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

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 v= -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 v2
 
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 | 21st Jun, 2021, 08:41: AM