Question

A bullet of mass 10gram travelling horizontally with a velocity 150 m/s ,strikes a stationary wooden block and comes to rest in 0.03 second. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force of the force exerted by the wooden block on the bullet.

plz help !!!!

Answer 1

hello friend...!!

⇒ according to the question, we should calculate the distance penetrated by the bullet into the block 

given that :

⇒ u = 150 m/s

⇒ v = 0 

⇒ t = 0.03 sec 

⇒ m = 10 g = 0,01 Kg

we know that , S = ( v² - u²) / 2a  ------------( 1 )

we should calculate acceleration ,  v = u + at 

a =  ( v - u ) / t

a = ( 0 - 150 ) / 0.03

a = -5000 m/s² ---------------( 2 ) 

substituting equation ( 2 ) in ( 1 ) gives ,

S = { 0 - (150)² } / - ( 2 x 5000 )

S= 22500 / 10000

S = 2.25 m 

therefore, the distance penetrated by the bullet is 2.25 m

to calculate force ,we know

F = m x a 

⇒ F = 0.01 x 5000

⇒ F = 50 N 

the force exerted by the wooden block on the bullet  is 50 N 

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hope it helps..!!

Answer 2

Answer:

Explanation: here we can use various formula so plese look carefully on the give answers on image