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A 40-kg soccer player going through the air at 4m/s collides with a 60-kg player who is headed toward her at 3m/s, in the air. What is the speed and direction of the entangled players? Please explain how to do it

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Answer: The entangled players would move with a speed of 0.2 m/s in the direction direction of the motion of second soccer player.

Step-by-step explanation:

From law of conservation of momentum,

Initial momentum of the closed system = final momentum of the closed system

Since, no external force is acting on the players, total momentum would be conserved.

m₁u₁+m₂u₂=(m₁+m₂)v

where, m₁ is the mass of one soccer player = 40 kg

u₁ is its initial velocity before collision = 4 m/s

m₂ is the mass of the other soccer player = 60 kg

u₂ is the initial velocity before collision = - 3 m/s (∵opposite direction)

After collision, the two players get entangled and hence move with the same final velocity v.

40 kg × 4 m/s - 60 kg × 3 m/s = (40 kg + 60 kg ) v

160 kg m/s - 180 kg m/s = 100 kg v

v = -0.2 m/s

Hence, the entangled players would move with a speed of 0.2 m/s in the direction direction of the motion of second soccer player.

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