Question

A bullet of mass 0.006 kg traveling at 180m/s.penetration deeply into a target and it is brought to rest in 0.01sec calculate.the distance of penetration of the target

Answer 1

Answer:

0.9 m

Explanation:

As per the provided information in the given question, we have :

• Mass of bullet, m = 0.006 kg
• Initial velocity, u = 180 m/s
• Time taken, t = 0.01 s
• Final velocity, v = 0 m/s [Comes to rest]

We have to calculate the distance of penetration of the target.

In order to find the distance travelled, firstly we'll have to calculate the acceleration.

★ Acceleration :

Acceleration is the rate of change in velocity.

$\longrightarrow\boxed{\bf{a = \dfrac{v - u}{t}}}$

» a denotes acceleration

» v denotes final velocity

» u denotes initial velocity

» t denotes time

Substitute the values.

$\longrightarrow\sf{a = \dfrac{0 - 180}{0.01}}$

$\longrightarrow\sf{a = \dfrac{- 180}{0.01}}$

$\longrightarrow\sf{a = \dfrac{-18000}{1}}$

$\longrightarrow\underline{\bf{a = - 18000 \; ms^{-2}}}$

★ Distance Travelled :

Now, by using the third equation of motion,

$\longrightarrow\boxed{\bf{v^2 - u^2 = 2as }}$

» v denotes final velocity

» u denotes initial velocity

» a denotes acceleration

» s denotes distance

$\longrightarrow\sf{(0)^2 - (180)^2 = 2 \times (-18000) \times s}$

$\longrightarrow\sf{ 0 - (180)^2 = -36000\times s}$

$\longrightarrow\sf{0 - 32400 = -36000\times s}$

$\longrightarrow\sf{ - 32400 = -36000\times s}$

$\longrightarrow\sf{ \cancel{\dfrac{ - 32400 }{-36000}} = s}$

$\longrightarrow\underline{\boxed{\sf{\blue{ 0.9 \; m = s}}}} \;$

Therefore, distance travelled is 0.9 m.

Answer 2

Answer:

Given :-

• A bullet of mass 0.006 kg travelling at 180 m/s penetration deeply into a target and it is brought to rest in 0.01 seconds.

To Find :-

• What is the distance of penetration of the target.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

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

$\clubsuit$ Third Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{v^2 =\: u^2 + 2as}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken
• s = Distance Travelled or Displacement

Solution :-

☆ First, we have to find the acceleration :

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 180 m/s
• Time Taken = 0.01 seconds

According to the question by using the formula we get,

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

$\implies \sf 0 =\: 180 + a(0.01)$

$\implies \sf 0 - 180 =\: 0.01a$

$\implies \sf - 180 =\: 0.01a$

$\implies \sf \dfrac{- 180}{0.01} =\: a$

$\implies \sf - 18000 =\: a$

$\implies \sf\bold{\underline{\underline{a =\: - 18000\: m/s^2}}}$

Hence, the acceleration is - 18000 m/s² .

☆ Now, we have to find the distance travelled of penetration of the target :

Given :

• Initial Velocity = 180 m/s
• Final Velocity = 0 m/s
• Acceleration = - 18000 m/s²

According to the question by using the formula we get,

$\implies \bf v^2 =\: u^2 + 2as$

$\implies \sf (0)^2 =\: (180)^2 + 2(- 18000)s$

$\implies \sf (0 \times 0) =\: (180 \times 180) + (- 36000)s$

$\implies \sf 0 =\: 32400 - 36000s$

$\implies \sf 0 - 32400 =\: - 36000s$

$\implies \sf {\cancel{-}} 32400 =\: {\cancel{-}} 36000s$

$\implies \sf 32400 =\: 36000s$

$\implies \sf \dfrac{32400}{36000} =\: s$

$\implies \sf 0.9 =\: s$

$\implies \sf\bold{\underline{s =\: 0.9\: m}}$

$\therefore$ The distance of penetration of the target is 0.9 m .