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A 0.0137-kg bullet is fired straight up at a falling wooden block that has a mass of 4.08 kg. The bullet has a speed of 795 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

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Answer:

t = 0.1360 s

Explanation:

Using the principles of conservation

[tex]m_1u_1+m_2u_2 = (m_1+m_2)v[/tex]

where

v = common velocity

replacing [tex]m_1[/tex] = 0.0137 - kg , [tex]u_1[/tex] = 795 m/s , [tex]m_2[/tex] = 4.08 kg , v = [tex]u_2[/tex] ; we have:

(0.0137)(795) - (4.08) ( v) = (0.0137 + 4.08) v

10.8915 - 4.08 v = 4.0937 v

10.8915 = 4.0937 v + 4.08 v

10.8915 = 8.1737 v

v = 1.3325 /s

But v = u + at ⇒ v = 0 + gt

v = gt

t = v/g

t = [tex]\frac{1.3225 \ m/s}{9.8\ m/s^2}[/tex]

t = 0.1360 s

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