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The masses of blocks $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{m}$ and $\mathrm{M}$ respectively. Between $\mathrm{A}$ and $\mathrm{B}$, there is a constant frictional force $\mathrm{F}$ and $\mathrm{B}$ can slide on a smooth horizontal surface. A is set in motion with velocity while $\mathrm{B}$ is at rest. What is the distance moved by A relative to B before they move with the same velocity?
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Verified Answer
The correct answer is:
$\frac{\mathrm{mMv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{M}+\mathrm{m})}$
For the blocks A and B FBD as shown below
Equations of motion
$$
\begin{array}{l}
\mathrm{a}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }-\mathrm{x} \text { direction }) \\
\mathrm{a}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }+\mathrm{x} \text { direction) }
\end{array}
$$
Relative acceleration, of A w.r.t. B,
$$
\begin{aligned}
\mathrm{a}_{\mathrm{A}, \mathrm{B}} &=\mathrm{a}_{\mathrm{A}}-\mathrm{a}_{\mathrm{B}}=-\frac{\mathrm{F}}{\mathrm{m}}-\frac{\mathrm{F}}{\mathrm{M}} \\
&=-\mathrm{F}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mm}}\right)(\text { along }-\mathrm{x} \text { direction })
\end{aligned}
$$
Initial relative velocity of A w.r.t. B, $\mathrm{u}_{\mathrm{AB}}=\mathrm{v}_{0}$
using equation $v^{2}=u^{2}+2$ as
$$
0=v_{0}^{2}-\frac{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M}) \mathrm{S}}{\mathrm{Mm}} \Rightarrow \mathrm{S}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}
$$
i.e., Distance moved by A relative to B
$$
\mathrm{S}_{\mathrm{AB}}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}
$$
Equations of motion
$$
\begin{array}{l}
\mathrm{a}_{\mathrm{A}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }-\mathrm{x} \text { direction }) \\
\mathrm{a}_{\mathrm{B}}=\frac{\mathrm{F}}{\mathrm{M}}(\text { in }+\mathrm{x} \text { direction) }
\end{array}
$$
Relative acceleration, of A w.r.t. B,
$$
\begin{aligned}
\mathrm{a}_{\mathrm{A}, \mathrm{B}} &=\mathrm{a}_{\mathrm{A}}-\mathrm{a}_{\mathrm{B}}=-\frac{\mathrm{F}}{\mathrm{m}}-\frac{\mathrm{F}}{\mathrm{M}} \\
&=-\mathrm{F}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{Mm}}\right)(\text { along }-\mathrm{x} \text { direction })
\end{aligned}
$$
Initial relative velocity of A w.r.t. B, $\mathrm{u}_{\mathrm{AB}}=\mathrm{v}_{0}$
using equation $v^{2}=u^{2}+2$ as
$$
0=v_{0}^{2}-\frac{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M}) \mathrm{S}}{\mathrm{Mm}} \Rightarrow \mathrm{S}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}
$$
i.e., Distance moved by A relative to B
$$
\mathrm{S}_{\mathrm{AB}}=\frac{\mathrm{Mmv}_{0}^{2}}{2 \mathrm{~F}(\mathrm{~m}+\mathrm{M})}
$$
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