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