# Force during collision between a body moving with a constant velocity and an accelerating body [closed]

Lets us assume that a train of huge mass is approaching me with a huge constant velocity of the order of hundreds of kilometres per hour. And little me is standing on the tracks but moving with acceleration of $1\ \rm ms^{-2}.$ Does that mean that I will exert a greater force and escape unhurt (Although this does not seem to happen)?

## closed as unclear what you're asking by Kyle Kanos, Steeven, Jon Custer, sammy gerbil, stafusaSep 27 '18 at 6:51

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• The second law tells you what force is acting on you causing you to accelerate. It doesn't tell you what force you apply on the train when you collide with the train. – npojo Sep 26 '18 at 12:35
• @vigneshwaran: $1 \; \mathrm{m/s^2}$ in which direction? Greater force compared to what, compared to you moving but without the acceleration? You need to make it clear which two situations you are comparing. – user7777777 Sep 26 '18 at 12:50
• @npojo Newton's 3rd law tells us that those two forces are equal. – Steeven Sep 26 '18 at 13:08
• Why would you derail the train? Sure, a big force will be applied on you, but the force that the train feels is relatively little for its size. The answer is no, you will not derail the train, no matter what - but why would you assume so in the first place? Does it come from Newton's 2nd law or something else? – Steeven Sep 26 '18 at 13:10
• @Steeven the OP is not assuming the train is derailed. The OP knows it doesn't derail, but is using N2L (incorrectly) to reason that the train will derail. Hence the posting of the question. – Aaron Stevens Sep 26 '18 at 13:26

## 3 Answers

You are confused about what the variables mean in Newton's second law.

Newton's second law is $$\vec F_{net}=m\vec a$$

What this means is that a net force (the sum of all forces acting on an object) produces an acceleration. This acceleration is determined by how strong the force is as well as the mass of the object.

What Newton's second law does not mean is that if an object has an acceleration a then it will produce a force consistent with $$F=ma$$ onto another object. All we know of an object with an acceleration $$a$$ is that a net force is acting on this object consistent with $$F=ma$$.

Certainly, if a massive fast train hits you it will exert a large force on you, which will then cause you to have a large acceleration.

This is a great lesson in using equations in physics. You have to know what each variable in the equation means before you start trying to understand how to apply that equation to physical systems. Mathematically there is nothing wrong with saying "more acceleration means more force with constant mass". But what force and what acceleration are we talking about here? Without knowing the physical meaning of the equation, you could essentially say anything.

• your answer makes sense. like the same ive confused for a long time on bernoulli's theorem. i couldnt understand what energy the equation talks about – vigneshwaran Sep 26 '18 at 13:12
• but what force would a body exert if it moves with a accelaration a – vigneshwaran Sep 26 '18 at 13:13
• @vigneshwaran If you are asking about a collision then the accelerations of the objects before the collision doesn't determine the result of that collision directly. Of course if the forces causing said accelerations still persist after the collision then those forces will influence what happens, so what you want to focus on is right before and after the collision. This depends on the momentum of each object, i.e. the mass and velocity of each object, rather than the accelerations at the time of impact. – Aaron Stevens Sep 26 '18 at 13:21

As we know, according to newton's law that force is rate of change of linear momentum . So when train will hit you the rate of change of momentum will be equal for both train and you. This change is so much huge that within no time your velocity will become equal to that of train and due to which you will face a huge impulse which will be very dangerous for you. On the other hand train's interia is very large, so the force with which you will hit it barely make any difference in its motion. So the conclusion is that forces applied by both bodies on eachother will be equal and opposite (according to third law). But due to large inertia of train , it will feel nearly 0 effect. Just like sand particle hitting your windsheild

• what do you call 'distorted space and time' ? and why are you talking about mass/energy conversion in a Newtonian mechanic problem? – sailx Sep 26 '18 at 12:48
• Sorry, it was just a mistake. – Sourabh Sep 26 '18 at 12:54

No.

Before the collision, you are accelerating due to a net force of magnitude $$F=m_\text{you}a$$ where $$a=1\text{ m s}^{-2}$$ applied on yourself (e.g. by running or driving a car or whatever). The train if it is moving at a constant velocity experiences no net forces, though that does not mean there are no forces involved (in this case, the main forces would be the train pushing itself forward vs the friction of the tracks).

Now cut to the collision. When the train hits you, your velocity will change, and likely very dramatically in a very short period of time (i.e. you will be going from jogging towards the train to flying away from the train). The acceleration from this is what you need to consider when working out the force exerted on yourself from the train (which will be huge because your momentum change will be very fast). This is why you'll likely end up in the hospital.

Now by Newton's third law, you also exert an equal and opposite force on the train, but because it is much more massive than you, its velocity will not be affected as much (probably by a negligible amount practically), and so it is unlikely you will derail the train.

• By Newton's Third Law, an object cannot exert a nonzero net force on itself. – user7777777 Sep 26 '18 at 13:02
• Yeah strictly speaking you are correct - e.g. in the case of the runner, it would be your legs pushing the ground / the ground pushing you forward causing you to accelerate. My wording was perhaps not the best, but I was just trying to make clear that the acceleration OP is referring to in the question has nothing to do with hitting the train. – Garf Sep 26 '18 at 13:07