A bomb projected from ground at an angle θθ≠90° with horizontal explodes into two fragments of equal mass at top most point of  its trajectory. If one of the fragment returns to point of projection then ratio of  de-Broglie wavelength of second fragment just after explosion to bomb just before explosion is

# A bomb projected from ground at an angle $\text{θ}\left(\text{θ}\ne 90°\right)$ with horizontal explodes into two fragments of equal mass at top most point of  its trajectory. If one of the fragment returns to point of projection then ratio of  de-Broglie wavelength of second fragment just after explosion to bomb just before explosion is

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### Solution:

From the law of conservation of momentum

$\begin{array}{l}mu\mathrm{cos}\theta =-\frac{m}{2}u\mathrm{cos}\theta +\frac{m}{2}{v}_{1}\\ {v}_{1}=3u\mathrm{cos}\theta \dots \left(1\right)\\ {\lambda }_{D}=\frac{h}{mv}\dots \left(2\right)\end{array}$

The ratio of De-Broglie wavelength

$⇒\frac{{\left({\lambda }_{D}\right)}_{2}}{\left({\lambda }_{D}\right)}=\frac{\frac{h}{\frac{m}{2}\left(3u\mathrm{cos}\text{θ}\right)}}{\frac{h}{mu\mathrm{cos}\text{θ}}}=\frac{2}{3}$

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