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Mdhiggenz

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## Homework Statement

A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.48m long. The block is initially at rest. A bullet with mass 0.0118kg is fired at the block with a horizontal velocity V0. 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 0.875m , the tension in the cord is 4.94N .

## Homework Equations

## The Attempt at a Solution

So what I first did was I created the equation for a completely inelastic collition

MAVA=(MA+MB)vf

Vf= final velocity MA(bullet)= 0.0118kg

VA= Initial velocity of bullet MB(Block)= 0.750 kg

I have to find a way to find the final velocity in order to get the initial speed of the bullet.

So I used the equations

U1=K2+U2 which is :

Mgy1=1/2mv^2+mgy2

squareroot(2(mgy1-mgy2))/m

The masses cancel and you are left with

squareroot (2[gy1-gy2]) = 3.44m/s

Plugging the final velocity into the completely inelastic equations gives 222.08m/s.

Not sure if I did the problem right, and its for mastering physics so some feedback would be greatly appreciated (: thank you