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 : $(g =10ms^{-2})$

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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