question_answer

1 kg body explodes into three fragments. The ratio of their masses is 1:1:3. The fragments of same mass move perpendicular to each other with speeds 30 m/s, while the heavier part remains in the initial direction. The speed of heavier part is:

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            \[{{v}_{3}}=\frac{30\sqrt{2}}{3}=10\sqrt{2}\,m/s\]