Question

A 20gm bullet moving at 300m/s stop s after penetrating 3cm of bone. calculate the average force exerted by the bullet.

Answer 1

Average force exerted by the bullet = 30 KN

Given:

A 20 gm. bullet moving at 300 m/s.

Penetrating 3 cm of bone.

To find:

The average force exerted by the bullet.

Formula used:

v²- u² =2as

Where v = Final velocity

             u = Initial velocity

             a = acceleration of object

             s = Distance traveled by the object

Force = mass × acceleration of object

Explanation:

u = initial velocity =  300 m/s.

v = Final velocity =0

s = Distance traveled by the object = 3 cm =0.03 m

v²- u² =2as     Put all the value in this equation

0² -300² = 2×a×0.03

           a  = $\frac{300^{2} }{2\times 0.03}$ = 1.5× $10^{6}$ m/s²

Mass of object =  20gm = 20×$10^{-3}$ kg.

Force = mass × acceleration of object

Force = 20×$10^{-3}$  ×  1.5× $10^{6}$

Force = 30 × $10^{3}$ N = 30 KN

Average force exerted by the bullet = 30 KN

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