A target glider, whose mass m2 is 400 g, is at rest on an air track, a distance

Question

A target glider, whose mass m2 is 400 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 600 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

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However doing the question as per my understanding.
v1i, the initial velocity of the projectile (1) is -75 cm/s
v1f, the final velocity of the projectile(1) is
(m1-m2) v1i /( m1 +m2) = - (600 – 400)*75 / (600 + 400) = -15cm/s.

The final velocity of the target (2) is (v1i + v1f) = (- 75 -15) = -90 cm /s
=========================

The distance moved by the projectile after impact in time t is - 15 *t. = a (say)

In this time the distance moved by the target is (d + b) where b is the return distance of the target.
(d + b) = 90 * t (magnitude alone)

a + (d + b) = 2 d

15 *t + 90 * t = 2*53

t ~ 1s

Hence a = 15 cm

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