A 0.15kg glider is moving to the right on a frictionless horizontal air track with a speed of 0.80m/s .It has a head on collision with a 0.300 kg glider that is moving to the left with a speed of 2.2m/s .Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

### Asked by arjunsah797 | 21st Jun, 2021, 06:57: AM

### In elastic collision , momentum and kinetic energy are conserved.
Let m_{1} = 0.15 kg is mass of glider moving towards right with velocity u_{1} = 0.8 m/s .
Let m_{2} = 0.3 kg is mass of glider moving towards left with velocity u_{2} = -2.2 m/s .
Total momentum before collision = m_{1} u_{1} + m_{2} u_{2}
Total momentum before collision = ( 0.15 × 0.8 ) - ( 0.3 × 2.2 ) = -0.54 kg m/s
Let v_{1} be the velocity of glider of mass m_{1} after collision .

Let v_{2} be the velocity of glider of mass m_{2} after collision .
Total momentum after collision = m_{1} v_{1} + m_{2} v_{2}
Total momentum after collision = 0.15 v_{1} + 0.3 v_{2 }= -0.54
From momentum conservation , 0.15 v_{1} + 0.3 v_{2 }= -0.54

Above expression is simplified as , v_{1} + 2 v_{2} = -3.6 or v_{1} = -3.6 - 2 v_{2} ........................(1)
_{}
Total kinetic energy before collision = (1/2) m_{1} u_{1}^{2} + (1/2) m_{2} u_{2}^{2}
Total kinetic energy after collision = (1/2) m_{1} v_{1}^{2} + (1/2) m_{2} v_{2}^{2 }
From kinetic energy conservation , we get the following expression after simplification
v_{1}^{2} + ( m_{2} / m_{1} ) v_{2}^{2} = u_{1}^{2} + ( m_{2} / m_{1} ) u_{2}^{2}
v_{1}^{2} + 2 v_{2}^{2} = 0.64 + 2 (2.2 )^{2} = 10.32
If we substituite v_{1} from eqn.(1) , above expression is written as
( 3.6 + 2 v_{2} )^{2} + 2 v_{2}^{2} = 10.32
Above expression is simplified as ,
6 v_{2}^{2} + 14.4 v_{2} + 12.96 = 10.32
v_{2}^{2} + 2.4 v_{2} + 0.44 = 0
we get v_{2} = -0.2 m/s from above quadratic equation
we get v_{1} = -3.2 m/s from eqn.(1)

_{1}= 0.15 kg is mass of glider moving towards right with velocity u

_{1}= 0.8 m/s .

_{2}= 0.3 kg is mass of glider moving towards left with velocity u

_{2}= -2.2 m/s .

_{1}u

_{1}+ m

_{2}u

_{2}

_{1}be the velocity of glider of mass m

_{1}after collision .

_{2}be the velocity of glider of mass m

_{2}after collision .

_{1}v

_{1}+ m

_{2}v

_{2}

_{1}+ 0.3 v

_{2 }= -0.54

_{1}+ 0.3 v

_{2 }= -0.54

_{1}+ 2 v

_{2}= -3.6 or v

_{1}= -3.6 - 2 v

_{2}........................(1)

_{}

_{1}u

_{1}

^{2}+ (1/2) m

_{2}u

_{2}

^{2}

_{1}v

_{1}

^{2}+ (1/2) m

_{2}v

_{2}

^{2 }

_{1}

^{2}+ ( m

_{2}/ m

_{1}) v

_{2}

^{2}= u

_{1}

^{2}+ ( m

_{2}/ m

_{1}) u

_{2}

^{2}

_{1}

^{2}+ 2 v

_{2}

^{2}= 0.64 + 2 (2.2 )

^{2}= 10.32

_{1}from eqn.(1) , above expression is written as

_{2})

^{2}+ 2 v

_{2}

^{2}= 10.32

_{2}

^{2}+ 14.4 v

_{2}+ 12.96 = 10.32

_{2}

^{2}+ 2.4 v

_{2}+ 0.44 = 0

_{2}= -0.2 m/s from above quadratic equation

_{1}= -3.2 m/s from eqn.(1)

### Answered by Thiyagarajan K | 21st Jun, 2021, 08:41: AM

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