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Lynna [10]
3 years ago
9

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?
Physics
1 answer:
Ierofanga [76]3 years ago
7 0

Answer:

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)

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

<em />

(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<em> = </em>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

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