Question

A car of masscollides with a truck of mass$\mathbf{2800}\mathbf{\hspace{0.33em}}\mathbf{kg}$, and just after the collision the car and truck slide along, stuck together. The car’s velocity just before the collision was$\mathbf{(}\mathbf{40}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$, and the truck’s velocity just before the collision was$\mathbf{(}\mathbf{-}\mathbf{14}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{29}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$. (a) What is the velocity of the stuck-together car and truck just after the collision? (b) In your analysis in part (a), why can you neglect the effect of the force of the road on the car and truck?

Short answer

a) the velocity of the stuck-together car and truck just after the collision is$19.185\hspace{0.33em}m/s$ and b) the reason for neglecting is the short collision time and also the negligible impulse that is compared to the large impulse between the truck and the car.

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the car is$2800kg$.
• The velocity of the car before collision is$(40,0,0)\hspace{0.33em}m/s$.
• The mass of the truck is$4700kg$.
• The velocity of the truck before the collision is$(-14,0,29)\hspace{0.33em}m/s$ .

significance of the law of conservation of momentum for the objects

This law states that the total momentum of a body before and after collision becomes equal if no external forces are involved.

The equation of the final velocity gives the velocity of the struck-together car and truck and also the effect of the forces.

Determination of the velocity of the struck-together car and truck along with the analysis

a) From the law of conservation of momentum, the equation of the velocity of the stuck-together car and truck is expressed as-

$v=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

Here, v is the final velocity, $m_{1}and{m}_2$are the masses of the car and the truck and $v_1\hspace{0.33em}and\hspace{0.33em}{v}_2$are the velocities of the car and the truck respectively.

$$\begin{gathered} v=\frac{(2800\mathrm{kg})\times \left(\right(40,0,0)\mathrm{m}/\mathrm{s})+(4700\mathrm{kg})\times \left(\right(-14,0,29)\mathrm{m}/\mathrm{s})}{(2800\mathrm{kg})+(4700\mathrm{kg})} \\ v=\frac{(2800\mathrm{kg})\times \left(\right(40,0,0)\mathrm{m}/\mathrm{s})+(4700\mathrm{kg})\times \left(\right(-14,0,29)\mathrm{m}/\mathrm{s})}{7500\mathrm{kg}} \\ \mathrm{v}=(6.16,0,18.17)\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, with the help of the coordinates, the velocity of the stuck-together car and truck is expressed as-

$$\begin{gathered} v=\sqrt{(6.16{)}^2+(0{)}^2+(18.17{)}^2}\mathrm{m}/\mathrm{s} \\ \mathrm{v}=19.185\hspace{0.33em}\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the stuck-together car and truck just after the collision is $19.185\hspace{0.33em}\mathrm{m}/\mathrm{s}$

b) The reason for neglecting the effect of the force of the road on the car and truck is the short collision time and also the negligible impulse that is compared to the large impulse between the truck and the car.

Thus, the reason for neglecting is the short collision time and also the negligible impulse that is compared to the large impulse between the truck and the car.