Centre Of Mass Ques 1
- A body of mass $1 \mathrm{~kg}$ falls freely from a height of $100 \mathrm{~m}$ on a platform of mass $3 \mathrm{~kg}$ which is mounted on a spring having spring constant $k=1.25 \times 10^6 \mathrm{~N} / \mathrm{m}$. The body sticks to the platform and the spring’s maximum compression is found to be $x$. Given that $g=10 \mathrm{~ms}^{-2}$, the value of $x$ will be close to
(JEE Main 2019, 11 Jan Shift I)
(a) $8 \mathrm{~cm}$
(b) $4 \mathrm{~cm}$
(c) $40 \mathrm{~cm}$
(d) $80 \mathrm{~cm}$
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Answer:
Correct Answer: 1.( * )
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
1 Initial compression of the spring,
$ \begin{aligned} m g & =k\left(\frac{x_0}{100}\right) \quad\left(x_0 \text { in } \mathrm{cm}\right) \\ \Rightarrow \quad x_0 & =\frac{3 \times 10 \times 100}{1.25 \times 10^6}=\frac{3}{1250} \end{aligned} $
Which is very small and can be neglected. Applying conservation of momentum before and after the collision i.e., momentum before collision $=$ momentum after collision.
$m \times \sqrt{2 g h}=(m+M) v(\because$ velocity of the block just before the collision is
$ \begin{aligned} v^2-0^2 & =2 g h \\ v & =\sqrt{2 g h}) \end{aligned} $
After, substituting the given values, we get
$ 1 \times \sqrt{2 \times 10 \times 100}=4 v \text { or } 4 v=20 \sqrt{5} $
so $ v=5 \sqrt{5} \mathrm{~m} / \mathrm{s} $
Let this be the maximum velocity, then for the given system, using
$ \begin{aligned} & \frac{1}{2} m v^2=\frac{1}{2} k x^2 \\ & \therefore \quad \frac{1}{2} \times 4 \times 125=\frac{1}{2} \times 1.25 \times 10^6 \times\left(\frac{x}{100}\right)^2 \\ & \Rightarrow \quad 4=10^4 \times \frac{x^2}{10^4} \text { or } x=2 \mathrm{~cm} \\ & \end{aligned} $
$\therefore$ No option given is correct.
BETA
