# Inelastic Momentum / Centripetal force problem

1. Aug 12, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
A small wooden block with mass .8kg is suspeded from the lower end of a light cord that is 1.6m long. The block is initially at rest. A bullet with mass 12g is fired at the block with a horizontal velocity vi. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of .8m, the tension in the cord is 4.8N. What is the initial speed vi of the bullet?

2. Relevant equations
conservation of energy
conservation of momentum
centripital acceleration = v^2/r

3. The attempt at a solution

first the simple part...
b subscript = bullet
x subscript = block
v2b = v2x
mbv1b = mbv2b + mxv2x
mbv1b = (mb+mx)v2x
v1b = ((mb+mx)v2x)/mb

Then i set up the centripetal force equation like this:
mx(v2x^2 / r) = T - mx*g*sin∅
sin∅ turns out to be .5

so v2x = ((T-mx*g*.5*r)/mx)^1/2

plugging this number into conservation of momentum i end up getting that the bullet was traveling at 90 m/s, but i'm supposed to get 330 m/s according to a website (which might be wrong but probably isn't).
I probably set up my centriptal force part wrong? help? The website that gives the answer 330 says TSin∅=ma... how could they leave gravity out? and why did they put the sin∅ onto the tension? Doesn't it point in the direction of the centripetal acceleration already?

2. Aug 12, 2013

### vela

Staff Emeritus
The equation is only good for when the block is moving with speed ${v_x}_2$, which is when it's at the bottom — in other words, when $\theta=90^\circ$. As the pendulum swings up, it slows down. The lefthand side is no longer equal to $\frac{m_x {v_x}_2^2}{r}$ because its speed is no longer ${v_x}_2$.

Instead of using F=ma here, try using conservation of energy.