Question

A bullet of mass 10g travelling horizontally with a velocity of 150ms-¹ strikes a stationery wooden block and comes to rest in 0.03s. 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

Answer:

According to the first equation of motion,

v= u + at

Acceleration of the bullet (a)

0 = 150 + (a — 0.03 s)

a = -150 / 0.03

a = – 5000 m/s2

Here the negative sign indicates that the velocity of the bullet is decreasing.

According to the third equation of motion,

v2= u2+ 2as

0 = (150)2+ 2 (-5000)

s = 22500 / 10000

s = 2.25 m

Hence, the distance of penetration of the bullet into the block is 2.25 m.

From Newton’s second law of motion,

Force (F) = Mass × Acceleration

Mass of the bullet (m) = 10 g

m = 0.01 kg

Acceleration of the bullet (a) = 5000 m/s2

F = m × a

F = 0.01× -5000

F = -50 N

Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.

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] .