question_answer

A vessel at rest explodes into three pieces. Two pieces having equal masses fly off perpendicular to one another with the same velocity 30 meter per second. The third piece has three times mass of each of other piece. The magnitude and direction of the velocity of the third piece will be       [AMU (Engg.) 1999]

A)             \[10\sqrt{2}\,m/second\] and \[135{}^\circ \] from either

B)             \[10\sqrt{2}\,m/second\] and \[45{}^\circ \] from either

C)             \[\frac{10}{\sqrt{2}}\,m/second\] and \[135{}^\circ \] from either

D)             \[\frac{10}{\sqrt{2}}\,m/second\] and \[45{}^\circ \] from either

Correct Answer: A

Solution :

                Let two pieces are having equal mass m and third piece have a mass of 3m. According to law of conservation of linear momentum. Since the initial momentum of the system was zero, therefore final momentum of the system must be zero i.e. the resultant of momentum of two pieces must be equal to the momentum of third piece. We know that if two particle possesses same momentum and angle in between them is 90° then resultant will be given by \[P\sqrt{2}=mv\sqrt{2}=m30\sqrt{2}\]             Let the velocity of mass 3m is V. So \[3mV=30m\sqrt{2}\]             \ \[V=10\sqrt{2}\] and angle 135° from either.                         (as it is clear from the figure)