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

[sensesfail]

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

 

[chaoseverlasting]

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 ([tex]s=ut-\frac{1}{2}a_{av}t^2[/tex]). 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.

[tex]t=1.267\times 10^{-6}[/tex]
[tex]F_{av}=189.423\times 10^{3}[/tex]

 

[ogb p]

Based on the given information, the average force of impact exerted by the car to stop the hail stone is 1200 Newtons. This force is significant enough to cause a dent of 0.20 cm in the car's hood, demonstrating the impact and force of the hail stone. It is important to note that this calculation assumes a perfectly elastic collision between the hail stone and the car, and factors such as air resistance and other external forces may affect the actual force of impact. Further experimentation and analysis would be needed to determine the most accurate average force of impact in this scenario.