Recent content by BC2210

Delete.. Delete..

I know momentum is conserved in each case. In the case of the wood block, the two objects stick together and move with a common velocity. But is that velocity the initial v of the bullet? However, kinetic energy is not conserved when the bullet hits the wood block correct? The kinetic energy...

All the information for the two scenarios are exactly the same. The mass of the block and bullet, initial v, and all is the same. Only difference is one block is steel and one is wood. One is elastic and one is inelastic collision. But do the V's of the blocks after the collision equal each...

Homework Statement 2 bullets of the same mass are fired at the same velocities. One hits a wood block, and embeds itself. One hits a steel block, and bounces off. Both blocks are on a frictionless table. Compare the two blocks speed after. Homework Equations The Attempt at a...

N - W = (mv2)/r but in the book it says the normal force is greater than the weight, resulting in a force pointing towards the center...

So right now I have F = N - (1.5mg) ... which is the only forces acting on the car... but I thought the force has to point to the middle of the circle?

Homework Statement A roller coasters weight increases 50% through the center of a dip in the track. The radius of the dip is 30m. What is the speed at the bottom? Homework Equations This is a uniform circular motion problem correct? But how do I calculate if I don't know the weight...

Thanks so much for yalls help...I really appreciate it! It makes sense... I was able to most all of my HW and get the correct answers but I am not very sure about this last one I did...(Ya know, if you get bored and want to help) :) A block of mass 1.50 kg is attached to one end of a...

Combined mass of the 2 blocks * Gravity * The height the block is dropped from... Or should it not be where the block is dropped from since they are moving as one object now? Im thinking that's my potential energy (mgy) ? Or instead of 1m I should have x on that side of the equation...

Wow that helps out a lot...makes sense: I came up with: (1/2)(.2)(2.21) + (.2)(9.8)(1m) = (1/2)(2000)x^2 x = .047m That look right to you guys? I really appreciate the fast help!

Where does the 1+x come from? Because of the 1m height? Or was that not correct Bright Wang, I saw you removed it... Trying out rl.bhat's method now...

Homework Statement A 0.10 kg flat wooden block is sitting on a vertical massless spring of spring constant 2000 N/m. An identical block is dropped from a height of 1.00 m above the first block. The collision between the two blocks is perfectly inelastic. How far down is the the spring...

A) (1/2)*(2000)*(-.1m)^2 ?

F of spring = -k\Deltax? So -2000 * 10cm? And then integrate from 0 to 10cm of the product?

Homework Statement A block of mass 1.50 kg is attached to one end of a horizontal spring, the other end of which is fixed to a vertical wall. The spring has a stifffness constant of 2000 N/m. The block slides without friction on a horizontal table, set close to the wall. Find the work done...