Question

A bullet moving with a velocity of 25 m/s penetrates into
a wooden log and comes to rest 0.05 s. If the mass of the
bullet 30 g, find the retardation force exerted by the log
on the bullet and the depth to which the bullet penetrates.



please help me ✌️

Answer 1

Answer :

• Retarding force exterted by the log on the bullet = 1500 N

• Depth to which the bullet penetrates = 0.625 m

Given :

• Initial velocity of the bullet = 25 m/s

• Time taken by the bullet to come to rest after penetrating into a wooden log = 0.05 s

• Final velocity of the bullet = 0 m/s [The final velocity will be zero becomes it comes to rest.]

• Mass of the bullet = 30 g

To find :

• Retarding force exterted by the log on the bullet

• Depth to which the bullet penetrates

Concept :

Here, we have to find the retarding force exerted by the log. So, in order to find the retarding force firstly, we will calculate the acceleration by using the first equation of motion. Then by using the formula of force and after substituting the values we will get the required answer.

To find the depth to which the bullet penetrates we will use the third equation of motion.

First equation of motion :-

• v = u + at

Third equation of motion :-

• v² - u² = 2as

Formula to find force = ma

where,

• v denotes the final velocity

• u denotes the initial velocity

• a denotes the acceleration

• t denotes the time taken

• s denotes the distance/displacement

• m denotes the mass of the body

Solution :

Calculating the acceleration of the bullet :-

⠀⠀⠀⇒ First equation of motion :-

⠀⠀⠀⇒ v = u + at

⠀⠀⠀⇒ Substituting the given values :-

⠀⠀⠀⇒ 0 = 25 + a × 0.05

⠀⠀⠀⇒ - 25 = a × 0.05

⠀⠀⠀⇒ - 25/0.05 = a

⠀⠀⠀⇒ - 500 = a

⠀⠀⠀⇒ The value of a = - 500

Acceleration = - 500 m/s²

Now, calculating the retarded force exerted by the log :-

⠀⠀⠀⇒ Force = m × a

⠀⠀⠀⇒ Substituting the given values :-

⠀⠀⠀⇒ Force = 30 × (-500)

⠀⠀⠀⇒ Force = - 1500

Force exerted by the log = - 1500 N

Hence, the retarded force = 1500 N

Now, calculating the depth to which the bullet penetrates :-

⠀⠀⠀⇒ Third equation of motion :-

⠀⠀⠀⇒ v² - u² = 2as

⠀⠀⠀⇒ Substituting the given values :-

⠀⠀⠀⇒ (0)² - (25)² = 2(-500)s

⠀⠀⠀⇒ - (25)² = - 1000 × s

⠀⠀⠀⇒ - 625/- 1000 = s

⠀⠀⠀⇒ 0.625 = s

Therefore, the depth to which the bullet penetrates is 0.625 m