Find Initial Velocity of a Bullet

[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.
m_bulletv_{bulletinitial}+m_blockv_blockinitial=m_bulletv_bulletfinal+m_blockv_blockfinal
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.
1/2mv_initial²+mgh_o=1/2mv_final²+mgh_final
Last: tension in the string (and here is where I'm really not sure I understand)
Tension is given as 4.8N
F_{g(centripetal)}=mg*cos(θ)
F_{g(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:
a_centripetal*m=T-mg*cos(θ)
my reasoning is because A_centripetal=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.

I really appreciate any help you all can provide. Thanks!

 

[mfb]

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

 

[cntchds]

Doing the last step again:

m_bulletv_{bulletinitial}+m_blockv_blockinitial=(m_block+m_bullet)v_final.
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.

 

[mfb]

Correct.

 

[HalfLostAndSearching]

Hello,

Thank you for sharing your thought process and equations. It seems like you have a good understanding of the concepts involved in solving this problem. I will try to provide some guidance and clarification on how to approach finding the initial velocity of the bullet.

First, let's review the equations you have used so far. The conservation of momentum equation you used is correct, as the total momentum of the system (bullet + block) should be equal before and after the collision. However, for the conservation of energy equation, there are a few things to note. The initial potential energy should include the potential energy of the block (mgh) and the bullet (0), as the bullet is initially at the same height as the block. The final potential energy should only include the potential energy of the combined system (mblock + mbullet) at a height of 0.8m. Additionally, the total kinetic energy should include the kinetic energy of both the block and the bullet after the collision (1/2(mblock + mbullet)vfinal^2). This will give you the correct equation:

1/2(mbulletvbulletinitial^2 + mblockvblockinitial^2) = 1/2(mblock + mbullet)vfinal^2 + (mblock + mbullet)ghfinal

Now, let's move on to the tension in the string. The tension in the string is due to both the centripetal force and the tangential force. The centripetal force is responsible for keeping the block and bullet moving in a circular path, while the tangential force is responsible for slowing down the motion of the block and bullet. In this case, the tension in the string is providing the centripetal force, and the tangential force is provided by the force of gravity (mg*cos(θ)). So, your equation for the tension should be:

Tension = Fg(centripetal) + Fg(tangential)

Now, we can substitute in the equations for the forces:

Tension = (mblock + mbullet)vfinal^2 / r + (mblock + mbullet)g*cos(θ)

Since we know the tension (4.8N) and the values for mass, height, and angle, we can solve for the final velocity (vfinal). This final velocity can then be used in the conservation of energy equation to solve for the initial velocity of the bullet (vbulletinitial).

I hope this