91Ó°ÊÓ

The given data can be listed below as-

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

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.

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 $v_1 and v_2$are the velocities of the car and the truck respectively.

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

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

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

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

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.