Question

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

Initial velocity of the bullet, u=150 ms

−1

Final velocity of the bullet, v=0

Mass of the bullet, m=10 g=0.01 kg

Time taken by the bullet to come to rest, t=0.03 s

Let distance of penetration be s

v=u+at

0=150+a(0.03)

⟹a=−5000 ms

−2

(- sign shows retardation)

v 2 =u 2 +2as

0=150

2

+2×(−5000)×s

s=2.25 m

Magnitude of the force exerted by the wooden block, ∣F∣=m∣a∣

∣F∣=0.01×5000=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] .