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Two spherical bodies of masses $M$ and $5 M$ and radii $R$ and $2 R$ respectively are released in free space with initial separation between their centres equal to $12 R$. If they attract each other due to gravitational force only,then the distance covered by the smaller body just before collision, is
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Verified Answer
The correct answer is:
7.5R
Let at $O$ there will be a collision. If smaller sphere moves $x$ distance to reach at $O$, then bigger sphere will move a distance of $(9 R-x)$.
$\begin{aligned}
F & =\frac{G M \times 5 M}{(12 R-x)^2} \\
a_{\text {small }} & =\frac{F}{M}=\frac{G \times 5 M}{(12 R-x)^2} \\
a_{\mathrm{big}} & =\frac{F}{5 M}=\frac{G M}{(12 R-x)^2} \\
x & =\frac{1}{2} a_{\text {small }} t^2=\frac{1}{2} \frac{G \times 5 M}{(12 R-x)^2} t^2 \ldots(i)\\
(9 R-x) & =\frac{1}{2} a_{\text {big }} t^2=\frac{1}{2} \frac{G M}{(12 R-x)^2} t^2\ldots(ii)
\end{aligned}$
Thus, dividing Eq. (i) by Eq. (ii), we get
$\frac{x}{9 R-x}=5$
$\begin{aligned}
& \Rightarrow \quad x=45 R-5 x \\
& \Rightarrow \quad 6 x=45 R \\
& \therefore \quad x=7.5 R \\
&
\end{aligned}$
$\begin{aligned}
F & =\frac{G M \times 5 M}{(12 R-x)^2} \\
a_{\text {small }} & =\frac{F}{M}=\frac{G \times 5 M}{(12 R-x)^2} \\
a_{\mathrm{big}} & =\frac{F}{5 M}=\frac{G M}{(12 R-x)^2} \\
x & =\frac{1}{2} a_{\text {small }} t^2=\frac{1}{2} \frac{G \times 5 M}{(12 R-x)^2} t^2 \ldots(i)\\
(9 R-x) & =\frac{1}{2} a_{\text {big }} t^2=\frac{1}{2} \frac{G M}{(12 R-x)^2} t^2\ldots(ii)
\end{aligned}$
Thus, dividing Eq. (i) by Eq. (ii), we get
$\frac{x}{9 R-x}=5$
$\begin{aligned}
& \Rightarrow \quad x=45 R-5 x \\
& \Rightarrow \quad 6 x=45 R \\
& \therefore \quad x=7.5 R \\
&
\end{aligned}$
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