Question

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

Answer 1

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 , 2aS = v² - u²

=> 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

Answer 2

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