A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Given, mass of the bullet (m) = 10 g = 0.01 kg

Initial velocity of the bullet (u) = 150 m/s

Terminal velocity of the bullet (v) = 0 m/s

Time period (t) = 0.03 s

We know that, \( v = u +at \)
\(\Rightarrow 0 = 150 + a(0.01) \)
\(\Rightarrow a = - 5000 \ m/s^2 \)

Now, \( v^2 - u ^2 = 2 as \)
\(\Rightarrow 0^2 - 150^2 = 2 × (-5000) s \)
\(\Rightarrow s = 2.25 \ m \)

Now, force exerted by wooden block on bullet = mass of bullet × acceleration of bullet

\( \Rightarrow F = 0.01 × (-5000) = -50 N \)