A) \[4.5\sqrt{2}m{{s}^{-1}}\]
B) \[5\,\,m{{s}^{-1}}\]
C) \[5\sqrt{32}m{{s}^{-1}}\]
D) \[1.5\,\,m{{s}^{-1}}\]
Correct Answer: A
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 \[3m\times 1.5-m\times v\cos \theta =0\] ... (i) and \[3m\times 1.5-m\times v\sin \theta =0\] ... (ii) These give\[mv\cos \theta =mv\sin \theta \] 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 \[4.5\,\,m=mv\cos {{45}^{o}}=\frac{mv}{\sqrt{2}}\] \[\therefore \] \[v=4.5\sqrt{2}m{{s}^{-1}}\] The third piece will go with a velocity of \[4.5\sqrt{2}m{{s}^{-1}}\] 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 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 \[m{{v}_{3}}=\sqrt{{{(3m\times 1.5)}^{2}}+{{(3m\times 1.5)}^{2}}}\] or \[{{v}_{3}}=4.5\sqrt{2}m{{s}^{-1}}\]You need to login to perform this action.
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