A gun of mass M fires a bullet of mass m with a horizontal speed V. The gun is fitted with a concave mirror of focal length f facing tpwards the receding bullet. Find the speed of separation of the bullet and the image just after the gun was fired.
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Recoil velocity of gun $=\text{V}_{\text{g}}=\frac{\text{mV}}{\text{M}}$
At any time ‘t’, position of the bullet w.r.t. mirror $=\text{V}_{\text{t}}=\frac{\text{mV}}{\text{M}}{\text{t}}=\Big(1+\frac{\text{m}}{\text{M}}\Big)\text{Vt}$
For the mirror, $=\text{u}=-\Big(1+\frac{\text{m}}{\text{M}}\Big)\text{Vt}=\text{kVt}$
v = position of the image
From lens formula,
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\Rightarrow\frac{1}{\text{v}}=\frac{1}{-\text{f}}+\frac{1}{\text{kVt}}=\frac{1}{\text{kVt}}-\frac{1}{\text{f}}=\frac{\text{f}-\text{kVt}}{\text{kVtf}}$
Let $\Big(1+\frac{\text{m}}{\text{M}}=\text{k}\Big),$
So, $\text{v}=\frac{\text{kVtf}}{-\text{kVt}+\text{f}}=\Big(\frac{\text{kVtf}}{\text{f}-\text{kVt}}\Big)$
So, velocity of the image with respect to mirror will be,
$\text{v}_1=\frac{\text{dv}}{\text{dt}}=\frac{\text{v}}{\text{dt}}\Big[\frac{\text{kVtf}}{\text{f}-\text{kVt}}\Big]=\frac{(\text{f}-\text{kVt})\text{kVf}-\text{kVtf}(-\text{kV})}{(\text{f}-\text{kVt})^2}=\frac{\text{kVt}^2}{(\text{f}-\text{kVt})^2}$
Since, the mirror itself is moving at a speed of $\text{m}\frac{\text{V}}{\text{M}}$ and the object is moving at ‘V’, the velocity of separation between the image and object at any time ‘t’ will be,
$\text{v}_{\text{s}}=\text{V}+\frac{\text{mV}}{\text{M}}+\frac{\text{kVf}^2}{(\text{f}-\text{kVt})^2}$
When, t = 0 (just after the gun is fired),
$\text{v}_{\text{s}}=\text{V}+\frac{\text{mV}}{\text{M}}+\text{kV}=\text{V}+\frac{\text{m}}{\text{M}}\text{V}+\Big(1+\frac{\text{m}}{\text{M}}\Big)\text{V}=2\Big(1+\frac{\text{m}}{\text{M}}\Big)\text{V}$
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