Question

a bullet mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationery wooden block and comes to rest in 0.03 seconds . 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

Answer 1

ur answer is attached.......

Answer 2

Answer :-

• Distance of penetration of the bullet into the block is 2.25 metres .

• Force exerted by the wooden block on the bullet is a retarding force of 50 Newtons .

Explanation :-

We have :-

→ Mass of the bullet (m) = 10 g = 0.01 kg

→ Initial velocity (u) = 150 m/s

→ Time taken (t) = 0.03 s

→ Final velocity (v) = 0

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Firstly, let's calculate acceleration of the bullet by using the 1st equation of motion .

v = u + at

⇒ 0 = 150 + (a)0.03

⇒ -150 = 0.03a

⇒ a = -150/0.03

⇒ a = -5000 m/s²

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Now, we can calculate distance of penetration of the bullet into the block by using the 3rd equation of motion.

v² - u² = 2as

⇒ 0 - (150)² = 2(-5000)(s)

⇒ -22500 = -10000s

⇒ s = -22500/-10000

⇒ s = 2.25 m

Finally, we will calculate the required force by using Newton's 2nd law of motion .

F = ma

⇒ F = 0.01(-5000)

⇒ F = -50 N

[Here, -ve sign represents retarding force] .