Question

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

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Data:

mass of the bullet (m) = 10 g = (10/1000) kg =0.01 kg

Initial velocity (u) = 150 m/s

Final velocity (v) = 0 m/s

Time taken (t) = 0.03 sec

i. Calculation of the distance of penetrationof the bullet into the block:

Calculation of the distance of penetrationof the bullet into the block = Displacement

Displacement = average velocity × time

Displacement = [(u + v) / 2] × time

Displacement = [(150 + 0) / 2] × 0.03

Displacement = 2.25 m

ii. Calculation of the magnitude of the force exerted by the wooden block on the bullet:

the magnitude of the force exerted by the wooden block on the bullet = Retarding force

Retarding force = mass × reatardation

Retarding force = mass ×[ (Intitial velocity - Final velocity) / t ]

Retarding force = 0.01 ×[ (150 - 0) / 0.03 ] N

Retarding force = [150/3] N

Retarding force = 50 N

Answer 2

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

⇒ $\boxed{ \bf{v=u+at}}$

⇒ 0 = 150 + a × 0.03

⇒ a = $\bf \frac{-150}{0.03}=-5000 \:m/s^2$

Distance travelled by the bullet before coming to rest is given by $\boxed{ \bf{v^2 =u^2 +2as}}$

⇒ 0 = (150)² + 2 × (-5000) × s

⇒ s = $\bf \frac{22500}{10000} = 2.25 m$

Magnitude of the force applied by the bullet on the block

F = ma

F = 0.01 × -5000

F = -50 N