Conservation of Momentum/Collisions

1. Mar 10, 2013

happycreature

Hi everyone, this is my first post to the forums. Nice to meet you all and thank you in advance.

1. The problem statement, all variables and given/known data
A bullet of mass 6.00g is fired horizontally into a wooden block of mass 1.21kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.210. The bullet remains embedded in the block, which is observed to slide a distance 0.260m along the surface before stopping.

What is the initial velocity of the bullet?

Mass of bullet = 6.00g = 0.006kg
Mass of block = 1.21kg (+bullet) = 1.216
Coeff. of friction µk = 0.210
∆d = 0.260m

2. Relevant equations
∆KE + ∆PE = -Wnc
KE = 1/2mv^2
Work = F∆d
F(kf) = µk(N)
*F∆t = m(vf - vi)

3. The attempt at a solution

F(sf) = Coeff. friction x normal force = (.210)(1.21)(9.8) = 2.49 N

Work = F∆d
Work done by friction = (2.49N)(0.260m) = -0.647 J

No potential energy here so...
KEf - KEi = -0.647
KEf = 0 (the block slid to a stop)
KEi = -0.647
1/2mvi^2 = -0.647
vi = √(2)(0.647)/0.006 = 14.64 m/s

...Which is incorrect.

*Because this problem is in the chapter on momentum it seems like they want me to use an equation involving momentum, but I don't know 1)if this problem can or cannot be solved with the work-energy theorem and 2) when I would use the momentum equation in the above scenario.

For the record I also tried finding the initial velocity of the block (1.03 m/s) and putting that as my KEf for the bullet but the difference was negligible.

Any advice greatly appreciated - thank you!

Last edited: Mar 10, 2013
2. Mar 10, 2013

Pagan Harpoon

When the bullet hits the block, not all of its kinetic energy goes into moving the block. A lot of it will be lost in breaking up the chemical bonds in the wood to embed the bullet in there.

The analysis you did assumed that the initial kinetic energy of the block was equal to the kinetic energy of the bullet, which I do not think is correct.

I suggest finding the velocity of the block of wood that would be required for it to slide 0.26 m and then using conservation of momentum to find the bullet's velocity.

Edit - I am assuming that you're looking for the speed of the bullet.

Last edited: Mar 10, 2013
3. Mar 10, 2013

ehild

Welcome to PF!

What is the question in the problem?

ehild

4. Mar 10, 2013

happycreature

Sorry, forgot to add the question when I was writing it out! It's there now.

Harpoon, you're right. Since this is a completely inelastic collision KEf does not equal KEi. Thank you for the correction. Is this why the work-energy theorem would not work here? Because energy is not conserved?

It looks like I have to use initial and final momentums with KE = 1/2mv^2 to solve it. I'll take a second look.

5. Mar 10, 2013

ehild

There are two parts of your problem. First you have an inelastic collision between bullet and wood. You can apply conservation of momentum to get the final velocity of the block with embedded bullet. Then you have the block, the bullet inside, travelling with the velocity gained in the collision and decelerating because of friction. In that second stage, you can use Work-Energy Theorem.

ehild

6. Mar 10, 2013

happycreature

...found the initial velocity of the block with the work-energy theorem, and then plugged that into m1v1 = (m1+m2)vf to solve for v1, which was 208.75 m/s (and correct).

It looks like I had some conceptual issues with work-energy and inelastic collisions.