Centre Of Mass Ques 74
- Two point masses $m _1$ and $m _2$ are connected by a spring of natural length $l _0$. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity $v _0$ along positive $x$-axis. When the system reaches the origin the string breaks
$(t=0)$. The position of the point mass $m _1$ is given by
$x _1=v _0 t-A(1-\cos \omega t)$ where $A$ and $\omega$ are constants.
Find the position of the second block as a function of time. Also, find the relation between $A$ and $l _0$.
$(2003,4$ M)
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Answer:
Correct Answer: 74.$x _2=v _0 t+\frac{m _1}{m _2} A(1-\cos \omega t), l _0=\frac{m _1}{m _2}+1 A$
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
- (a)
$$ v=10 m / s $$
$$ \begin{aligned} & x _1=v _0 t-A(1-\cos \omega t) \\ & x _{CM}=\frac{m _1 x _1+m _2 x _2}{m _1+m _2}=v _0 t \\ & \therefore \quad x _2=v _0 t+\frac{m _1}{m _2} A(1-\cos \omega t) \\ & a _1=\frac{d^{2} x _1}{d t^{2}}=-\omega^{2} A \cos \omega t \end{aligned} $$
The separation $x _2-x _1$ between the two blocks will be equal to $l _0$ when $a _1=0$
or $\cos \omega t=0$
$$ \begin{aligned} x _2-x _1 & =\frac{m _1}{m _2} A(1-\cos \omega t)+A(1-\cos \omega t) \\ \text { or } l _0 & =\frac{m _1}{m _2}+1 \quad A(\because \cos \omega t=0) \end{aligned} $$
Thus, the relation between $l _0$ and $A$ is,
$$ l _0=\frac{m _1}{m _2}+1 A $$
BETA
