# CBSE Class 11-science Answered

**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 | 21 Jun, 2021, 06:57: AM

Expert Answer

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 }= -0.54From momentum conservation , 0.15 v

_{1}+ 0.3 v_{2 }= -0.54Above 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}= 10.32If we substituite v

_{1}from eqn.(1) , above expression is written as( 3.6 + 2 v

_{2})^{2}+ 2 v_{2}^{2}= 10.32Above expression is simplified as ,

6 v

_{2}^{2}+ 14.4 v_{2}+ 12.96 = 10.32v

_{2}^{2}+ 2.4 v_{2}+ 0.44 = 0we get v

_{2}= -0.2 m/s from above quadratic equationwe get v

_{1}= -3.2 m/s from eqn.(1)
Answered by Thiyagarajan K | 21 Jun, 2021, 08:41: AM

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