Solving Momentum and Impulse Homework Problem

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SUMMARY

The discussion focuses on a physics problem involving momentum and impulse related to a shell launched at an angle of 55 degrees with an initial velocity of 150 m/s. Upon reaching its highest point, the shell explodes into two fragments: M_1 (9 kg) and M_2 (3 kg). The heavier fragment lands back at the launch point, while the lighter fragment's landing position is determined by the conservation of momentum, leading to the conclusion that M_1's speed is three times that of M_2.

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Homework Statement



A 12 kg shell is launched at an angle of 55 degrees above horizontal with v_O = 150 m/s. When at highest point, it explodes into two fragments, M_1 with mass 9 kg and M_2 with mass 3 kg. The fragments reach the ground at the same time. The heavier fragment lands back at the same point - where does the lighter fragment land?


The Attempt at a Solution



Ok, I know that the speed at the top is V_x = 86 m/s. For the heavier fragment to land at the same point, it must have the same v_initial (150 m/s) but opposite direction, so v_x must be changed -2 times (so it's -86 m/s) and v_y = -122 m/s.
THen I say that momentum of single fragment before explosion equals total momentum of the two fragments - but I get that the speed of the lighter fragment is very high!

Where am I wrong?
 
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I got it.. m_1 = 3*m_2 and v_2 = 3*v_1
 

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