Question

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

• A. 30 m
• B. 10 m
• C. 40 m
• D. 20 m

Answer 1

Answer: (C)

40 m

Answer 2

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 = $40\, m$