A bullet of mass 0.03 kg moving with a speed of 400 m/s penetrates 12 cm into a fixed block of wood. The average force exerted by the wood on the bullet will be

Concept:

Equations of motion

v = u +at

s = ut + 0.5at² 

v² - u² = 2as

Where, u = initial velocity, v = final velocity, a = acceleration/retardation, s = displacement

Calculation:

The bullet strikes the wooden block (with velocity ‘u’) and then tries to enter it. The block applies resisting force on the bullet causing some sort of retardation (a) to its motion. Finally the bullet comes to rest (v) after travelling some distance(s) as a result of this retardation offered by the block.

Given, u = 400m/s, s = 12 cm, m = 0.03kg

Using 3^rd equation of motion

v² - u² = 2as

- (400)² = 2a × 0.12

\(a = \frac{{160000}}{{2 \times 0.12}} = \; - 666666.67\;m/s^2\)

Negative sign shows that it is retardation

Force = ma

⇒ F = 0.03 × (-666666.667)

⇒ F = 20000N = 20 kN