A bullet is fired through an apple 5.6 cm thick through the line of it\'s path.

Question

A bullet is fired through an apple 5.6 cm thick through the line of it's path. If the bullet enters with a speed of 413 m/s and emerges with a speed of 380 m/s, what is its acceleration as it passes through the apple?

Imagine the same bullet with the same entry speed fired into a bullet-proof vest of thickness 1.25 cm. The bullet pierces it completely and stops. What is the acceleration in this case?

Using g = 10 m/s2, how many g's does the acceleration you found for 2 a) correspond to? NOTE: Quote your answer as an absolute value

Explanation / Answer

using v^2 - u^2 = 2ad
380^2 - 413^2 = 2 x a x 0.056
a = - 233651.79 m/s2

using v^2 - u^2 = 2ad
0 - 413^2 =2 a x 0.125
a =-6822760 m/s2

divide it by g u will get your answer.

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