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 m1 = 0.15 kg is mass of glider moving towards right with velocity u1 = 0.8 m/s .
Let m2 = 0.3 kg is mass of glider moving towards left with velocity u2 = -2.2 m/s .
Total momentum before collision = m1 u1 + m2 u2
Total momentum before collision = ( 0.15 × 0.8 ) - ( 0.3 × 2.2 ) = -0.54 kg m/s
Let v1 be the velocity of glider of mass m1 after collision .
Let v2 be the velocity of glider of mass m2 after collision .
Total momentum after collision = m1 v1 + m2 v2
Total momentum after collision = 0.15 v1 + 0.3 v2 = -0.54
From momentum conservation , 0.15 v1 + 0.3 v2 = -0.54
Above expression is simplified as , v1 + 2 v2 = -3.6 or v1 = -3.6 - 2 v2 ........................(1)
Total kinetic energy before collision = (1/2) m1 u12 + (1/2) m2 u22
Total kinetic energy after collision = (1/2) m1 v12 + (1/2) m2 v22
From kinetic energy conservation , we get the following expression after simplification
v12 + ( m2 / m1 ) v22 = u12 + ( m2 / m1 ) u22
v12 + 2 v22 = 0.64 + 2 (2.2 )2 = 10.32
If we substituite v1 from eqn.(1) , above expression is written as
( 3.6 + 2 v2 )2 + 2 v22 = 10.32
Above expression is simplified as ,
6 v22 + 14.4 v2 + 12.96 = 10.32
v22 + 2.4 v2 + 0.44 = 0
we get v2 = -0.2 m/s from above quadratic equation
we get v1 = -3.2 m/s from eqn.(1)
Answered by Thiyagarajan K | 21 Jun, 2021, 08:41: AM
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