Okay--- Here's the question: 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? I understand that this is an elastic question and I'll need to use conservation of momentum and probably conservation of kinetic energy... But the whole spring thing throws me off? What does it mean? And how do I go about solving this problem when NO velocities are given? Accept for the block being initially at rest..... I'm confused. Help. Please Help.
Try treating this as two colisions: First the bullet has a (perfectly inelastic) colision with the block. Then the block hits the spring conserving energy.
Okay, that's easy to picture.... but then, isnt m1v1=m2v2? (cons. of momentum) do you solve for one of these velocities and then plug it into the cons of kinetic energy equation?
I am still confused with this problem, 2 days later.. so if anyone is willing to help, i'm willing to be helped!
Use conservation of mechanical energy and conservation of momentum. Conservation of momentum [tex] m_{bullet}v_{bullet} = (m_{bullet} + m_{block})v_{both} [/tex] Conservation of Mechanical energy [tex] \frac{1}{2}(m_{bullet} + m_{block})v_{both}^{2} = \frac{1}{2}kx_{max}^{2} [/tex] do you need more help than that?
Well if you really want to know.... Being patient, practicing alot!!!, making sure you understand the concepts and the LIMITATIONS!!! of the theories.