If a bullet of mass 5 gm moving with velocity 100 m/sec, penetrates the wooden block upto 6 cm. Then the average force imposed by the bullet on the block is

The correct option is A. 417 N.

Given,

Mass of bullet = m = 5 mg = $5\times {10}^{-3}$ kg

Initial velocity of bullet = u = 100 m/s,

Final velocity of bullet = v = 0,

Distance of penetration of the bullet = s = 0.06 m

We know that, the retardation experienced by the bullet is given by,

Retardation = $a=\frac{u^2}{2s}$

$\Rightarrow$ Retardation = $a=\frac{u^2}{2s}=\frac{(100{)}^2}{2\times 0.06}=\frac{1\times {10}^4}{12\times {10}^{-1}}$
$\therefore$ Force = ma = $\frac{5\times {10}^{-3}\times 1\times {10}^5}{12}=\frac{5000}{12}=417N$.

Thus, the average force imposed by the bullet on the block is $417N$.