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A bomb is projected vertically up with a velocity of $100 m/s.$ After $8\; s$ it blasted into two pieces of mass in the ratio $1:2$ . The lighter piece goes it with the velocity $40 \;m/s$ then what is the velocity of the heavier piece just after blasting?

\[\begin {array} {1 1} (a)\;10\;m/s\;up & \quad (b)\;10\;m/s\;down \\ (c)\;20\;m/s & \quad  (d)\;20 m/s\;down \end {array}\]

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Velocity of bomb after 8 seconds
$V= u-gt$
$\quad=100-80$
$\quad= 20 m/s$
By momentum conservation
$m \times 40 +2mv=20 \times 3m$
$40+ 2v=60$
$2v=+20$
$v= +10 m/s$
The correct answer is $v=+10\;m/s\;upwards$
answered Nov 22, 2013 by meena.p
edited Jun 17, 2014 by lmohan717
 

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