Solving Momentum and Impulse Homework Problem

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The problem involves a 12 kg shell launched at 150 m/s at a 55-degree angle, which explodes into two fragments at its highest point. The heavier fragment (9 kg) lands back at the launch point, while the lighter fragment (3 kg) must be analyzed for its landing position. The initial horizontal speed at the top is calculated as 86 m/s, and the heavier fragment's velocity is reversed to maintain momentum. The user realizes that the momentum conservation equation indicates a relationship between the masses and velocities of the fragments, leading to an unexpectedly high speed for the lighter fragment. The key takeaway is correctly applying momentum conservation to determine the landing position of the lighter fragment.
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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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