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This may be a really stupid question, but I am learning mechanics and asked myself with this question:

What exerts more force when it hits an object: a car driving at a constant speed of 40mph, or the same car driving at a constant speed of 30mph?

If F=ma, then surely the force is always 0, but then the force with which the car hits an object when it crashes is the same no matter what the speed is, as long as it was going at a constant speed?

But this makes no sense because you will obviously feel a bigger force hit you if the car is faster...

Sorry if this is really obvious.

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    $\begingroup$ What's the acceleration when your car goes from 30 mph to 0 mph? (I'll give you a hint, it's not always 0). $\endgroup$
    – JMac
    Commented Jan 6, 2020 at 22:14
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    $\begingroup$ And if the car goes from 40mph to 0, is the acceleration higher or lower? $\endgroup$
    – TimWescott
    Commented Jan 6, 2020 at 22:15
  • $\begingroup$ if the car hitting you is the force and you are the mass which impact will accelerate you more? $\endgroup$
    – Adrian Howard
    Commented Jan 6, 2020 at 23:02
  • $\begingroup$ @TimWescott, If a car decelerates from 30 to 0 by hitting a cement bridge pier, the peak acceleration experienced by any given part of the car may be considerably higher than for the corresponding part of an identical car that decelerates from 40 to 0 by crashing through dense bushes. $\endgroup$
    – Solomon Slow
    Commented Jan 6, 2020 at 23:11
  • $\begingroup$ @SolomonSlow I'm fairly certain Tim's response had an implied collision with the same object at 30 mph and 40 mph in his comment. You can always play this game with impulse no matter what the change in velocity, the point was to illustrate that all other things considered the same, a greater change in velocity would mean a greater force $\endgroup$
    – Triatticus
    Commented Jan 6, 2020 at 23:42

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Basically, your confusion arises due to a misunderstanding of when to apply Newton's law. In your scenario you have two different phases:

  • the driving phase and
  • the impact phase.*

You tried to apply Newton's law to the driving phase but then used the result to say something about the impact phase. This will not work. You must apply the law within the phase you are investigating.

In detail

The force is zero while you are driving 30 mph and while you are driving 40 mph. Because there is no acceleration $a$ (the speed is unchanged):

$$\sum F =ma\quad \Leftrightarrow \quad \sum F =0$$

But, when you stop (when you start braking, start slowing down), then you are reducing your speed to zero. Then you do have a acceleration (negative acceleration, or deceleration), which slows you down. If your speed is high, your deceleration must also be large because you must slow down to zero faster in order to make it in the short duration of the impact.

* As @TimWescott pointed out in the comments, the impact phase may be complex and out to be split into several more phases. Still, in many, at least theoretical, problems, assumptions/simplifications are typically made for the entire impact phase such as constant acceleration

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    $\begingroup$ Now that we have that out of the way: it's more complicated when a car is involved, first because there's not just one collision (the first is car vs. object, the second is occupants vs. car), and second because part of the first collision (car vs. object) involves the car deforming. The car isn't a perfectly rigid body, and various phases of the collision may tend to have the same deceleration from different starting speeds, just for different durations. $\endgroup$
    – TimWescott
    Commented Jan 7, 2020 at 0:15
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    $\begingroup$ @TimWescott : And moreover, the occupants' bodies also behave in an analogous way when they contact the front of the crew cabin, and this is what causes the injury. If every part of their body came to rest simultaneously, there would be no injury. That's why a gravitational field - alone, without hitting anything, and suitably uniform - will not kill you no matter how strong. But because the front of both the moving car and moving bodies stops before the back, it creates deformation and thus damage. $\endgroup$
    – The_Sympathizer
    Commented Jan 7, 2020 at 11:05
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The a in that formula is acceleration, which is change in velocity per time. When you're going 30mph and hit a brick wall, you change your speed by 30mph in a fraction of a second (from 30mph to 0mph). When you hit it going 40mph, you're now changing your speed by 40mph in (roughly) the same amount of time, which results in a 33% larger acceleration 33% and a larger force.

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    $\begingroup$ The kinematic equation (suvat form) to use here is $v^2 - v_0^2 = 2 a \Delta x$ ($v^2 - u^2 = 2 a s$). If you assume the wall is solid enough and the car uses the full length of it's crush zone in both cases (neither is exact but each is a decent first approximation for some speeds) then the acceleration go by the square of the initial speed. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jan 6, 2020 at 22:20
  • $\begingroup$ ... or, equivalently, force x distance = change in kinetic energy. $\endgroup$
    – gandalf61
    Commented Jan 7, 2020 at 13:10
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This may be a really stupid question, but I am learning mechanics and asked myself with this question:

There is no such think as a stupid question. Only stupid people who don't ask questions. That being said,

What exerts more force when it hits an object: a car driving at a constant speed of 40mph, or the same car driving at a constant speed of 30mph?

For the same car (meaning same mass) the car with the higher speed experiences the greater force when it hits an object if the stopping distance of the two cars is the same. To simplify things, let's say each car hits a fixed wall. The average force that each car experiences is based on the work energy theorem which states that the net work done on an object equals its kinetic energy. Assuming each car is brought to a stop, that means

$$F_{ave}d=-\frac{mv^2}{2}$$

where $F_{ave}$ is the average force experienced by the car, $d$ is its stopping distance, $m$ is its mass, $v$ is its velocity just prior to to when the car hits the object. So, all other things being equal, the car at the higher speed experiences a greater force, as you would expect.

If F=ma, then surely the force is always 0, but then the force with which the car hits an object when it crashes is the same no matter what the speed is, as long as it was going at a constant speed?

Yes, if the car is going at constant speed the net force on the car is zero. But when the car crashes it decelerates. That requires an net force to reduce its speed and that work be done to take away the kinetic energy it had before the crash.

But this makes no sense because you will obviously feel a bigger force hit you if the car is faster...

Yes your gut feeling is correct. And the reason it will feel a bigger force is due to the work-energy theorem I described above. The greater the kinetic energy (velocity of the same mass car) before impact, the greater the stopping force given the same stopping distance.

Hope this helps.

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If F=ma, then surely the force is always 0

No, it's not, because car is stopped from initial speed to rest, yes ? Then it HAD negative acceleration. $$ F=ma=m\frac{dv}{dt}=m\frac{\Delta v}{\Delta t}=m\frac{v_e-v_0}{\Delta t} $$ $v_e=0$ (final speed), in your case because object stops car fully, so : $$ F = m \frac{-v_0}{\Delta t} $$

Thus put 40 mph and 30 mph into $v_0$ and you will get an answer. Btw, minus sign means that force vector is directed in opposite direction that car initial speed was directed at.

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  • $\begingroup$ You should define what $v_e$ is as standard usage usually uses $v_f$ for final velocity $\endgroup$
    – Triatticus
    Commented Jan 6, 2020 at 23:47
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    $\begingroup$ Did it. It's just a notation, so it can be in principle any symbol like $v_1$ as well $\endgroup$
    – Agnius Vasiliauskas
    Commented Jan 7, 2020 at 7:08
  • $\begingroup$ That is indeed true, no real issue there, mostly just a suggestion. $\endgroup$
    – Triatticus
    Commented Jan 7, 2020 at 7:09

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