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
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
$3v\cos \theta $
$2v\cos \theta $
$\frac{3}{2}v\cos \theta $
$\frac{{\sqrt 3 }}{2}v\cos \theta $
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
Shell is fired with velocity $v$ at an angle $\theta$ 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
$mv\cos \theta = \frac{{ - m}}{2}v\cos \theta + \frac{m}{2}V$==> $V = 3v\cos \theta $
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