Is the answer 5 m/s at 53° south of west?
If all 3 children push with different forces, wouldn’t the change in momenta of each vector or component vector be different? If only child 3 pushed, the internal forces would be equal changes in momenta per vector and component vectors by Newton’s second and third law. Given each dimension has different vector...

answers:
Andrew Smith say: Momentum is conserved.
Find the change of momentum for the two children that the information is specified for.
= 50 * 4 North + 30 * 5 East
Now as every action has an equal AND OPPOSITE reaction the third child gains the same magnitude but in the opposite direction.
ie 50*4 South + 30 *5 West
Velocity = momentum / mass = 4 south + 3 West
which is a 3,4,5 triangle. ie the magnitude of the velocity is 5 m/s
And the direction is given by theta = atan( opposite / adjacent) = atan( 4 south / 3 west) = 53 degrees south of west.
"Internal forces" are misleading. Without any external force the net force on the system must be zero.
Which is why the third skater MUST have an equal and opposite reaction to the momenta of the other two.

oubaas say: Mr = √M1^2+M2^2 = 100√2^2+1.5^2 = 250 kg*m/sec
heading = arctan 200/150 = 53.13° NoE
M3 = Mr = 250 kg*m/sec
V3 = M3/m3 = 250/50 = 5.0 m/sec
heading = 180+arctan (200/150 = 53.13° SoW

BigBird say: The answer is correct. The net impulse (force * time) on each child must account for his change in momentum. Since they started at rest the only requirement is that the three momentum vectors must add to zero just after the separation.

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