Question

An 8.0g bullet, moving at 400 m/s, goes through a stationary block of wood in 4.0 x 10^-4s, emerging at a speed of 100 m/s. (a) what average force did the wood exert on the bullet? (b) how thick is the wood?

Answer 1

Step-by-step explanation:

Initial velocity (u) of the Bullet = 400 m/sec

Final velocity (v) of the Bullet = 100 m/sec

Bullet passed through the block in (t) = 0.0004 sec

Using 1st Equation of motion :

400 m/s = 100 m/s - a (0.0004)

Deceleration of Bullet = 750,000 m/sec^2

(a) F (force exerted by the wooden block on the bullet) = F (force exerted by the bullet on the wooden block)

F = m * a = 0.008 * 750,000 = 6000 N

(b) Using 3rd Equation of motion :

`v^(2) = u^(2) - 2aS`

10000 = 160000 - 2 * 750,000 * S

Thickness of wood (S) = 0.1 m