Question

A bullet of mass 12 g travelling with a velocity of 150 m/s penetrates
deeply into a fixed target and is brought to rest in 0.02 s.
Calculate : (i) depth of penetration in the target
(ii) average retarding force exerted on the bullet.​

Answer 1

Given :

• Mass of the bullet (m) = 12 g or 0.012 kg.

• Intial velocity (u) = 150 m/s.

• Time of contact (t) = 0.02 s.

To find :

• Depth of penetration.

• Average Retarding force.

Solution :

First let us find the Acceleration Produced by the bullet :

We know the First Equation of Motion i.e,

$\boxed{\bf{v = u + at}}$

Where :

• v = Final Velocity
• u = Intial velocity
• a = Acceleration
• t = Time Taken

Now using the First Equation of Motion and substituting the values in it, we get :

$:\implies \bf{v = u + at}$

$:\implies \bf{0 = 150 + a \times 0.02}$

$:\implies \bf{0 = 150 + \dfrac{2}{100}a}$

$:\implies \bf{-150 = \dfrac{2}{100}a}$

$:\implies \bf{-150 \times 100 = 2a}$

$:\implies \bf{\dfrac{-150 \times 100}{2} = a}$

$:\implies \bf{\dfrac{-15000}{2} = a}$

$:\implies \bf{-7500 = a}$

$\boxed{\therefore \bf{a = -7500\:ms^{-2}}}$

Hence the acceleration produced by the bullet is - 7500 m/s².

Depth of the Penetration of the target :

We know the second Equation of Motion.i.e,

$:\implies \bf{S = ut + \dfrac{1}{2}at^{2} }$

Where :

• S = Distance
• a = Accelaration
• t = Time Taken
• u = Initial Velocity

Using the Second equation of motion and substituting the values in it, we get :

$:\implies \bf{S = 150 \times 0.02 + \dfrac{1}{2} \times (-7500) \times 0.02^{2}}$

$:\implies \bf{S = 3 + \dfrac{1}{2} \times (-7500) \times 0.0004}$

$:\implies \bf{S = 3 + \dfrac{1}{2} \times (-7500) \times 0.0004}$

$:\implies \bf{S = 3 + \dfrac{1}{2} \times (-7500) \times 0.0004}$

$:\implies \bf{S = 3 + \dfrac{1}{2} \times (-7500) \times 0.0004}$

$:\implies \bf{S = 3 + (-3750) \times 0.0004}$

$:\implies \bf{S = 3 + (-1.5)}$

$:\implies \bf{S = 3 - 1.5}$

$:\implies \bf{S = 1.5}$

$\boxed{\therefore \bf{S = 1.5\:m}}$

Hence the Depth of Penetration in the target is 1.5 m.

Retarding Force exerted on the bullet :

We know that ,

F = ma

Where :

• F = Force Exerted
• m = Mass
• a = Acceleration.

Using the above equation and substituting the values in it , we get :

F = 0.012 × (-7500)

=> F = -90 N

=> Retarding F = 90 N

Hence the Retarding Force exerted on the bullet is 90 N .