Posted by Confused on Monday, December 5, 2011 at 5:02pm.
I 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.
1) 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?
oops its 1 kg
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