Question

A bullet hits the wall with velocity of 20m/s and penetrates it up toa distance of 5cm. Find deceleration of the bullet in the wall

Answer 1

Given :

• Bullet hits the wall with velocity (u) = 20m/s
• It penetrates it up to a distance (s) = 5cm or 0.05m

To Find :

• Deceleration of the bullet (a)

Formula Used :

Third law of motion -

$\bullet\underline{\boxed{\sf v^2=u^2+2as}}$

Solution :

$\implies{\sf a = \dfrac{v^2-u^2}{2s}}$

$\implies{\sf a = \dfrac{0-(20)^2}{2\times 0.05} }$

$\implies{\sf a = \dfrac{-400\times 10^2}{10}}$

$\implies{\bf a = -4000\: m/s^2}$

Answer :

Deceleration of the bullet in the wall is -4000m/s²

✏ Some Related Formula -

First law of motion - v = u + at

Second law of motion - s = ut + ½at²

Third law of motion - v² = u² + 2as

Answer 2

Answer:

$\large\boxed{\sf{-4000\;m{s}^{-2}}}$

Explanation:

Given that,

A bullet hits the wall with velocity of 20 m/s.

Penetration caused in the wall = 5 cm

To find the deceleration of the bullet.

We know that,

$\bold{{v}^{2} = {u}^{2} + 2as}$

Where,

• u = Initial velocity
• v = Final velocity
• a = acceleration
• s = distance covered

Now, from the question, we have,

• u = 20 m/s
• v = 0 m/s
• s = 5 cm = 0.05 m
• a = To find it.

Substituting the values, we get,

$= > {0}^{2} = {(20)}^{2} + 2a(0.05) \\ \\ = > 0.1a + 400 = 0 \\ \\ = > 0.1a = - 400\\ \\ = > a = - \frac{400}{0.1} \\ \\ = > a = - 4000$

Hence, deceleration is $-4000\;m{s}^{-2}$