6. A bullet is shot into a block of plastic. The bullet penetrates the block 0.6

Question

6. A bullet is shot into a block of plastic. The bullet penetrates the block 0.65 m. The mass of the bullet is 12 g. It is traveling with a speed of 505 m/s before it hits the block.

(a) Use kinematic equations to find the magnitude of the acceleration on the bullet as it is penetrating the block (ignore gravity, and assume that the force on the bullet as it penetrates the block is constant).

_________ m/s^2

(b) Use Newton's Second Law to find the magnitude of the force exerted on the bullet by the plastic block.

________N

Explanation / Answer

a) According to third equation of motion of a body V2 - u2 = 2as

v = final velocity of the body , u = initial velocity of the body , a = linear acceleration of the body ,

s = displacement covered by the body

After the body comes to rest v= 0 m/s then -u2 = 2as

a = - u2 / 2s

   = - ( 505 )2  / 2 x0.65

   = - 255025 / 1.3

   = - 196173.07692 m/s2   here ' - ' indicates the deceleration of the bullet while it is penetrating.

b) According to Newton's second law, the force F = ma

   = 0.012 Kg x 196173.07692 m/s2

   = 2354.07692 N

SECOND METHOD:

The kinetic energy of the bullet before it hits the plastic block

K.E = 1/2.m .v2

   = 0.5 x 0.012 Kg x ( 505 )2

   = 0.5 x 0.012 Kg x 255025

= 0.5 x 3060.3

   = 1530.15 J

work done by the plastic on the bullet after hits = work done by the bullet before it hits the block of plastic

i.e F.S = K.E

   F x 0.65 m = 1530.15 J

   F = 1530.15 J / .65 m

   = 2354.07692 N

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