question_answer

1 kg body explodes into three fragments. The ratio of their masses is 1:1:3. The fragments of same mass move perpendicular to each other with speeds 30 m/s, while the heavier part remains in the initial direction. The speed of heavier part is :

A)  \[\frac{10}{\sqrt{2}}\] m/s

B)  \[10\sqrt{2}\]m/s

C)  \[20\sqrt{2}\]m/s     

D)  \[30\sqrt{2}\] m/s

Correct Answer: B

Solution :

Key Idea: Equate the momenta of the system along two perpendicular axes. Let u be the velocity arid Q the direction of the third piece as shown.                 Equating the momenta of the system along OA and OB to zero, we get                 \[m\times 30-3m\times v\cos \theta =0\] ... (i) and        \[m\times 30-3m\times v\sin \theta =0\] ... (ii) These give \[3\,mv\cos \theta =3\,mv\sin \theta \]                 or \[\cos \theta =\sin \theta \] Thus, \[\angle AOC=\angle BOC={{180}^{o}}C-{{45}^{o}}={{135}^{o}}C\] Putting the value of \[\theta \] in Eq. (i), we get                 \[30\,m=3mv\cos {{45}^{o}}=\frac{3mv}{\sqrt{2}}\] \[\therefore \] \[v=10\sqrt{2}\,m/s\] The third piece will move with a velocity of\[10\sqrt{2}\]m/s in a direction making an angle of \[{{135}^{o}}\] with either piece. Alternative: Key Idea: The square of momentum of third piece is equal to sum of squares of momentum of first and second piece. As from key idea,                 \[{{p}_{3}}^{2}={{p}_{1}}^{2}+{{p}_{2}}^{2}\] or            \[{{p}_{3}}=\sqrt{{{p}_{1}}^{2}+{{p}_{2}}^{2}}\] or            \[3\,m{{v}_{3}}=\sqrt{{{(m\times 30)}^{2}}+{{(m\times 30)}^{2}}}\] or \[{{v}_{3}}=\frac{30\sqrt{2}}{3}=10\sqrt{2}\,m/s\]