A 5.00 g bullet moving with an initial speed of 390 m/s is fired into and passes

Question

A 5.00 g bullet moving with an initial speed of 390 m/s is fired into and passes through a 1.00 kgblock. The block, initially at rest on a frictionless, horizontalsurface, is connected to a spring of force constant 850 N/m. (a) If the block moves 5.10 cm tothe right after impact, find the speed at which the bullet emergesfrom the block.
1 m/s

(b) If the block moves 5.10 cm to theright after impact, find the mechanical energy converted intointernal energy in the collision.
2 J A 5.00 g bullet moving with an initial speed of 390 m/s is fired into and passes through a 1.00 kgblock. The block, initially at rest on a frictionless, horizontalsurface, is connected to a spring of force constant 850 N/m. (a) If the block moves 5.10 cm tothe right after impact, find the speed at which the bullet emergesfrom the block.
1 m/s

(b) If the block moves 5.10 cm to theright after impact, find the mechanical energy converted intointernal energy in the collision.
2 J

Explanation / Answer

Kinetic energy of the bullet just befire it hits the spring =potential energy of the spring at maximum compression            (1/ 2) mv ^ 2 = ( 1/ 2) k x^ 2 from this speed of the block just before hit thespring   v = x [ k / m ]                                                                  = ( 5.1 / 100) [ 850 / 1 )                                                                  = 1.4868 m / s from law of conservationof momentum , m ' U+ m u= m ' V + mv from this speed of the bullet after emerges from theblock   V = ( m' U + mu -mv ) / m ' where m = mass of block = 1 kg           m ' =mass of bullet =   5 g            U= 390 m / s            u= 0           v= 1.4868 m / s plug the values we get    V value                 

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