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I've been looking at this question:

"Weight" of moving object in a car collision

but to be honest I can't get my head around how to work it out myself (I'm not a physics student!). What I would like to do is create a calculator in Excel that does the workings in the questions so that I can very easily and simply plug in the figures I want, and have it work out the resultant weight during deceleration. This would allow me to very simply run several scenarios.

The posted answer is,

From $dp/dt\simeq \Delta p/\Delta t=F$. A 2 kg object at 50 km/hr has an initial momentum of $$ p=mv = 2\,{\rm kg}\cdot13.9\,{\rm m/s}=27.8\,{\rm kg\,m/s} $$ If we naively assume that the crash is over the course of a 0.05 seconds, then the force is $F=556\,{\rm N}\to m_{eff}\simeq58\,{\rm kg}$ (final momentum being 0 because the velocity is zero). Letting $\Delta t=0.03\,{\rm s}$ gives $m_{eff}=94.6\,{\rm kg}$.

So what I'm trying to do is calculate the weight of an item in the event of an emergency stop. So I want to have a spreadsheet where I can simply enter in the weight of a product, initial speed, resultant speed, and time delay between the two. The spreadsheet should then give me the resultant weight of the item as it decelerates.

Can someone help me simplify the equation so that it can be put in a series Excel compatible formulae?

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  • $\begingroup$ Kinda funny to me that the text looked familiar to me, lo & behold, I wrote it! Pretty sure I used $m_{eff}=F/9.81\,\text{m/s}^2$, but what is it about the equations I wrote that are confusing to write it in Excel (or any other software)? $\endgroup$ – Kyle Kanos Mar 15 at 13:12
  • $\begingroup$ @KyleKanos Are you mocking the op your comment isn't making much sense to me. $\endgroup$ – Avnish Kabaj Mar 15 at 14:02
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    $\begingroup$ @AvnishKabaj good heavens no. I found it amusing that I recognized something I wrote 5+ years after the fact. Then I asked for clarification as to what wasn't clear with what I'd written. $\endgroup$ – Kyle Kanos Mar 15 at 14:03
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    $\begingroup$ $\Delta$ means a change, so $\Delta p=p(t_2)-p(t_1)$ (i.e., change of momentum between two times). The $\simeq$ means "approximately equal". $m_{eff}$ is the effective mass of the object using the acceleration due to Earth's gravity. $\endgroup$ – Kyle Kanos Mar 15 at 14:41
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    $\begingroup$ I may be able to write an actual answer in about 2 hours (currently at work), but if you get to it in the meantime, I can help edit it. $\endgroup$ – Kyle Kanos Mar 15 at 16:07
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What my answer in the linked question is attempting to do is determine the apparent weight of an object that is rapidly decelerating. The relation, $$\frac{\Delta p}{\Delta t}=\frac{p(t_2)-p(t_1)}{t_2-t_1}\simeq F\tag{1}$$ tells you that the rate of change momentum is equal to the force. So what I do in my answer is take this force and set it equal to the gravitational force, $$ F = mg,\tag{2} $$ to determine what the effective mass of this object is, to try verifying the claim in the question. $g$ here is the acceleration due to gravity (9.81 m/s$^2$). where $g=9.81\,\text{m/s}^2$.

So my answer sets (1) and (2) equal and solves for $m$ (calling it $m_\text{eff}$ for the effective mass): $$ m_\text{eff}=\frac{1}{g}\cdot\frac{p(t_2)-p(t_1)}{t_2-t_1} $$ So then what you need are the initial and final velocities and the time it takes to go from the starting velocity to the final velocity:

m     = 2       // mass of object, in kg
v_t2  = 0       // ending velocity, in m/s
v_t1  = 13.9    // starting velocity, in m/s
dt    = 0.05    // time it takes for object to go from v_t1 to v_t2, in sec
g     = 9.81    // acceleration due to gravity, in m/s/s

// solve for Force and effective Mass
F     = m * (v_t2 - v_t1) / dt
m_eff = F / g

which should give you $F=556\,\text{N}$ and $m=58\,\text{kg}$.--note that the above formula can return a negative value for the mass, that just indicates the deceleration instead of acceleration so you can just ignore it.

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  • $\begingroup$ Thanks @KyleKanos. Interestingly, that results in m = 56.6769 (556 / 9.81). Just tried with Δt=0.03s which results in 94.4614. However that's only an FYI, it's close enough for me. Thanks very much for your help! $\endgroup$ – 5Diraptor Mar 18 at 6:57

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