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Since $\text{force = mass}\times\text{acceleration}$,

  1. is it right to say that an object traveling at a high constant velocity (zero acceleration), exerts zero force upon impact with a stationary object?

  2. I understand that upon impact, the projectile decelerates rapidly from initial velocity down to zero. Could force then be computed using integrals and differential equations?

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Thanx for both answers, which confirmed my hunch, albeit I did not know the details as these two answers presented them. –  user22097 Mar 17 '13 at 17:26

2 Answers 2

  1. Nope. Initially, your object has no net force on it, and thus zero acceleration. Once it hits the wall, the wall exerts a force on it (and it exerts an equal/opposite reaction force back) which decelerates it according to $F=ma$

Note that $F=ma$, expanded, means "net force on a body is equal to its mass times its acceleration at any instant". It makes no promises on the force the body exerts on others (If there is only one more body in the system we can easily find this out by action-reaction, but this net reaction force can be split up in multibody systems).

  1. Well, force will be a function of time. $F_{instant}=ma_{instant}=m\frac{dv_{instant}}{dt}$ So, while we can't find the force (a meaningless quantity), we can find how the force varies over time. For example, if the body the object crashes into obeys Hooke's law/Young's modulus (and the object is rigid), we can calculate force by solving: $$m\frac{d^2x}{dt^2}=F=-kx^2$$ with the initial condition $\frac{dx}{dt}=v_0$. This will give us $F=\frac{mv_0}{k}\sin(\sqrt{\frac{k}{m}}t)$ by solving the differential equation.

However, there is only one average force (which won't vary over time), defined as $\langle F\rangle=\frac{\int F dt}{\int dt}=\frac{m\Delta v}{\Delta t}$

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First of all just to refine your notion of force and acceleration by `force = mass X acceleration' what we mean is the relationship between the vectors:

$\vec{F}_{net} = m \vec{a}$

that is, its the net force that appears in Newton's second law, not just any force.

  1. The velocity of the traveling object will change, hence there is an acceleration $\vec{a} = \frac{d\vec{v}}{dt}$, hence by Newton's second law there is a force applied to the traveling object $\vec{F}_{net} = \vec{F}_{S \hspace{1mm}on\hspace{1mm} T} = m \vec{a}$. Where 'S on T' denotes that the force is applied to the traveling object by the stationary object. By Newton's third law, the traveling object applies an equal and opposite force to the stationary object: $\vec{F}_{S \hspace{1mm}on\hspace{1mm} T} = -\vec{F}_{T \hspace{1mm}on\hspace{1mm} S}$. So there is a force on the stationary object by the traveling object.

  2. If you have velocity as a function of time you can find the instantaneous force

$ \hspace{10mm}\vec{F}_{S \hspace{1mm}on\hspace{1mm} T} = \vec{F}_{net} = m \vec{a} =m \frac{d\vec{v}}{dt} $

$\hspace{10mm}$or if you just have the initial and final velocity you can find the average force:

$\hspace{10mm}\vec{F}_{S \hspace{1mm}on\hspace{1mm} T, ave} = \vec{F}_{net, ave} = m \vec{a}_{ave} = m \frac{\vec{v}_f-\vec{v}_i}{t_f-t_i}$

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