Crash!

By Newton's Third Law, the forces are equal in magnitude but opposite in direction. Therefore the requested ratio must be $\boxed{-1}.$

All forces are equal but opposite in this scenario therefore the ratio of them is -1, as they are in opposite directions but the same magnitude.

We take initial direction of motion of car and truck to be positive direction for vector. Also, units for mass, distance and time are assumed to be kg, km and hr .

Let $m_{c}, v_{c}$ be mass, velocity of car while $m_{t}, v_{t}$ be mass, velocity of truck.

$m_{c} = 1000, v_{c} = 100, m_{t} = 5000, v_{t} = 90$

By conservation of momentum, momentum before = momentum after.

$m_{c} \cdot v_{c} + m_{t} \cdot v{t} = (m_{c} + m_{t}) \cdot v_{f}$ , where $v_{f}$ is final velocity.

Substituting values from above and solving, we have $v_{f} = \frac{550}{6}$ .

As $v_{t} < v_{f} < v_{c}$ , the car's velocity decreases and the truck's velocity increases.

• Car undergoes negative acceleration. so $F_{2}$ opposes the initial direction.
• Truck experiences positive acceleration so $F_{1}$ is in the initial direction.

We use the formula $F = \frac{m(v_{f}-v_{i})}{t}$ from Newton's 2nd law of motion.

1. $F_{1} = \frac{1000(100 - \frac{550}{6})}{T} > 0$
2. $F_{2} = \frac{5000( 90 - \frac{550}{6})}{T} < 0$ , where $T$ is time of collision.

So $\frac{F_{1}}{F_{2}} = \frac{1000(\frac{50}{6})}{5000(\frac{-10}{6})} = -1$ .

Note: From Newton's 3rd law of motion , force exerted by object A on object B has equal magnitude and opposite direction as force exerted by object B on object A.

This shows that $F_{2} = - F_{1}$ which yields $\frac{F_{1}}{F_{2}} = -1$ .

By Newton's third law, F1=-F2. Thus, F1/F2=-1

According to Newton's Third Law of Motion: Every action has an equal, opposite and instantaneous reaction. Hence the forces are equal and opposite. If one force is 'x' then the other is '-x'. x/-x = -1

By Newton 's third law,

F {1} = -F {2} => F {1} / F {2} = \boxed{-1}