Answer:
Suppose the explosion of the rock occurs at O. One piece of mass 100 kg goes along go off, other piece of mass 200 kg goes along \[OB\bot OA\]. Let the third piece of mass m fly off at a speed of \[25\,m/s\]along OC, at \[\angle \theta \]with OA, Fig. 3 (HT). 14. Applying principle of conservation of linear momentum along OA \[100\times 12-m\upsilon \cos \theta =0\] \[1200=m\upsilon \cos \theta =25m\,\cos \theta \] \[\therefore \] \[m\cos \theta =\frac{1200}{25}=48\] .... (i) Again, applying principle of conservation of linear momentum along OB \[200\times 8-m\upsilon \sin \theta =0\] \[1600=m\upsilon \sin \theta =25\,m\,\sin \theta \] \[\therefore \] \[m\,\sin \theta =\frac{1600}{25}=64\] .... (ii) Square (i) and (ii),attd add \[{{m}^{2}}={{48}^{2}}+{{64}^{2}}\] m = 80 kg Divide (ii) by (i), \[\tan \theta =\frac{64}{48}=\frac{4}{3}\] \[\theta ={{\tan }^{-1}}\left( 4/3 \right)={{53}^{o}}\]
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