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A 0.026 kg bullet is fired straight up at a falling wooden block that has a mass of 5.0 kg. The bullet has a speed of 750 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 occured. 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.

Respuesta :

Answer:

0.198 s

Explanation:

Consider the motion of the block  before collision

[tex]v_{o}[/tex] = initial velocity of block as it is dropped = 0 m/s

[tex]a[/tex] = acceleration = - g

[tex]t[/tex] = time of travel

[tex]v_{f}[/tex] = final velocity of block before collision

Using the kinematics equation

[tex]v_{f} = v_{o} + at \\v_{f} = 0 + (-g)t\\v_{f} = - gt[/tex]

[tex]m[/tex] = mass of the bullet = 0.026 kg

[tex]v_{b}[/tex] = velocity of block just before collision = 750 m/s

[tex]M[/tex] = mass of the block = 5 kg

[tex]V[/tex] = final velocity of bullet block after collision = gt

Using conservation of momentum

[tex]m v_{b} + Mv_{f} = (m + M) V\\(0.026) (750) + (5) (- gt) = (0.026 + 5) (gt)\\19.5 - 49 t = 49.2548 t\\t = 0.198 s[/tex]

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