# Q. A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity $100 \; ms^{-1}$, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is : $(g =10ms^{-2})$

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Solution:

## Time taken for the particles to collide, $t = \frac{d}{V_{rel }} = \frac{100}{100} = 1 \sec$ Speed of wood just before collision = gt = 10 m/s & speed of bullet just before collision v-gt = 100 - 10 = 90 m/s Now, conservation of linear momentum just before and after the collision - -(0.02) (1v) + (0.02) (9v) = (0.05)v $\Rightarrow$ 150 = 5v $\Rightarrow$ v = 30 m/s Max. height reached by body $h = \frac{v^2}{2g}$ $h = \frac{30 \times 30}{2 \times 10} = 45 m$ $\therefore$ Height above tower = 40 m

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