backslash \left(\frac{1\text{ mph}}{1.47\text{ ft}·{\text{s}}^{-1}}\right)\\ & =& 13\text{ mph}\end{array}$

Hence, the speed of wreckage before hitting the tree is $\begin{array}{rcl}& & 13\text{ mph}\end{array}$.