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July 26, 2014

Posted by **Confused** on Monday, December 5, 2011 at 5:02pm.

1) If the maximum height of the block is ,h, is 35 cm, how fast was the bullet originally moving?

(can you explain the steps i'm really confused)

- PHYSICS HELP MEEE SO CONFUSED -
**Anonymous**, Monday, December 5, 2011 at 5:08pmI think there's more to this question.

Are the cables there to provide resistance to motion of the block? If so, what parameters?

Is the bullet fired straight up?

conserve momentum as the moving mass changes.

- PHYSICS HELP MEEE SO CONFUSED -
**drwls**, Monday, December 5, 2011 at 5:34pm1) From the height H that the block+bullet rises, assuming conservation of energy for the process, calculate the initial velocity V' of the block after the bullet is embedded.

(m+M) V'^2/2 = (m+M) g H

V' = sqrt(2 g H) = 2.62 m/s

2) Momentum (but not mechanical energy) is conserved during the bullet-embedding process. If V is the initial bullet velocity,

mV = (m+M) V'

Solve for V.

m is the bullet mass and M is the block's mass.

Are you sure the block is lighter that the bullet?

- PHYSICS HELP MEEE SO CONFUSED -
**confused**, Monday, December 5, 2011 at 9:51pmoops its 1 kg

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