Question

a bullet hit a wall with a velocity of 20 m per second and penetrates it up to a distance of 5 cm find the declaration of the bullet in the wall

Answer 1

Given initial velocity (u) = 20m/s.

Final velocity v = 0m/s.

Distance s = 5cm.

                   = 0.05m

By the third law of motion:

=> v^2 - u^2 = 2as

= > (0)^2 - (20)^2 = 2a(0.05)

= > -400 = 0.1a

= > a = -4000m/s^2.


Therefore negative sign represents retardation.


Hope this helps!

Answer 2

Here we go,

A bullet hits a wall with a velocity of 20 m/s , i.e. a bullet is coming towards a wall with the velocity of 20 m/s and penetreats upto a distance of 5 cm ,i.e the bullet covers the distance of 5 cm. After covering the distance of 5 cm, it finally comes to rest.

Here,

Initial velocity ( u ) = 20 m/s

Final velocity ( v ) = 0

Distance covered ( s ) = 5 cm = 0.05 cm

Using equation of motion,

=> v² = u² + 2as

=> 0² = ( 20 m/s )² + 2 a ( 0.05 m )

=> 0 = ( 400 m²/s² ) + 0.1 m a

=> 0 - 400 m²/s² = 0.1 m a

=> -400 m²/s² = 0.1 m a

=> [ ( -400 m²/s² ) ÷ 0.1 m ] = a

=> [ ( -400 × 10) m/s² ] = a

•°• a = -4,000 m/s²

Hence, retardation = 4,000 m/s²

Note : Retardation is the negative acceleration.