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Two astronauts, of masses 54 kg and 84 kg, are initially at rest in outer space. they push each other apart. what is their separation after the lighter astronaut has moved 12 m?

Answer :

In a case where two astronauts push each other apart, the separation after the lighter astronaut has moved 12 meters is determined by the conservation of momentum. Assuming no external forces, the momenta of the astronauts are equal and opposite. The total separation is 19.71 meters when the lighter astronaut is 54 kg and the heavier one is 84 kg.

The question involves the concept of conservation of momentum in physics, more specifically within the context of two astronauts pushing each other apart in the vacuum of space. Since there are no external forces acting on the system of two astronauts, the total momentum of the system must remain constant.

At the initial state, both astronauts are at rest, thus the total momentum is zero. After they push each other, they will move in opposite directions with their momenta being equal in magnitude but opposite in direction to maintain the total momentum at zero.

If the lighter astronaut, with a mass of 54 kg, has moved 12 meters, we can infer the separation between the two astronauts based on their masses. They exert the same force on each other for the same duration, so their momentums are equal and opposite.

The momentum of each astronaut is given by the product of their mass and their velocity (p = mv). Hence, if the 54 kg astronaut moves 12 m, the heavier astronaut (84 kg) must move a shorter distance, since his greater mass will result in a lower velocity for the same amount of momentum.

To calculate the separation, we let x be the distance the heavier astronaut moves. Since momentum is conserved, 54 kg imes 12 m = 84 kg imes x. Solving for x, we find that x = 54 kg imes 12 m / 84 kg = 7.71 m (approximate). Therefore, the total separation after the lighter astronaut has moved 12 m is 12 m + 7.71 m = 19.71 m.

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