a) $(1+ \sqrt 3)P$ in direction $120 ^{\circ}$ with that of A

b) $(1+ \sqrt 3)P$ in direction $150 ^{\circ}$ with that of A

c) $2P$ in direction $150 ^{\circ}$ with that of A

d) $2P$ in direction $150 ^{\circ}$ with that of B

a) $(1+ \sqrt 3)P$ in direction $120 ^{\circ}$ with that of A

b) $(1+ \sqrt 3)P$ in direction $150 ^{\circ}$ with that of A

c) $2P$ in direction $150 ^{\circ}$ with that of A

d) $2P$ in direction $150 ^{\circ}$ with that of B

using conservation of momentum

momentum of A+momentum of B+momentum of C = 0

Therefore momentum of C=-(momentum of A+momentum of B)

$|A+B|=\bigg(\sqrt {(1^2+3}\bigg)P$

$\qquad\quad=\sqrt {4} P=2P$

$\tan \theta=\sqrt 3 =>\theta=60 ^{\circ}$

Making an angle $60 ^{\circ}$ with $\overrightarrow i $ axis

$-(\bar A+\bar B)=C$

makes an angle of $(90 +60)$ from B. ie $150 ^{\circ}$ with that of B

Hence d is the correct answer.

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