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A small car collides with a large truck. Why do both vehicles experience the same magnitude of force? Wouldn't the large vehicle experience less force than the small one?

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3 Answers 3

The two vehicles experience a force of the same magnitude due to Newton's third law:

If object $A$ exerts a force $\mathbf F_{AB}$ on object $B$, then object $B$ will exert a fore $\mathbf F_{BA}$ on object $A$ and $$ \mathbf F_{BA} = -\mathbf F_{AB} $$

However, what you're probably thinking about is that motion of the car is more drastically affected by the collision. This can be explained by Newton's second law. Let's say the truck has mass $M$ and the car has mass $m$. If the magnitude of the force that both vehicles experience is $F$, then the magnitudes of their respective accelerations are $$ a_\mathrm{truck} = \frac{F}{M}, \qquad a_\mathrm{car} = \frac{F}{m} $$ and combining these we get $$ \frac{a_\mathrm{truck}}{a_\mathrm{car}} = \frac{m}{M} $$ So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.

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Thanks but how do you know the forces are the same? – user1530249 Feb 16 '13 at 23:31
Does equal but opposite reaction mean equal but opposite force? – user1530249 Feb 16 '13 at 23:33
Yup! See the edit where I wrote down Newton's third law. – joshphysics Feb 16 '13 at 23:49

The forces on the car and truck are equal and opposite. However, the force on someone sitting in either vehicle is not the same (assuming the frame around them stays rigid enough). As joshphysics explained, the equal forces cause accellerations of the truck and car inversely proportional to their mass. Assuming you sitting in one of these vehicles are a relatively small portion its overall mass, then what you feel is proportional to the accelleration of the vehicle around you.

Even putting aside getting physically crushed, you are therefore better off in a more massive vehicle in a collision. Think of a extreme case. You are driving a long and a bug gets splatted on the windshield. Your car and the bug both got pushed with the same force, but the resulting change in velocity (the accelleration) of the bug was much higher. As a result, you sitting in the car had a rather better experience than the bug.

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You asked:

Why do both vehicles experience the same magnitude of force?

The larger principle at work is conservation of momentum. (Noether's Theorem, symmetry, and all that jazz.) During the small time frame of the collision we generally assume that there is no transfer of momentum into or out of the system of the car and truck. Changes of momentum are called impulses and are calculated by $$\vec{\Delta p} = \int_t^{t+\Delta t} \vec{F}(t) dt$$ For momentum to be conserved we must have $\vec{\Delta p}_{total}=0$, so $$\vec{\Delta p}_{car}=-\vec{\Delta p}_{truck} $$ The duration of the collision is $\Delta t$ so the force of the car on the truck lasts exactly as long as the force of the truck on the car. Replacing each impulse with the integral we get $$ \int_t^{t+\Delta t} \vec{F}_{c\rightarrow t}(t) dt = -\int_t^{t+\Delta t} \vec{F}_{t\rightarrow c}t(t) dt $$ $$ \int_t^{t+\Delta t} \left(\vec{F}_{c\rightarrow t}(t)+ \vec{F}_{t\rightarrow c}t(t)\right) dt = 0 $$ If neither of these forces is generally zero then $$\left(\vec{F}_{c\rightarrow t}(t)+ \vec{F}_{t\rightarrow c}t(t)\right)=0$$ Which means the forces are equal in magnitude and opposite in direction.

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protected by Qmechanic Feb 5 at 23:46

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