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:                               [AIPMT 2001]

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