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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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An object initially at rest explodes into three fragments. A,B and C The momentum of A is P$ \overrightarrow i $ that of B is $\sqrt 3$ P $\overrightarrow j$ where P is positive number. The momentum of C into be

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

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1 Answer

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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.
answered Jul 5, 2013 by meena.p
edited May 26, 2014 by lmohan717

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