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 \[{{135}^{o}}\] 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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