Question

A bullet hits a wall with a Velocity of 20m/s and penetrates it up to distance 5cm find the deceleration of the bullet in the wall

Answer 1

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Answer 2

✬ Deceleration = –4000 m/s² ✬

Explanation:

Given:

• Bullet hits wall with velocity (u) = 20 m/s
• Wall is penetrated up to distance (s) = 5 cm or 0.05 m.

To Find:

• What is the deceleration (a) of the bullet in wall?

Formula to be used:

• Third law of motion
• v² = u² + 2as

Solution: Since, the bullet will stop after penetrating the wall. Therefore bullet's final velocity (v) = 0 m/s

Here we have

• v = 0 m/s ( Final Velocity )
• u = 20 m/s ( Initial Velocity )
• a = ?
• s = 0.05 m ( Distance )

Putting the value on formula :-

$\implies{\rm }$ v² = u² + 2as

$\implies{\rm }$ 0² = (20)² + 2 $\times$ a $\times$ 0.05

$\implies{\rm }$ 0 = 400 + 2a $\times$ 0.05

$\implies{\rm }$ –400 = 0.1a

$\implies{\rm }$ –400/0.1 = a

$\implies{\rm }$ –400/1 $\times$ 10 = a

$\implies{\rm }$ –4000 m/s² = a

Here, Negative (–) sign shows the deceleration. Hence, the deceleration of bullet in wall is –4000 m/s².

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★ Some related terms ★

➟ First law of motion = v = u + at

➟ Second law of motion = v = ut + 1/2at²

➟ Third law of motion = v² = u² + 2as

➟ The distance covered by an object in given interval of time is called it's final velocity.

➟ The velocity of an object just before the acceleration causes a change into it is called initial velocity.