Question

a bullets hits a sand box with a velocity of 20m/s and it penetrates its up to a distance of 6 cm. find the decelaration of the bulletvin the sand box

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Initial velocity of bullets, u = 20 m/s

Final velocity of bullets, v = 0

Distance covered, c = 6 cm = 0.06 m

To Find,

Deceleration of the bullet, a = ?

According to the 3rd law of motion,

We know that,

v² - u² = 2as

So, putting all the values, we get

⇒ v² - u² = 2as

⇒ (0)² - (20)² = 2 × a × 0.06

⇒ 400 = 0.12a

⇒ 400/0.12 = a

⇒ a = 3333.33 m/s²

Hence, the deceleration of the bullet is 3333.33 m/s².

Answer 2

GIVEN THAT:

• Bullet hits sand box with Velocity 20 m/s and penetrates it upto a distance of 6 cm.

To Find:

• Deceleration of bullet in the sand box.

Formula used:

• v^2 = u^2 + 2aS.

SOLUTION:

ATQ, we have

initial velocity (u) = 20 m/s

Distance (S) =6 cm = 0.06 m

Final Velocity (v) = 0 [bullet stops]

Acceleration (a) = ? [To be calculated]

So, Using the above formula, we have

0^2 = 400 + 2a*(0.06)

-400 = 0.12a

a = -3333.33 m/s^2

So, deceleration is -3333.33 m/s^2.

Hope this helps