A 5.00-g bullet moving with an initial speed of vi = 400 m/s is fired into and p

Question

A 5.00-g bullet moving with an initial speed of vi = 400 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 900 N/m. The block moves d = 5.00 cm to the right after impact before being brought to rest by the spring. (a) Find the speed at which the bullet emerges from the block. (b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in the bullet-block system during the collision.

Explanation / Answer

E in spring: E=0.5*900N/m*0.05²m²=1.125 J E in Bullet before impact: 0.5*0.005 kg*400²m²/s² = 400 J Energydifference: 1,125 J (from elastic potential energy in the spring, the momentum of the block is 0, since all its kinetic energy (taken from the bullet) is converted into elastic potential) Speed at emergence: v= squareroot((2*(400J-1,125J))/0.005kg=399… m/s

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