Question

A Bullet of mass 10g Travelling horizontally with a velocity of 150m/s strikes a stationary 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:

Given :-

mass of the bullet (m) = 10g = 0.01 kg [as 1000g = 1 kg]

Initial velocity (u) = 150 m/s

Final velocity (v) = 0 [ as bullet comes to rest ]

Time (t) = 0.003 seconds

Now,

From Newton's First Law of Motion

⇒ v = u + at

⇒ 0 = 150 + a × 0.03

= -150/0.03 = (-5000 m/s²)

Distance travelled by the bullet before coming to rest is given by

v² - u² = 2as

⇒ 0 = (150)² + 2 × (-5000) × s

⇒ s = 22500 / 10000

= 2.25 m

Magnitude of the force applied by the bullet on the block=

F = ma

F = 0.01 × -5000

F = -50 N

Answer 2

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