Pendulum Question

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How would you solve for the horizontal displacement of a 3.6 kg pendulum that hangs from a string 2.8 meters high, which is shot (and embeded) by a 0.018 kg bullet traveling at a velocity of 230 m/s?

I solved for the velocity of the objects together...
mv + mv = mv + mv
18(230) + 3.6(0) = 21.6v
v = 191.7 m/s

If the pendulum swings upward, how do i solve for the horizontal displacement?
 

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  • #2
Doc Al
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Format said:
How would you solve for the horizontal displacement of a 3.6 kg pendulum that hangs from a string 2.8 meters high, which is shot (and embeded) by a 0.018 kg bullet traveling at a velocity of 230 m/s?
Solve it in two steps:
(1) Apply conservation of momentum for the collision
(2) Apply conservation of energy after the collision to see how high it swings. (Then use trig to find the horizontal displacement.)

I solved for the velocity of the objects together...
mv + mv = mv + mv
18(230) + 3.6(0) = 21.6v
v = 191.7 m/s
Huh? Conservation of momentum: [itex]m_1v_1 + m_2v_2 = (m_1 + m_2)v_f[/itex]
 
  • #3
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k ill give that a try, thx.

And the 21.6 is just the 2 masses added, didnt show all my work lol.


Edit: Well theres the problem lol...i put 18-g not 0.018-kg. I got it, thx
 
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  • #4
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K, well i lied...i thought i had it but i didnt lol. Can someone help a bit further with this?
 
  • #5
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Initially, the bullet and pendulum have some momentum (the pendulums happens to be zero). Because of conservation of momentum, the momentum of the pendulum and bullet combination will be the same as the initial momentum of the bullet. That way you can get your velocity for the pendulum+bullet.

Next, mechanical energy must be conserved after that. So find your initial kinetic energy. The bob will keep going until all the kinetic energy has been converted into potential energy. See if you can take it from there.
 
  • #6
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K well this is what i've got...

mv + mv = v(m + m)
0.018(230) + 3.6(0) = v(3.618)
v = 1.144 m/s <-- This is the velocity of both the objects together


Then i did...

1/2mv² + mgh = 1/2mv² + mgh <-- masses cancel out
1/2(1.44)² + 0 = 0 + (9.8)h <-- h=0 initially and v=0 at top
h = 0.105 m

Im guessin there is something wrong with this part...
 
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  • #7
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Format said:
K well this is what i've got...

mv + mv = v(m + m)
0.018(230) + 3.6(0) = v(3.618)
v = 1.144 m/s <-- This is the velocity of both the objects together


Then i did...

1/2mv² + mgh = 1/2mv² + mgh <-- masses cancel out
1/2(1.44)² + 0 = 0 + (9.8)h <-- h=0 initially and v=0 at top
h = 0.105 m

Im guessin there is something wrong with this part...
that looks right
you can figure out the x displacement with pythagorean theorem, maybe drawing a diagram it will help
 
  • #8
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I did that, but it doesnt give me the correct answer :grumpy:

2.8² - 0.105² = x²

x = 2.7

The answer says 0.61....maybe it was a typo, but thats pretty far off.
 
  • #9
Doc Al
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Typo: wrong speed

Format said:
1/2mv² + mgh = 1/2mv² + mgh <-- masses cancel out
1/2(1.44)² + 0 = 0 + (9.8)h <-- h=0 initially and v=0 at top
h = 0.105 m

Im guessin there is something wrong with this part...
You plugged in the wrong speed. You used 1.44 instead of 1.144.
 

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