Question

A piece of wood of mass $0.03kg$ is dropped from the top of a $100m$ height building. At the same time, a bullet of mass $0.02kg$ is fired vertically upward, with a velocity $100m{s}^{-1}$, 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 : $(g=10m{s}^{-2})$.

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

Answer 1

Answer: (C)

$40m$

Time taken for the particles to collide,
$t=\frac{d}{V_{rel}}=\frac{100}{100}=1sec$
Speed of wood just before collision $=gt=10m/s$ and speed of bullet just before collision $v-gt=100-10=90m/s$
Now, conservation of linear momentum just before and after the collision -
$-\left(0.02\right)\left(1v\right)+\left(0.02\right)\left(9v\right)=\left(0.05\right)v$
$\Rightarrow 150=5v$
$\Rightarrow v=30m/s$
Max. height reached by body $h=\frac{v^2}{2g}$
Before : $0.03kg\downarrow 10m/s$
$0.02kg\uparrow 90m/s$
After : $v0.05kg$
$h=\frac{30\times 30}{2\times 10}=40m$.