Good evening, I'm trying to calculate what kind of impact force a falling object would have once it hit something. This is my attempt so far:

Because $x= \frac{1}{2} at^2$, $t=\sqrt{2x/a}$
$v=at$, therefore $v=a \sqrt{2x/a}$
$E_k=\frac{1}{2} mv^2$, so $E_k=\frac{1}{2} m(2ax)=m \cdot a \cdot x$
Since $W=E_k=F_i s$, $F_i=E_k/s=(m \cdot a \cdot x)/s$

For an object weighing about as much as an apple, $0.182$ kg, falling $2.00$ m straight down and creating a dent of $0.00500$ m, this would result in:

$$F_i=(m \cdot a \cdot x)/s$$

$$F_i=(0.182 \cdot 9.81 \cdot 2.00)/0.00500=706 \, \text{N}$$

Does this make any sense? I wouldn't be surprised at all to find out I'm missing something here.

Any input would be appreciated,

thanks in advance!

  • $\begingroup$ You forget air resistance. $\endgroup$ – Hawk Aug 9 '12 at 19:38
  • $\begingroup$ Regarding Qmechanic's edit: While I know this is high-school level physics, please note that this is not, in fact, homework, as the tag would suggest. I do not have any assignment to go from nor any way to look up the answer. $\endgroup$ – Chris Aug 9 '12 at 20:06
  • $\begingroup$ Your calculations are ok and the model you used (constant hitting force) is reasonable. Though remember that's just a model, real apples hit the ground in a more complicated way, but I think for what you might use your calculations, your model is accurate enough. Unless, of course, you are not allow for the apple to bounce. $\endgroup$ – Yrogirg Aug 10 '12 at 8:20
  • $\begingroup$ Read all replies and marked an answer, thanks all for your help! $\endgroup$ – Chris Aug 10 '12 at 9:29

If your apple falls $2m$ it's velocity is calculated using the equation you give:

$$ v^2 = 2as $$

and you get $v^2 = 39.24 \space m^2s^{-2}$ (I've haven't taken the square root for reasons that will become obvious). You know the apple is slowed to rest in $0.005m$, so you just need to work out what acceleration is needed when $v^2 = 39.24$ and $s = 0.005$. A quick rearrangement of your equation gives:

$$ a = \frac{v^2}{2s} $$

and plugging in $v^2 = 39.24$ and $s = 0.005$ gives $a = 3925 \space ms^{-2}$. To get the force just use Newton's equation:

$$ F = ma $$

where $m$ is the mass of the apple, $0.18 kg$, and you get $F = 706.32N$. So you got the correct answer (my answer differs from yours only because I used $g = 9.81 \space ms^{-2}$).

To get a more general result substitute for $v^2$ in the second equation to get:

$$ F = ma = m\frac{2gs_1}{2s_2} = mg\frac{s_1}{s_2}$$

where $s_1$ is the distance the apple falls and $s_2$ is the distance it takes to stop.

  • $\begingroup$ Exactly what I was looking for, explanation and confirmation. Thank you very much! $\endgroup$ – Chris Aug 10 '12 at 9:26

I'm not sure where you're going with the .0050 being treated as time, unless you meant to say it's in contact with the ground for .005 seconds. In that case, what you have would work algebraically.


I would have done this calculus if there was a uniform constant breaking force applied all all along the trajectory and if at impact, the ball had a zero velocity. The work done by such a force must exactly balance the initial mechanical energy of the ball. For a shock, I would rather use something like: d(mV)/dt=mg-F and use for instance a contact time dt=0.001s

Before impact: v=sqrt(2*g*x) After impact: v=0 Duration of the impact: dt=0.001s


F=1140 N


Calculate Potential energy using the following formula $PE = mgh = 0.182 \cdot 9.81 \cdot 2\: \mathrm{Joules}$

Average Impact Force = $0.182 \cdot 9.81 \cdot 2 \cdot 0.005\: \mathrm{Newtons}$ (that is the answer)


protected by Qmechanic May 12 '13 at 0:00

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