A Ballistic Pendulum

1. Feb 16, 2008

Arejang

[SOLVED] A Ballistic Pendulum

1. The problem statement, all variables and given/known data

A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block? The length of the string is 2 meters.

2. Relevant equations

Momentum Conservation:

$$m_{a}v_{a1}+m_{b}v_{b1}=(m_{a}+m_{b})*v_{2}$$

Energy Conservation:

$$1/2mv^{2}=mgy$$

3. The attempt at a solution

Since the block is at rest before the bullet hit, if we use the momentum conservation formula, we only have to deal with the initial speed of the bullet. The resultant formula is

$$m_{b}v_{b}=(m_{w}+m_{b})*v_{2}$$

Once the bullet is embedded in the block, it will have a potential energy of zero and a kinetic energy of

$$K=1/2(m_{b}+m_{w})v_{2}^{2}$$

and the block with the bullet travels up a height .40m, and comes to a rest. At this point, the block/bullet unit has a kinetic energy of zero, and a potential energy of

$$U=(m_{b}+m_{w})gy$$

Using energy conservation we get:

$$1/2(m_{b}+m_{w})v_{2}^{2}=(m_{b}+m_{w})gy$$

we can solve for velocity here and get the speed after the bullet hit the block. The masses should cancel out, leaving:

$$v_{2}=\sqrt{2gy}$$

We sub in this expression for v back into the first momentum formula, getting:

$$m_{b}v_{b}=(m_{b}+m_{w})*\sqrt{2gy}$$

solving for the initial velocity $$v_{b}$$, we get $$v_{b}=(m_{b}+m_{w})/m_{b}*\sqrt{2gy}$$

At this point I just plugged and chugged, using the given values in the problem and came up with 423 m/s, but it turned out to be the wrong answer. Can anyone help me figure out what I did wrong? Much thanks in advance!

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2. Feb 16, 2008

Gear300

If the bullet lodged into the wood, you cant use conservation of Kinetic Energy. Rather than that...try using what you know about centripetal forces and acceleration.

3. Feb 16, 2008

Arejang

How could you use centripetal force if the pendulum isn't moving in a circle?

4. Feb 16, 2008

Staff: Mentor

Your solution looks good to me. Who says it's wrong?

5. Feb 16, 2008

Arejang

Our assignments are given online and it grades it the instant you input an answer. In this case, the answers are given in multiple choice, so I'm usually cautious to choose an answer as the chances to get it right are obviously limited to the amount of choices available. But I did choose the 423 m/s and it was shown as incorrect.

250 m/s
423 m/s
66.7 m/s
646 m/s
366 m/s

And yes, I felt the same way too, doc. I'm pretty sure I didn't fumble on any of the calculations and that I used the correct formula to derive the speed, unless "the initial speed of the bullet" and "the speed of the bullet just before it hits the block" mean totally different things.

6. Feb 16, 2008

Gear300

Wait...I see what you did there...nevermind...everything looks to be alright...dont know whats wrong with your answer.

Last edited: Feb 16, 2008
7. Feb 16, 2008

Staff: Mentor

That's exactly what Arejang did!

8. Feb 16, 2008

Staff: Mentor

I doubled checked your arithmetic; I would have chosen the same answer.

9. Feb 16, 2008

Arejang

I just guessed all the answers and the right one was 66.7 m/s......What the heck?

10. Feb 17, 2008

Arejang

Sorry, this problem has been irking me to no end. Would the length of the string, 2m, play any part of this problem? That would be the only factor I could see that might alter the answer somewhat. But if that were the case, then I don't know how else to approach this problem.

11. Feb 17, 2008

Staff: Mentor

Not that I can see. Your solution is perfectly correct.

12. Feb 17, 2008

Arejang

thanks, I'm going to go ahead and marked this solved for the time being. If I find anything else, I'll post it back up again. Thanks for all your help!

13. Feb 17, 2008

Staff: Mentor

If this homework is graded (or even if it's not), make sure your instructor sees your solution.

14. Feb 17, 2008

Arejang

I plan on doing so when I see him tomorrow.