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 and \[\theta \] the direction of the third piece as shown.
Equating the momenta of the system along OA and OB to zero, we get \[m\times 30-3\pi \times v\cos \theta =0\] ...(i) and \[m\times 30-3m\times v\sin \theta =0....(ii)\] These give \[3mv\cos \theta =3mv\sin \theta \] or \[\cos \theta =\sin \theta \] \[\therefore \theta ={{45}^{o}}\] Thus, \[\angle AOC=\angle BOC={{180}^{o}}-{{45}^{o}}={{135}^{o}}\] Putting the value of \[\theta \]in Eq. (i), we get \[30\,m=3\,mv\cos {{45}^{o}}=\frac{3mv}{\sqrt{2}}\] \[\therefore v=10\sqrt{2}\,m/s\] The third piece will go with a velocity of \[10\sqrt{2}\]m/s in a direction making an angel of \[{{135}^{\text{o}}}\] with either piece. Alternative: Key Idea: The square of momentum of third piece is equal to sum of squares of momentum first and second pieces. 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 \[{{v}_{3}}=\frac{30\sqrt{2}}{3}=10\sqrt{2}\,m/s\]
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