Question

A bullet initially moving with a velocity of 20 m/s strikes a target and comes to reset after penetrating a distance of 0.01 m in the target calculate the retardation producer by the target.

please help me ​

Answer 1

Answer :-

Retardation produced by the target is 20000 m/s² .

Explanation :-

We have :-

→ Initial velocity = 20 m/s

→ Distance travelled = 0.01 m

________________________________

As the bullet finally comes to rest, so it's final velocity will be 0 m/s.

By using the 3rd equation of motion, we get :-

v² - u² = 2as

Where :-

• v is the final velocity.

• u is the Initial velocity.

• a is acceleration.

• s is distance.

⇒ 0 - (20)² = 2(a)(0.01)

⇒ -400 = 0.02a

⇒ a = -400/0.02

⇒ a = -20000 m/s²

∵ Retardation is negative acceleration .

∴ Retardation :-

= -(-20000)

= + 20000 m/s²