# Find Initial Velocity of a Bullet

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1. Nov 2, 2014

### cntchds

This problem was given verbally during a class period, so I will set it up in my own words.

A bullet (.012kg) is fired at a block (.800kg) hanging on a strong, massless string of length 1.6m. After the collision the bullet is embedded in the block. When the block is .8m above its original location the tension in the string is 4.8N. What is the initial velocity of the bullet?

It appears there are a lot of different things going on.
First: the bullet's interaction with the block is conservation of momentum.
mbulletvbulletinitial+mblockvblockinitial=mbulletvbulletfinal+mblockvblockfinal
Second: the combination of the bullet and block then have conservation of energy from the time they are combined to the point where they are at heigh .8m.
1/2mvinitial2+mgho=1/2mvfinal2+mghfinal
Last: tension in the string (and here is where I'm really not sure I understand)
Tension is given as 4.8N
Fg(centripetal)=mg*cos(θ)
Fg(tangential)=mg*sin(θ)

My fear is that I have a major logical flaw in how I'm thinking of the tension in the string.

First thing I did was subtracted the tension in the string provided by gravity.

So I did:
acentripetal*m=T-mg*cos(θ)
my reasoning is because Acentripetal=v2/r , and I assumed that the tension added by the gravity in the same direction does not play a part in that equation.

After I found v at that point to be 1.27m/s it became an energy conservation problem to find velocity when potential energy was 0 which I found to be 4.16m/s. Then I moved on to conservation of momentum to find the velocity of the bullet to have an inelastic collision with the block to have the momentum stay the same before and after. For this I found the initial velocity to be 2810m/s for the bullet.

I feel very confident with my work from having the velocity of the block/bullet combo at the height in the swing back to the velocity at the time directly after the impact and for the conservation of momentum between the bullet and block from before to after the collision. What I really have a hard time grasping is how to deal with finding the velocity of the bullet/block combo at height .8m with the tension. There was no answer given in class, and we aren't turning it inso it won't be graded, but I want to be ready for an upcoming midterm.

2. Nov 2, 2014

### Staff: Mentor

Check the last step, where you calculate the bullet speed. Everything up to the velocity of block+bullet at the bottom is correct.

3. Nov 3, 2014

### cntchds

Doing the last step again:

mbulletvbulletinitial+mblockvblockinitial=(mblock+mbullet)vfinal.
or
.012kg*v+.8kg*0m/s=.812kg*4.16m/s
then
v=(.812kg*4.16m/s)/.012kg=281.5m/s

Different than my initial answer by a factor of ten. I probably misplaced a decimal place. Does this look correct?

Thank you again for your time.

4. Nov 3, 2014

Correct.