Basically, I understand this question but I keep coming up with the wrong answer!! Please show me where I am going wrong! A 12.0-g bullet is fired horizontally into a 100-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 80.0cm, what was the speed of the bullet at impact with the block? Book answer: 273 m/s My work: Conservation of Mechanical Energy (KE + PEg + PEs)i = (KE + PEg + PEs)f 1/2(12)(Vi^2) = 1/2(112)(Vf^2) + 1/2(150)(80^2) soo..... 6(Vi^2) = 56(Vf^2) + 480,000 and from Conservation of Momentum: 12V1i =112V2f 0= 56Vf^2 - 6 Vi^2 + 480,000 plugging in for Vi.. I get 87 x 6 (I think I'm wrong here..) so 46Vf2 = 480,000 Vf = 32.0 m/s 12Vi = 112(32) Vi = 299 m/s Please, please, please help me! I'm going to go crazy!