question_answer

A shell is fired from a cannon with velocity v m/sec at an angle \[\theta \] with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is             [IIT 1984; RPET 1999, 2001; UPSEAT 2002]

A)             \[3v\cos \theta \]

B)               \[2v\cos \theta \]

C)             \[\frac{3}{2}v\cos \theta \]

D)               \[\frac{\sqrt{3}}{2}v\cos \theta \]

Correct Answer: A

Solution :

                Shell is fired with velocity v at an angle q with the horizontal.             So its velocity at the highest point             = horizontal component of velocity =\[v\cos \theta \]             So momentum of shell before explosion = \[mv\cos \theta \] When it breaks into two equal pieces and one piece retrace its path to the canon, then other part move with velocity V. So momentum of two pieces after explosion                 \[=\frac{m}{2}(-v\cos \theta )+\frac{m}{2}V\]             By the law of conservation of momentum             \[mv\cos \theta =\frac{-m}{2}v\cos \theta +\frac{m}{2}V\]Þ \[V=3v\cos \theta \]