question_answer

A stationary body of mass m explodes into the  three parts having masses in the ratio 1:3:3. The two fractions with equal masses move at right angles to each other with a velocity of \[1.5\,m{{s}^{-1}}.\] The velocity of the third part is:

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 \[{{v}_{3}}=4.5\sqrt{2}\,m{{s}^{-1}}\]