A target glider, whose mass m2 is 380 g, is at rest on an air track, a distance

Question

A target glider, whose mass m2 is 380 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 620 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? How much more time elapses before the gliders collide again?

Explanation / Answer

I broke it down into 2 seperate stages, a t1 from when m2 goes from its starting point to the wall (a distance of d) and a t2 from when m2 rebounds from the wall and collides with m1 again.

v1f=(m1m2)/(m1+m2)v1i

240/1000×0.75m/s=0.18m/s

v2f=2m2 / (m1+m2)v1i

=(760/1000)×.75m/s=.57m/s

t1= x / v02= 0.53m / 0.57m/s= 0.929s

x1=v01t=(0.18)(.929)=.16722m
So in time interval t1 the collision occurs and accelerates m2 from rest to 0.57 m/s and m1 is still moving at 0.18 m/s. It takes 0.929 seconds for m2 to go d and reach the end of the track and in this time m1 moves 0.16722m. Then m2 has an elastic collision with the short spring and now has a velocity of v2f.

Now:

x2=v02t

x1=v01t + x01

Setting these equal when they collide and solving for t:
t2=x01 / v02(v01) = 0.36278 / 0.57m/s+ 0.18m/s=0.48378s

Calculations maybe wrong, logic used is correct.

Hope this helps :)

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