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# Conservation of momentum

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How fast must a 10.5g bullet be traveling when it strikes a ballistic pendulum that consist of a block of wood of mass 3.0kg that is suspended by a cord? The bullet gets embedded in the block. The block rises by .22m after the impact.

a. 143 m/s
b. 595 m/s/s
c. 595 m/s
d. 2.67 m/s

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Answer: c. 595 m/s
From the conservation of momentum
mv = (m+M) V

m= mass of bullet= 10.5 g= 0.0105 Kg
M= mass of wood= 3 Kg
(m+M)= 3.0105 Kg

But V= square root of (2gh)
g= 9.81 m/s^2
h= 0.22 m

Therefore V= 2.08 m/s

mv = (m+M) V
or v= (m+M) V/ m= 596 m/s