A) \[\frac{2}{3}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{3}\]
D) \[2\]
Correct Answer: D
Solution :
[d] From question, \[{{m}_{A}}=M;m=\frac{m}{2}\] \[{{u}_{A}}=V\text{ }{{u}_{B}}=0\] Let after collision velocity of \[A={{V}_{1}}\] and velocity of \[B={{V}_{2}}\] Applying law of conservation of momentum, \[mu=m{{v}_{1}}+\left( \frac{m}{2} \right){{v}_{2}}\text{ or, 2}\mu \text{=2}{{v}_{1}}\text{ +}{{\text{v}}_{2}}\] ?.(i) By law of collision \[e=\frac{{{v}_{2}}-{{v}_{1}}}{u-1}\text{ or }u={{v}_{2}}-{{v}_{1}}\] ?(ii) \[\left[ \because \text{collision is elastic, }e=1 \right]\] using eqns (i) and (11) \[{{v}_{1}}=\frac{1}{3}\mu \text{ and }{{v}_{2}}=\frac{4}{3}u\] De-Broglie wavelength \[\lambda =\frac{h}{p}\] \[\therefore \frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{{{P}_{B}}}{{{P}_{A}}}=\frac{\frac{m}{2}\times \frac{4}{3}u}{m\times \frac{1}{3}}=2\]
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