27DQ - University Physics with Modern Physics

Question

A large truck and a small compact car have a head-on collision. During the collision, the truck exerts a force ${\overrightarrow{\mathbf{F}}}_{\mathbf{T}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{C}}$ on the car, and the car exerts a force ${\overrightarrow{\mathbf{F}}}_{\mathbf{C}\mathbf{ }\mathbf{on}\mathbf{ }\mathbf{T}}$ on the truck. Which force has the larger magnitude, or are they the same? Does your answer depend on how fast each vehicle was moving before the collision? Why or why not?

Step-by-Step Solution

Answer

It’s not the $\mathrm{v}=\mathrm{u}+\mathrm{at}$ that hurts you; it’s the sudden change in the momentum, i.e., $\mathrm{F}=-\frac{\mathrm{mu}}{\mathrm{t}}$ at the bottom.

Step 1: Newton’s 3 rd law

The first body exerts a force on the second body, and the second body exerts an equal and opposite force on the first body. 

$\overrightarrow{{\mathbf{F}}_{\mathbf{12}}}\mathbf{=}\mathbf{-}\overrightarrow{{\mathbf{F}}_{\mathbf{21}}}$

This action and reaction of forces act on different bodies. These forces act on each other as long the objects are in contact. Hence, the force on a car by truck and truck by car are equal in magnitude and opposite in direction.

${\overrightarrow{\mathrm{F}}}_{\mathrm{T} \mathrm{on} \mathrm{C}}=-{\overrightarrow{\mathrm{F}}}_{\mathrm{C} \mathrm{on} \mathrm{T}}$

Step 2: Newton’s Second law

A body accelerates when a net force acts on it and travels in the same direction as the net force. The product of mass and acceleration vector is equal to the force vector.

$\sum \overrightarrow{\mathrm{F}}=\mathrm{m}\overrightarrow{\mathrm{a}}$

Step 3: Kinematic equation for acceleration, velocity, and initial velocity.

For constant acceleration, the final velocity v is related to the initial velocity u by the expression given below,

$\mathbf{v}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{at}$

Here, the final velocity will be zero, and the initial velocity will be the relative velocity of the car wrt the truck or vice versa. The force acted on the body at the end of the collision will depend on the de-acceleration and the time it took to stop the vehicles completely.

$\begin{array}{rcl}\mathrm{F}& =& \mathrm{ma}\\ & =& \frac{\mathrm{m}\left(\mathrm{v}-\mathrm{u}\right)}{\mathrm{t}}\\ & =& \frac{\mathrm{m}\left(0-\mathrm{u}\right)}{\mathrm{t}}\\ & =& -\frac{\mathrm{mu}}{\mathrm{t}}\end{array}$

The relative velocity will be the same for both vehicles before the collision, and the time of deacceleration will be the same too, which is intuitive. Eventhough the magnitude of the force depends on the relative velocity. Still, the conclusion that force on each vehicle is equal is purely deduced from Newton’s third law without the need for velocity.

Thus, it’s not the $\mathrm{v}=\mathrm{u}+\mathrm{at}$ that hurts you; it’s the sudden change in the momentum, i.e., $\mathrm{F}=-\frac{\mathrm{mu}}{\mathrm{t}}$ at the bottom.