- #1
PsychonautQQ
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Homework Statement
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?
Homework Equations
conservation of energy
conservation of momentum
centripital acceleration = v^2/r
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?