Question

You toss a 0.90kg M-80 into the sky with a velocity 3.9m/s i^+ 9.8m/s j^ at...

You toss a 0.90kg M-80 into the sky with a velocity 3.9m/s i^+ 9.8m/s j^ at t= 0s . At t= 0.60s the M-80 explodes into two unequal chunks of m1= 0.3kg and m2= 0.6kg . Ignore any air resistance.

What is the  x -component of the first chunk's velocity immediately after the explosion if the second chunk's velocity immediately after the explosion is −12m/s  i^+  33.8m/s  j^ ?

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Answer #1

apply momentum conservation

=  

0.9* ( 3.9 i +9.8 j ) = 0.3 *( ) + 0.6* ( -12 i +33.8 j )

0.3 *( ) = 0.9* ( 3.9 i +9.8 j ) - 0.6* ( -12 i +33.8 j )

= ( 0.9*3.9+0.6*12) i +( 0.9*9.8 -0.6*33.8)j

0.3 *( ) = 10.71 i   -11.46 j

= 3.57 i -3.82 j

x -component of the first chunk's velocity = 35.7 m/s answer  

in adition

y -component of the first chunk's velocity = -3.82 m/s

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