a bullet of mass 10g travelling horizontally with a velocity of 150 metre per second -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.
Answers
distance =speed×time
150×0.03
4.5 m
f=ma
v²-u²=2as
o-150×150/2×4.5=a
22500/9=a
2500=a
f=ma
f=0.01×2500
f=25 n
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] .