Conservation of Linear Momentum of exploding object

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The discussion revolves around a 3.0-kilogram object that explodes into three fragments, with two fragments' masses and velocities provided. The first fragment, weighing 0.50 kg, moves at 2.8 m/s along the negative x-axis, while the second fragment, weighing 1.3 kg, moves at 1.5 m/s along the negative y-axis. To find the speed and direction of the third fragment, conservation of momentum is applied, noting that the initial momentum is zero, thus the final momentum must also equal zero. The third fragment's momentum must counterbalance the first two fragments, resulting in negative x and y components. This analysis leads to the conclusion that the third fragment's velocity is determined by the vector sum of the first two fragments' momenta.
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A 3.0-kilogram object initially at rest explodes and splits into three fragments. One fragment has a mass of 0.50 kg and flies off along the negative x-axis at a speed of 2.8 m/s, and another has a mass of 1.3 kg and flies off along the negative y-axis at a speed of 1.5 m/s. What are the speed and direction of the third fragment?
 
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I've been trying to figure this problem out for the past hour can anyone please help me?!
 
Consider the horizontal and vertical components of the linear momentum separately (since these components are perpendicular, they can be analyzed independently).
 
That's what i thought but i don't know how to incorporate that into the equation mv = mv0
 
The initial linear momentum is 0. Thus the final linear momentum is also 0, because momentum is conserved.
 
The total momentum of the three vectors need to be zero. This means that the third vector must be the negative of the resultant of the first two. Which means that the third vector has the negative x and y-components of the first two.
 
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