# Meteorite falling on a Car...

by PhyzicsOfHockey
Tags: falling, meteorite
 P: 42 1. The problem statement, all variables and given/known data A 27 pound meteorite fell and struck a car, creating a dent about 28 cm deep. If the initial speed of the meteorite was 530 m/s, what was the magnitude of the average force exerted on the meteorite by the car? 2. Relevant equations Vf^2=Vi^2+2ad F=ma 3. The attempt at a solution Just need someone to tell me if my work is right. I found the acceleration using Vf^2=Vi^2+2ad it came out to be-501,607 m/s^2 I to the mass of the meteorite 27/2.2= 12.273 kg and multiplied it my the acceleration and came up with 6.156 MN. Did I do this problem right?
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 Quote by PhyzicsOfHockey 1. The problem statement, all variables and given/known data A 27 pound meteorite fell and struck a car, creating a dent about 28 cm deep. If the initial speed of the meteorite was 530 m/s, what was the magnitude of the average force exerted on the meteorite by the car? 2. Relevant equations Vf^2=Vi^2+2ad F=ma 3. The attempt at a solution Just need someone to tell me if my work is right. I found the acceleration using Vf^2=Vi^2+2ad it came out to be-501,607 m/s^2 I to the mass of the meteorite 27/2.2= 12.273 kg and multiplied it my the acceleration and came up with 6.156 MN. Did I do this problem right?
Not quite. Using F =ma gives you the net average force acting on the meteorite. Threre is more than one force acting. You are asked to find the force of the car on the meteorite.
P: 42
 Quote by PhanthomJay Not quite. Using F =ma gives you the net average force acting on the meteorite. Threre is more than one force acting. You are asked to find the force of the car on the meteorite.

Isn't the force equal but opposite?

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## Meteorite falling on a Car...

 Quote by PhyzicsOfHockey Isn't the force equal but opposite?
Your calculation seems right to me (I did not check the numbers, just the method)

Patrick
 Sci Advisor P: 5,390 I suppose PhantomJay is referring to other forces on the meteorite such as its own weight. Its weight of 12.273*9.8 N = 0.00012 MN is negligible compared to the force caused by the impact, so it is reasonable to ignore it. The question doesn't say what direction the meteorite was travelling, and if you have ever seen a "shooting star", there is no reason to assume it would be falling vertically downwards. So even if you did want to include the weight, you don't know how to add the two force vectors.
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