A Ballistic Pendulum

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[SOLVED] A Ballistic Pendulum

Homework Statement



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.


Homework Equations



Momentum Conservation:

[tex]m_{a}v_{a1}+m_{b}v_{b1}=(m_{a}+m_{b})*v_{2}[/tex]

Energy Conservation:

[tex]1/2mv^{2}=mgy[/tex]

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

[tex]m_{b}v_{b}=(m_{w}+m_{b})*v_{2}[/tex]

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

[tex]K=1/2(m_{b}+m_{w})v_{2}^{2}[/tex]

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

[tex]U=(m_{b}+m_{w})gy[/tex]

Using energy conservation we get:

[tex]1/2(m_{b}+m_{w})v_{2}^{2}=(m_{b}+m_{w})gy[/tex]

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

[tex]v_{2}=\sqrt{2gy}[/tex]

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

[tex]m_{b}v_{b}=(m_{b}+m_{w})*\sqrt{2gy}[/tex]

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

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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Answers and Replies

  • #2
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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
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How could you use centripetal force if the pendulum isn't moving in a circle?
 
  • #4
Doc Al
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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!
Your solution looks good to me. Who says it's wrong?
 
  • #5
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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.

Here are my answer choices:

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
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Wait...I see what you did there...nevermind...everything looks to be alright...dont know whats wrong with your answer.
 
Last edited:
  • #7
Doc Al
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For inelastic collisions, energy is lost in the collision due to deformation. Because of that, Ki = Pf + Ediss...To avoid dealing with the dissipated energy, the best thing to do would be to use the conservation of momentum. What you do is set Pf = Ka, in which Ka is the energy right after collision (because gravitational energy is interconvertable with kinetic energy). So it would be (1/2)*(Ma + Mb)*(v2)^2 = (Ma+Mb)gh. Solve for v2, plug it into your momentum equation, and then solve for v1.
That's exactly what Arejang did!
 
  • #8
Doc Al
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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.
I doubled checked your arithmetic; I would have chosen the same answer.
 
  • #9
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I just guessed all the answers and the right one was 66.7 m/s......What the heck? o_O
 
  • #10
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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
Doc Al
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Would the length of the string, 2m, play any part of this problem?
Not that I can see. Your solution is perfectly correct.
 
  • #12
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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
Doc Al
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If this homework is graded (or even if it's not), make sure your instructor sees your solution.
 
  • #14
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I plan on doing so when I see him tomorrow.
 

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