Question

A 0.22 kg air track glider moving at 0.60 m/s a collides elastically into a 0.44 kg glider at rest. After collision the 0.22 kg glider is moving with velocity -0.20 ms. What is the speed of the 0.44 kg glider after collision

Answer 1

The glider collides elastically  an another glider at rest, in this case we have the coservation of quantity of movement 

m1 v1  + m2 v2 = m1V1 + m2 V2
v is the velocity before the shock, V the velocity after the shock
m1=0.22
v1=0.60m/s
m2=0.44
V1=-0.20m/s
remark that 0.44 kg glider is initially at rest, so v2= 0
the equation above can be written  m1 v1  = m1V1 + m2 V2 ( because v2=0 at rest)
m1 v1  = m1V1 + m2 V2 implies  V2= (m1v1 -m1V1) / m2

V2=0.22x0.60 -0.22x (-0.20) ] / 0.44= 0.4m/s