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.

I'll give 30 points if you answer this correct and also mark you as brainliest​

Answer 1

Answer :-

→ Retarding force exerted is 15 N .

→ Required depth is 0.625 m .

Explanation :-

We have :-

→ Initial velocity of the bullet = 25 m/s

→ Final velocity of the bullet = 0 m/s

→ Mass of the bullet = 30g = 0.03 kg

→ Time taken = 0.05 s

________________________________

Firstly, let's calculate acceleration of the bullet by using the 1st equation of motion :-

v = u + at

⇒ 0 = 25 + a(0.05)

⇒ -25 = 0.05a

⇒ a = -25/0.05

⇒ a = -500 m/s²

According to Newton's 2nd law of motion :-

F = ma

⇒ F = 0.03 × (-500)

⇒ F = -15 N

[Here, -ve sign represents retarding force]

________________________________

Now, from the 3rd equation of motion :-

v² - u² = 2as

⇒ 0 - (25)² = 2(-500)s

⇒ -625 = -1000s

⇒ s = -625/-1000

⇒ s = 0.625 m

Answer 2

Answer:

Given :-

• A bullet moving with a velocity of 25 m/s penetrates into a wooden log and comes to rest of 0.05 seconds.
• The mass of the bullet is 30 g.

To Find :-

• What is the retardation force exerted by the log on the bullet.
• What is the depth to which the bullet penetrates.

Formula Used :-

$\clubsuit$ First Equation Of Motion :

$\longmapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time

$\clubsuit$ Force Formula :

$\longmapsto \sf\boxed{\bold{\pink{F =\: ma}}}$

where,

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

$\clubsuit$ Third Equation Of Motion :

$\longmapsto \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Travelled

Solution :-

First, we have to find the retardation force exerted by the log on the bullet :

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 25 m/s
• Time = 0.05 seconds

According to the question by using the formula we get,

$\implies \sf 0 =\: 25 + a \times 0.05$

$\implies \sf 0 =\: 25 + 0.05a$

$\implies \sf 0 - 25 =\: 0.05a$

$\implies \sf - 25 =\: 0.05a$

$\implies \sf \dfrac{- 25}{0.05} =\: a$

$\implies \sf - 500 =\: a$

$\implies \sf \bold{\red{a =\: - 500\: m/s^2}}$

$\therefore$ The retardation exerted by the log on the bullet is - 500 m/s².

____________________________________

Now, we have to find the force :

Let's find the mass first :

$\leadsto \sf Mass = 30\: g$

$\leadsto \sf Mass =\: \dfrac{30}{1000}\: kg \: \bigg\lgroup \sf\bold{1\: g =\: \dfrac{1}{1000}\: kg}\bigg\rgroup$

$\leadsto \sf\bold{Mass =\: 0.03\: kg}$

Given :

• Mass = 0.03 kg
• Acceleration = - 500 m/s²

According to the question by using the formula we get,

$\implies \sf Force =\: 0.03 \times (- 500)$

$\implies \sf\bold{\green{Force =\: - 15\: N}}$

$\therefore$ The retardation force exerted by the log on the bullet is 15 N .

___________________________

Now, we have to find the depth to which the bullet penetrates :

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 25 m/s
• Acceleration = - 500 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf (0)^2 - (25)^2 =\: 2 \times (- 500) \times s$

$\longrightarrow \sf 0 - 625 =\: - 1000 \times s$

$\longrightarrow \sf {\cancel{-}} 625 =\: {\cancel{-}} 1000s$

$\longrightarrow \sf 625 =\: 1000s$

$\longrightarrow \sf \dfrac{625}{1000} =\: s$

$\longrightarrow \sf 0.625 =\: s$

$\longrightarrow \sf\bold{\red{s =\: 0.625\: m}}$

$\therefore$ The depth to which the bullet penetrates is 0.625 m .