Question

A 0.0135-kg bullet is fired straight up at a falling wooden block that has a mass of 4.02 kg. The bullet has a speed of 809 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.

Answer 1

Answer on Question #75464-Physics-Other

A 0.0135-kg bullet is fired straight up at a falling wooden block that has a mass of $4.02\mathrm{kg}$. The bullet has a speed of $809~\mathrm{m/s}$ 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.

Solution

Before collision:

$$M g H = \frac {1}{2} M v _ {1} ^ {2} \rightarrow g H = \frac {1}{2} v _ {1} ^ {2}$$

The collision:

$$m v _ {0} - M v _ {1} = (m + M) v _ {2}.$$

After collision:

$$(M + m) g H = \frac {1}{2} (M + m) v _ {2} ^ {2} \rightarrow g H = \frac {1}{2} v _ {2} ^ {2}$$

So,

$$v _ {1} = v _ {2}$$

Thus,

$$m v _ {0} - M v _ {1} = (m + M) v _ {1}.$$ $$v _ {1} = \frac {m}{m + 2 M} v _ {0}$$

The time is

$$t = \frac {v _ {1}}{g} = \frac {m}{m + 2 M} \frac {v _ {0}}{g} = \frac {0 . 0 1 3 5}{0 . 0 1 3 5 + 2 (4 . 0 2)} \frac {8 0 9}{9 . 8 1} = 0. 1 4 s.$$

Answer: 0.14 s.

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