A) \[\frac{10}{\sqrt{2}}\text{m/s}\]
B) \[\text{10}\sqrt{2\,}\text{m/s}\]
C) \[\text{20}\sqrt{2\,}\text{m/s}\]
D) \[\text{30}\sqrt{2\,}\text{m/s}\]
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 \[m\times 30-3m\times v\cos \theta =0\] ?(i) and \[m\times 30-3m\times v\sin \theta =0\] ?(ii) These give \[3\,mv\,\cos \theta =3mv\sin \theta \] 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 \[30\,m=3mv\cos {{45}^{o}}=\frac{3mv}{\sqrt{2}}\] \[\therefore \] \[v=10\sqrt{2}\,m/s\] 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 \[3m{{v}_{3}}=\sqrt{{{(m\times 30)}^{2}}+{{(m\times 30)}^{2}}}\] or \[{{v}_{3}}=\frac{30\sqrt{2}}{3}=10\sqrt{2}\,m/s\]You need to login to perform this action.
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