Disclaimer: I took High School Physics over a couple decades ago.

"Hypothetically," suppose a 12oz (353.8 grams) paper cup from let's say Wendy's, filled with ice and water, was delivered from a car going one direction into the grill of, let's say a 1999 Jeep going the other direction. I'm interested in how one goes about computing the force that was rendered to the grill.

Let's further suppose that each vehicle was going 35 mph (56.33 meters/hour). Let's assume that the cup was launched with zero force (we can adjust speed of cars to account for launch force, right?). In case it matters, let's also assume that the cup was launched from 3-4 car lengths away. That might be ~540 (13.72 meters)-720 (18.29 meters) inches.

The ultimate question I'm trying to answer is if it is possible to break the grill and then deform the condenser behind it. I realize that we need to know the force necessary to break a grill, but if I can at least know how much force, preferable in some unit I understand like pounds, then I can make an educated guess.

I know that a 12oz cup of water is .78 lbs or 353.8 grams. I'm supposing that the ice would not increase its weight?


closed as off-topic by Jon Custer, sammy gerbil, honeste_vivere, Rory Alsop, Yashas Aug 13 '17 at 15:56

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  • 7
    $\begingroup$ If you're looking for legal help, it'd probably be better for you to find a physicist at a local university who'd be willing to help, rather than trying to outsource to the internet. $\endgroup$ – Kyle Kanos Aug 7 '17 at 19:55
  • 1
    $\begingroup$ My humble European mind would like to kindly ask for SI units :) Who on Earth measures car lengths in inches... $\endgroup$ – Steeven Aug 7 '17 at 20:02
  • $\begingroup$ @KyleKanos I'm not asking for legal help. I'm trying to determine if it is reasonable that damage was created by such an event. $\endgroup$ – Outfast Source Aug 7 '17 at 20:06
  • $\begingroup$ @Steeven SI units in the description now. $\endgroup$ – Outfast Source Aug 7 '17 at 20:09
  • $\begingroup$ And I'd not be comfortable answering whether it is reasonable or not if this is to be used for legal troubles. $\endgroup$ – Kyle Kanos Aug 7 '17 at 20:09

A long, long time ago I made a career out of studying liquid impact... although the scale was different (small drops, high velocity) some of the principles are probably valid.

Based on that, I have written a number of answers on this site, including
this question on water bombing
this one on "the pressure of falling water"
and this one on the force of a drop of water impacting.

Without wanting to repeat everything I said, and referring you to some of the diagrams in the other answers, the conclusions are:

  • When a liquid hits a surface, there can be a very high (but transient) initial pressure; the pressure can be as high as the water hammer pressure of the liquid, that is $\rho ~c ~v$, where $\rho$ is the density of water, $c$ is the velocity of sound in water, and $v$ is the velocity of the impact. For your scenario ($\Delta v = 70~\rm mph \approx 30~ m/s$), these numbers give you approximate pressure of $1000 \cdot 1500 \cdot 30 = 45 ~\rm MPa$. But that is the "belly flop" pressure, and it will only last for a very small fraction of a second (namely, until the contact point starts traveling at less than the speed of sound; this happens when the contact angle is about 0.02 rad, which means a contact strip that is about 2 mm wide for a 5 cm radius cup). If we assume the contact strip is 10 cm long (exceptionally well aligned with the grill), then the peak force due to the liquid, would be $F = P \cdot A = 45\cdot 10^6 \cdot 0.002 \cdot 0.1 = 9000 \rm ~N$. The distance moved would be 10 µm, and the time of this high force would be 0.3 µs. This would not be enough to damage the radiator.
  • If we assume the cup to be filled with solid ice instead, the problem is now one of solid particle impact; the entire momentum of the cup has to be absorbed (assuming inelastic collision, so the ice does not bounce off but it is relatively well contained by the cup). Then the math goes like this:

Momentum P = $m\cdot v = 0.33\rm kg \cdot 30 m/s = 10 ~Ns$. The impact is fully absorbed as the cup moves 10 cm closer to the car (again assuming a radius of 5 cm); with a relative velocity of 30 m/s this happens in 0.003 seconds. The mean force must have been

$$F = \frac{P}{\Delta t} = \frac{m \Delta v}{D / \Delta v} = \frac{m\Delta v^2}{D} = 0.33\cdot 30^2 / 0.1 \approx \rm 3000 ~N$$

That is a pretty significant force, and it would probably be enough to dent a grill and perhaps a radiator.

Note that the initial "liquid" impact force is higher - but it is transient. The "mostly solid" approach gives a lower peak force, but a greater over all impact - because the liquid can't get away.

In reality, the cup probably "exploded" (disintegrated) on impact, so the force would have been significantly lower than the 3000 N I estimated. The only way to be absolutely sure would be to do the experiment, with the grille instrumented with strain gages, and with a high speed camera running. In the '80s I had to use a piece of gear that cost upwards of $100,000 to do this experiment; today you might find that a good cell phone gives you 240 frames per second, which should show you fairly well what's happening.. Either you prove damage (and now you have another repair), or you prove nothing.

I wouldn't use my car to do this experiment.

  • $\begingroup$ Wow, I wish I understood all of the units as you obviously do, but you explained it very well and I understand it at a simplistic level. I appreciate your explanation a lot. You put math to what I was thinking. I don't feel qualified to mark any answer here as correct. If others want to weigh in on whether this should be marked correct, I'd appreciate it. $\endgroup$ – Outfast Source Aug 7 '17 at 23:31
  • $\begingroup$ @OutfastSource I'm glad if this cleared some things up - it's really rather complicated and it's hard to know whether I can get any useful information across (in this case: "it's really hard to come up with a reasonable estimate"!) Please don't worry about marking the answer as "correct"... note that the acceptance mark only means "this answer was most helpful to me". But by all means leave it blank and let's see what comments / votes / alternative answers show up. $\endgroup$ – Floris Aug 7 '17 at 23:36

So let's make a ton of assumptions. aka "A cow is a sphere"

<= "less than or equal to"

Cup Mass <= 0.35Kg Cup Speed <= 60km/hr = 16.67m/s

Kinetic energy

K = 0.5(m)(v)^2 k = 0.5(0.35)(16.67)^2
k = ~48J

This means that the upper bounds of force applied could not have exceeded anything more than about 10 pounds of force.

J = W = ΔE = N.m

Flexural strength, also known as modulus of rupture, or bend strength.

High Density Polyethylene have a Flexural Yield Strength 31.7 MPa

1 Pa = J/m^3

Pa = N.m/m^2

Assume it did break grill.

And it was able to move grill 10cm to break it, to test if that energy could break grill.

Pa = N.d/a.d = 48/(.002*.002*.1) = 120000000 = 12mPA

12 mPa < 31mPa

So using that energy and that displacement we rejects the possibility that water in a cup can provide enough energy to reach the required mPA to break the Grill.

Again this is basically assuming elasticity does not exist.

So even at the MAX imaginable force Energy of 48J its hard to believe this caused any damage.

It might have moved an already damaged grill. Thats harder equation on crack propagation.

  • 1
    $\begingroup$ You've calculated an energy (joules), not a force (newtons). $\endgroup$ – CDCM Aug 7 '17 at 21:35

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