Question

A bullet of mass 10g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. 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 bullet​

Answer 1

Answer:

Acceleration of the bullet (a) = 5000 m/s2

F = m × a

F = 0.01× -5000

F = -50 N

Hence, the magnitude of the force exerted by the wooden block on the bullet is 50 N.

Answer 2

⇒ 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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