- #1

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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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- #1

- 87

- 0

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?

- #2

Doc Al

Mentor

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Solve it in two steps: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?

(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.)

Huh? Conservation of momentum: [itex]m_1v_1 + m_2v_2 = (m_1 + m_2)v_f[/itex]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

- #3

- 87

- 0

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

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

Last edited:

- #4

- 87

- 0

K, well i lied...i thought i had it but i didnt lol. Can someone help a bit further with this?

- #5

- 298

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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

- 87

- 0

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...

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...

Last edited:

- #7

- 10

- 0

that looks rightFormat 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...

you can figure out the x displacement with pythagorean theorem, maybe drawing a diagram it will help

- #8

- 87

- 0

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

Mentor

- 45,093

- 1,397

You plugged in the wrong speed. You used 1.44 instead of 1.144.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...

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