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1.So the problem given is the following,

2. Right now the I can figure out a few equations from this. I know that because the bullet becomes embedded in the wooden block, it can be considered a perfect inelastic collision. Thus

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**"A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.58m long. The block is initially at rest. A bullet with mass 0.0134 kg is fired at the block with a horizontal velocity[UP][/UP] v**_{0}. 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 0.700 m , the tension in the cord is 4.92N."2. Right now the I can figure out a few equations from this. I know that because the bullet becomes embedded in the wooden block, it can be considered a perfect inelastic collision. Thus

**m1v1 + m2v2 = (m1 + m2)vf.**For this problem, I'm going to call the bullet m1 and the wooden block m2. Thus, I know that from this I can solve for the final velocity of the two embedded object. Also, I figure that because the block is on a string, and it is moving, that centripetal force/ acceleration must come into play. Therefore, should I use the equation**F**_{C}=ma_{c}=T - mg?3.

**v**However, I'm confused as to how to progress from here? I know that the tension when the object (combined) has reached 0.700 m is equal to 4.92 N, but I'm having difficult picturing this. Does this tension value represent the tension when the 1.58 m string is facing right, forming a triangle of sorts where 1.58 is the hypotenuse and the y-component is 0.88m? Please let me know how to progress from here, or at least if you know how I should be visualizing this that would be so helpful! Thanks!_{f}=(0.0134kg * v_{0}+ 0.775 kg * 0 m/s) / (0.0134 kg + 0.775 kg) = 0.016996 v_{0}.**EDIT: The question is what is the initial speed V**_{0}of the bullet?
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