A 10.0 g bullet is firedhorizontally into a 109 g woodenblock that is initially

Question

A 10.0 g bullet is firedhorizontally into a 109 g woodenblock that is initially at rest on a frictionless horizontalsurface and connected to a spring having springconstant 160 N/m. The bulletbecomes embedded in the block. If the bullet-block systemcompresses the spring by a maximum of 82.0cm, what was the speed of the bullet at impactwith the block?

Explanation / Answer

According to conservation of energy we have         KEb-b1 +PE1 = KEb-b2 + PE2        0.5 (m+M)V2 + 0 = 0+ 0.5kx2 ==> V = [ k x2 / (m+M)]              = [160 * 0.82*0.82 / (0.119)]              = 30.06 m/s Applying conservation of momentum to before and aftercollision we have          m v = (M+m) V            ==> velocity of bullet is    v = 0.119 * 30.06 / 0.01                       = 357.80 m/s

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