A target glider, whose mass m2 is 360 g , is at rest on an air track, a distance

Question

A target glider, whose mass m2 is 360 g, is at rest on an air track, a distance d = 53 cm from the end of the track. A projectile glider, whose mass m1 is 550 g, approaches the target glider with velocity v1i = -75 cm/s and collides elastically with it (see the figure). The target glider rebounds elastically from a short spring at the end of the track and meets the projectile glider for a second time.

How far from the end of the track does this second collision occur ( cm)?
37.4cm

How long does the target glider take to reach the end of the track?
.585 s

How much more time elapses before the gliders collide again?
??? I need the answer for this part thank you!

A target glider, whose mass m2 is 360 g, is at rest on an air track, a distance d = 53 cm from the end of the track. A projectile glider, whose mass m1 is 550 g, approaches the target glider with velocity v1i = -75 cm/s and collides elastically with it (see the figure). The target glider rebounds elastically from a short spring at the end of the track and meets the projectile glider for a second time. How far from the end of the track does this second collision occur ( cm)? How long does the target glider take to reach the end of the track? .585 s How much more time elapses before the gliders collide again?

Explanation / Answer

after collision,

v(1) = {m_1 - m_2}{m_1 + m_2} v(v1i) = -15.659
v(2) = {2 m_2}{m_1 + m_2} v(1i) = -59.3407

u have written that it takes 0.585 s to reach the end of the track

distance covered by m1 in this time = -15.659*0.585 = 9.160515

=> sparation between m1 and m2 at that istant = 75-9.160515 = 66.739485

separation when they collide = 0

=>relative distance to be covered by they collide = 66.739485

velocity of m2 after collision with spring = 59.3407

=> relative velocity = 59.3407+15.659 = 74.9997

=> more time required = relative distance/relative velocity = 66.739485/74.9997 = 0.8898643 s

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