Question

A bullet moving with a velocity of
$50\sqrt{2} \frac{m}{s}$
is fired into a target which it penetrates to the extent of
$d$
metre. If this bullet is fired into a target of
$\frac{d}{2}$
metre thickness with the same velocity, it will come out of this target with a velocity of [Assume that resistance to motion is similar and uniform in both the cases]

Answers:-

a)
$40 \sqrt{2} \frac{m}{s}$
b)
$25 \sqrt{ 2 } \frac{m}{s}$
c)
$50 \frac{m}{s}$
d)

$5.0 \sqrt{3}$
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Answer 1

Answer :-

The bullet will come out of the target with a velocity of 50 m/s [Option.(c)]

Explanation :-

Acceleration of the bullet will be same in both the cases .

1st case :-

→ Initial velocity (u) = 50√2 m/s

→ Final velocity (v) = 0

→ Distance (s) = d

Let's calculate the acceleration of the bullet by using the 3rd equation of motion .

v² - u² = 2as

⇒ 0 - (50√2)² = 2ad

⇒ -5000 = 2ad

⇒ a = -5000/2d

⇒ a = -2500/d ----(1)

2nd case :-

→ Initial velocity (u) = 50√2 m/s

→ Acceleration (a) = -2500/d

→ Distance (s) = d/2

So, the required velocity of the bullet (v) will be :-

⇒ v² - u² = 2as

⇒ v² - (50√2)² = 2(-2500/d)d/2

⇒ v² - 5000 = -2500

⇒ v² = -2500 + 5000

⇒ v = √2500

⇒ v = 50 m/s

Answer 2

Answer:

Given :-

• A bullet moving with a velocity of 50√2 m/s is fired into a target which it penetrates to the extent of d metres.
• This bullet is fired into a target of d/2 metre thickness with the same velocity.

To Find :-

• The bullet will come out of the target with a velocity.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

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

where,

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

Solution :-

${\small{\bold{\purple{\underline{\bigstar\: In\: first\: case\: :-}}}}}$

$\leadsto$ A bullet moving with a velocity of 50√2 m/s is fired into a target which it penetrates to the extent of d metres.

Given :

• Final Velocity = 0 m/s
• Initial Velocity = 50√2 m/s
• Distance Covered = d metres

According to the question by using the formula we get,

$\implies \sf (0)^2 =\: (50\sqrt{2})^2 + 2ad$

$\implies \sf 0 \times 0 =\: 50\sqrt{2} \times 50\sqrt{2} + 2ad$

$\implies \sf 0 =\: 2500 \times 2 + 2ad$

$\implies \sf 0 =\: 5000 + 2ad$

$\implies \sf 0 - 5000 =\: 2ad$

$\implies \sf - 5000 =\: 2ad$

$\implies \sf \dfrac{- \cancel{5000}}{\cancel{2}d} =\: a$

$\implies \sf \dfrac{- 2500}{d} =\: a$

$\implies \sf\bold{\green{a =\: \dfrac{- 2500}{d}}}$

Hence, the acceleration is - 2500/d .

${\small{\bold{\purple{\underline{\bigstar\: In\: second\: case\: :-}}}}}$

$\leadsto$ This bullet is fired into a target of d/2 metre thickness with the same velocity.

Given :

• Acceleration = - 2500/d
• Initial Velocity = 50√2 m/s
• Distance Covered = d/2 metre

According to the question by using the formula we get,

$\longrightarrow \sf v^2 =\: (50\sqrt{2})^2 + 2 \times \dfrac{- 2500}{d} \times \dfrac{d}{2}$

$\longrightarrow \sf v^2 =\: 50\sqrt{2} \times 50\sqrt{2} + \dfrac{- 5000}{d} \times \dfrac{d}{2}$

$\longrightarrow \sf v^2 =\: 2500 \times 2 + \dfrac{- 5000}{\cancel{d}} \times \dfrac{\cancel{d}}{2}$

$\longrightarrow \sf v^2 =\: 5000 + \dfrac{- 5000}{2}$

$\longrightarrow \sf v^2 =\: 5000 + (- 2500)$

$\longrightarrow \sf v^2 =\: 5000 - 2500$

$\longrightarrow \sf v^2 =\: 2500$

$\longrightarrow \sf v =\: \sqrt{2500}$

$\longrightarrow \sf\bold{\red{v =\: 50\: m/s}}$

$\therefore$ The bullet will come out of the target with a velocity is 50 m/s .

Hence, the correct options is option no (c) 50 m/s.