So... sorry if the question seems a bit stupid, but it's been nagging me all day and I couldn't find an answer on the net. It goes like this: If a body, say, a car is moving with constant speed. This means its acceleration is 0. And say that an object, like a wall or a stone is in the way. In this case, when the car and wall/stone collide, the car obviously exerts force on the stone. But, since F = ma, and a= 0, F also becomes 0. In this case, what am I missing? Shouldn't the force on the wall be 0?
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1$\begingroup$ You are missing the wall. Does $a=0$ when you hit it? $\endgroup$– JEBCommented Oct 5, 2019 at 17:13
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$\begingroup$ Also you are missing that continuous force is applied to keep a car moving at constant velocity, because it is continuously decelerated by the friction of the road on the tires. $\endgroup$– anna vCommented Oct 5, 2019 at 17:15
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2$\begingroup$ You are assuming the "trivial" answer that I have heard so many times in my physics class room. Everybody knows that the speed and acceleration of the car is zero AFTER the collision. The question is: what is the acceleration DURING the collision? $\endgroup$– David WhiteCommented Oct 5, 2019 at 17:21
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$\begingroup$ Why are you assuming no acceleration during the collision? $\endgroup$– BioPhysicistCommented Oct 5, 2019 at 17:32
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4 Answers
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Many are pointing out the importance of friction of the car against the road and the need to apply a constant force in the real situation to mantain the car moving uniformly.
I think this is not the point of the question, an object (let's say a micrometeoroid) moving in the vacuum of space and hitting a "wall" (let's say the solar panel of an artificial satellite) wouldn't have any important source of friction until the collision itself and still the question would hold.
The thing is that the change in speed from $v_0$ (previous to the collision) to $0\; m/s$ (when the wall finally stops you) is $\Delta v = 0-v_0=-v_0$ and since acceleration on average is that change in velocity between tome $t_0$ and time $t_1$ then there is indeed an acceleration $a_{average} = \Delta v/(t_0-t_1) = -v_0/\Delta t$. The thing is that you don't whan the average but the acelleration at each particular moment in time. For any time before and after the collision it is clear that $a=0$ as you said, but then the entirety of the acceleration (and thus the force) should appear during the collision itself.
In an idealistic world this change in motion is instantaneus ($\Delta t = 0$), so the acceleration must be $a=\Delta v/ \Delta t = \infty$. But the reality is that objects compress when they collide (both the wall and the object), that means that the contact between both (and thus the interaction between both) happens for a timespan greater than zero. As the object deforms and the wall absorbs the kinetic energy of the object both are stressing their structure by pressing, flexing, etc... This compression is the source of a reactive elastic force coming from the wall that will be the greatest in the moment of maximum compression/deformation and would fade as both objects lose contact until the force dissapears completely as they detatch from each other.
So, calculating the actual aceleration in a realistic scenario is extremely complicated. It involves knowing the mass and material properties of the impactor and the wall.
So I would think of this as if the impactor was heading towards a wall full of springs and then bouncing on them. The only difference is that the springs on the wall are extremely stiff, so stiff that you don't really see any compression and expansion of them. That's more or less what's going on. But why the wall acts like a wall of stiff springs? Because of the electric potentials of the molecules of the wall acting like that. I reccomend you reading this answer in case you want to go deeper on this last point.
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Friction and air resistance are irrelevant.
If a body is coasting through space (no friction and air resistance) at a constant speed relative to the wall, then hits the wall, the wall speeds up and the car slows down, so they both accelerate.
answered Oct 5, 2019 at 17:35
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F=dp/dt and we get F=mdv/dt, which is equal to Ma, whenever there is change in momentum the body will simultaneously applied and experience the force, that's what Newton third law describe, now coming to your question it is OK acceleration is zero before collision with wall, but
during the Collison momentum changes like this Final momentum - initial momentum which gives Mv(momentum of car before collision) + 0(momentum of wall/car2) - final. Momentum (let us take it as inelastic collision so both car stops) =dp now during collision dv/dt is non zero because v continuity decrease.
Assumption :since you have not mention about friction, so I assume track to be frictionless.
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The acceleration a = 0 is for the motion of the car when the wall doesn't exist, that it , there is no external force acting on the car which leads to it having a constant speed. When the car collides with the wall, it experiences a force opposite to the direction of it's velocity which causes it's velocity to slow down over a very small period of time. This force changes it's speed and hence imparts deceleration or negative acceleration. So, the acceleration of the car after collision is not 0
