How Much Force Did the Car Exert to Stop a Hail Stone?

AI Thread Summary
A 0.012 kg hail stone traveling at 20 m/s created a 0.20 cm dent in a car's hood, prompting a calculation of the average force exerted by the car to stop it. The average velocity of the hail stone was determined to be 10 m/s, and the time taken to stop was initially calculated as 0.0002 seconds. However, it was noted that this approach was incorrect due to the need for average acceleration and the impulse momentum theorem. The correct time to stop was found to be approximately 1.267 microseconds, leading to an average force of about 189,423 N. The discussion emphasizes the importance of using the appropriate physics principles for accurate calculations.
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Homework Statement



During an autumn storm, a 0.012 kg hail stone traveling at 20 m/s made a .20 cm deep dent in the hood of a person's car. What average force did the car exert to stop the damaging hail stone?

Homework Equations



d = vt
ft = mv


The Attempt at a Solution



Vav = (Vf + Vo)/2 = (0 + 20)/2 = 10 m/s

V = 10 m/s
d = 0.002 m

t = d/v = 0.002/10 = 0.0002 seconds

m = 0.012kg
delta v = 20 m/s
t = 0.0002 seconds

f = mdeltav / t = (0.012 * 20) / 0.0002 = 1200 N
 
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Thats not right. You can't use d=vt here as there is an average force applied resulting in an average acceleration. What you have to do is compute the average acceleration using the impulse momentum theorem, then use that to calculate the time required (s=ut-\frac{1}{2}a_{av}t^2). From there, you may calculate the average force using, again, the impulse momentum theorem. To help you along, I shall give you the answers, but you have to get there.

t=1.267\times 10^{-6}
F_{av}=189.423\times 10^{3}
 
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