Centre of Mass 3 Question 6
9. 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 (Take, $g=10 ms^{-2}$ )
(2019 Main, 10 Jan I)
(a) $20 m$
(b) $30 m$
(c) $10 m$
(d) $40 m$
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Answer:
Correct Answer: 9. (d)
Solution:
- Key Idea As bullet gets embedded in the block of wood so, it represents a collision which is perfectly inelastic and hence only momentum of the system is conserved.
Velocity of bullet is very high compared to velocity of wooden block so, in order to calculate time for collision, we take relative velocity nearly equal to velocity of bullet.
So, time taken for particles to collide is
$$ t=\frac{d}{v _{\text {rel }}}=\frac{100}{100}=1 s $$
Speed of block just before collision is;
$$ v _1=g t=10 \times 1=10 ms^{-1} $$
Speed of bullet just before collision is
$$ \begin{aligned} v _2 & =u-g t \\ & =100-10 \times 1=90 ms^{-1} \end{aligned} $$
Let $v=$ velocity of bullet + block system, then by conservation of linear momentum, we get
$$ \begin{aligned} -(0.03 \times 10)+(0.02 \times 90) & =(0.05) v \\ \Rightarrow \quad v & =30 ms^{-1} \end{aligned} $$
Now, maximum height reached by bullet and block is
$$ \begin{array}{rlrl} & h & =\frac{v^{2}}{2 g} \Rightarrow h=\frac{30 \times 30}{2 \times 10} \\ \Rightarrow \quad h & =45 m \end{array} $$
$\therefore$ Height covered by the system from point of collision $=45$ $m$
Now, distance covered by bullet before collision in $1 sec$.
$$ =100 \times 1-\frac{1}{2} \times 10 \times 1^{2}=95 m $$
Distance of point of collision from the top of the building $=100-95=5 m$
$\therefore$ Maximum height to which the combined system reaches above the top of the building before falling below $=45-5=40 m$
