According to Newton's second law, $F = ma$. If acceleration is zero then the force must be zero, but assuming you have an object moving with a constant velocity of say $2 \mathrm{ms^{-1}}$, and that object strikes you, then obviously some sort of 'force' would be felt by you, so my question is what do you call that 'force' since it actually is not a 'force' and is there an equation to calculate it?
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An object will slow down or change direction upon hitting something else. To test this, you could do an experiment where you throw something at a wall or any other object. You should see the thrown object slow down or change direction upon impact. Thus the force really is non-zero. – Brian Moths Nov 15 '13 at 04:20
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The "force" (actually called an impulse) depends on what it strikes. – John Alexiou Nov 15 '13 at 12:56
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Why do you feel it is "actually not a force"? Is it because you feel the force doesn't cause any acceleration to the body it hits? That is not because there is no force, but because all the forces acting on that body cancel out. – udiboy1209 Nov 15 '13 at 14:26
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If an object is accelerating that means someone applying force on the object, And if there is no force on the object then the object moves with constant velocity. now if the object is moving with a constant velocity and it strike me and object stop moving that mean I have applied a force on the object and the ball will also apply a force on me by Newtons 3rd law..thats why i will get hurt. – Paul Dec 19 '14 at 06:23
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Some students learning physics for the first time mistakenly think that objects that are accelerating have force.
Force is not a property possessed by an object, but rather something you do to an object that results in the object accelerating (changing its speed), given by the equation F = ma.
That is, Forces cause acceleration, not the other way around. This means that if you observe an object accelerating, then it implies a force is acting on the object to cause such an acceleration.
In this case, as the object strikes the hand, your hand applies a force to the ball causing it to slow down (decelerate), and the ball applies an equal and opposite force to your hand causing it to accelerate ever so slightly (Newton's third law), which is detected by your sensory neurons.
Kenshin
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'the ball applies an equal and opposite force to your hand causing it to accelerate ever so slightly' by 'it' do you mean the hand is accelerating ever so slightly? – fYre Nov 16 '13 at 04:56
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Lets just say the ball moving with constant velocity hits a wall, then according to you the wall must apply a force, but if the wall were to apply a force it must accelerate, but surely the wall is not accelerating? So what exactly is happening? How is the wall exerting a force? – fYre Nov 20 '13 at 01:10
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@HaniSayegh, the wall accelerates backwards, and the ball accelerates forwards (Newton's 3rd law). The wall doesn't have to accelerate forwards to apply a force. Acceleration results from force, you don't need forward acceleration to produce a force. – Kenshin Nov 20 '13 at 02:09
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@HaniSayegh, the wall doesn't appear to accelerate backwards because the acceleration is small, because of increased mass of the wall and supporting structures which provide an opposing force on the wall to resist its acceleration. (F=ma, so large mass means less acceleration). Locally however just where the ball hits the wall, you can see the wall move inwards a little bit with a slow motion camera. – Kenshin Nov 20 '13 at 02:11
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- Ball hits the wall. 2) Atoms in ball and wall oppose each other, which generates a force, causing the ball to reflect of the wall, and causing the wall to reflect off the ball.
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Let me see if I can get this right... 1) The ball hits the wall. 2) The wall accelerates backwards. 3) There is a force by the wall on the ball to the 'left' (assuming ball is coming from the right). 4) Ball accelerates backwards. – fYre Nov 20 '13 at 04:04
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Yes, except points (2) (3) and (4) occur simultaneously. That is, the electrons in the ball are pushing the wall back, while the electrons in the wall are pushing the ball back. Immagine I had just 2 electrons and fired them at each other, they will get to the middle and then both repel from each other simultaneously. You wouldn't say that one electron was the main culprit, both electrons affected each other. – Kenshin Nov 20 '13 at 04:05
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Also when taking the mass to calculate the force by the wall, you obviously wouldn't take the whole wall, so do you just take the mass of the area were the ball hits the wall? – fYre Nov 20 '13 at 04:08
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@HaniSayegh, now it becomes complex, because it depends upon the arrangments of the atoms in the wall. Technically you only would take the atoms that are in contact with the ball. Then these atoms will contact atoms behind it, causing a new collision, and a chain reaction of collisions within the wall. It is complicated because the wall is held in place by support structures, so ultimately the force will be transferred from the wall to the support structres in the ground. – Kenshin Nov 20 '13 at 04:10
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Wow thanks a lot for your help, I would just like to ask you what level of physics would I need to understand theses things? Advanced classical mechanics? – fYre Nov 20 '13 at 04:15
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Basic classical mechanics you should learn newton's laws, and Free Body Diagrams. To understand the forces acting upon a wall, an understanding of "free body diagrams" should enable you to do so, by labeling all the support structures applying forces to the wall itself. To master this, you must first master equillibrium and statics from classical mechanics, e.g. http://www.physicsclassroom.com/class/vectors/u3l3c.cfm – Kenshin Nov 20 '13 at 04:25
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When two objects (of mass $m_1$ and $m_2$) collide with relative speed $v_{rel}$ then there is an exchange of momentum (called an impulse) of magnitude $$ J = \frac{(\epsilon+1)\,v_{rel}}{\frac{1}{m_1}+\frac{1}{m_2}} $$
where $\epsilon$ is the coefficient of restitution and it describes if the objects bounce or stick.
This impulse changes the speed of the object by $\Delta v_1 = -\frac{J}{m_1}$ and $\Delta v_2 = \frac{J}{m_2}$.
The actual force cannot be found from this as it changes rapidly with time, but an average force can be computed if you know that the impact takes $\Delta t$ time to occur.
$$F_{average} = \frac{J}{\Delta t} $$
PS. The definition of the impulse is $J = \int F(t)\,{\rm d}t$.
John Alexiou
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When an object, moving at a constant velocity, hits something, it will either stop, decelerate (or accelerate in some cases), or bounce back. During this collision time, there is a change in velocity (acceleration/deceleration) of the moving object, hence force is exerted on the wall (or in this case, on you), due to change in velocity. Hence we can use Isaac Newton's 2nd Law: $$ F= ma = m(dv/dt) = m(d^2x/dt^2)$$
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Assume a ball 0.1 kg hits a wall with a speed of 2 m/s and bouce back in 0.2 sec, deceleration will be 2/0.2=10 m/sqsec, and force will be 1.0 N.
Arshad
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1This might be a better answer if you mentioned why the 0.2 sec is important. – Kyle Kanos Jan 28 '15 at 03:49
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Resistance or reaction that stopped a striking object is a (d'Alembert's) force that causes you pain on impact deceleration ( high velocity reduced to zero in a small distance) $ =v^2/( 2 s).$
Narasimham
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