# Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m s–1 while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m s–1 towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.

### Asked by sasikumarsarala2020 | 25th Jun, 2021, 04:37: PM

Expert Answer:

### Type of collision is Inelastic , only momentum is conserved .
Total momentum p_{i} before collision, p_{i} = m_{1} u_{1} + m_{2} u_{2}
where m_{1 }and m_{2 }are masses of player-1 and player-2 . Velocity of player-1 before collision is u_{1 }and that of player-2 is u_{2}
p_{i }= ( 60 × 5 ) - ( 55 × 6 ) = -30 kg m/s
Momentum p_{f }after collision = ( m_{1} + m_{2 }) v
where v is velocity of entangled players
By momentum conservation , ( m_{1} + m_{2 }) v = 115 v = -30
Hence , velocity v = ( -30 / 115 ) = -0.26 m/s
Hence , after collision, entangled players will move with velocity 0.26 m/s in the direction of initial velocity of player-2

_{i}before collision, p

_{i}= m

_{1}u

_{1}+ m

_{2}u

_{2}

_{1 }and m

_{2 }are masses of player-1 and player-2 . Velocity of player-1 before collision is u

_{1 }and that of player-2 is u

_{2}

_{i }= ( 60 × 5 ) - ( 55 × 6 ) = -30 kg m/s

_{f }after collision = ( m

_{1}+ m

_{2 }) v

_{1}+ m

_{2 }) v = 115 v = -30

### Answered by Thiyagarajan K | 25th Jun, 2021, 05:54: PM

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