Assuming the 20m particle is initially moving to the right at 50 m/s, from the point of view of the particle it is stationary and the lab is moving to the left, so its initial momentum is zero. After the explosion the 16m particles is moving upward and to the left (that's how it would appear to go vertical in the lab reference frame), so the 4m particle must be going down and to the right. For the 16m particle, its velocity in the x-direction __relative to the zero momentum frame__ is -50, so its momentum in the x-direction is -50 x 16 = -800. Consequently the 4m particle must have momentum in the x-direction of +800, which means its velocity in the x-direction is 800/4 = 200. We still have unknown y-components of velocity for both, but we know the total KE = E. If we let Vp= vertical velocity of the 16m particle and Vq = vertical velocity of 4m particle, then from vertical momentum:
16Vp + 4Vq = 0
and from energy:
(1/2)16(50^2 + Vp^2) + (1/2)4(200^2 + Vq^2) = E
So you have two equations in two unknowns. From this you can find the velocities of the two particles relative to the zero momentum frame.
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Last edited by ChipB; Nov 6th 2014 at 12:51 PM.
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