A target glider, whose mass m2 is 310 Â g , is at rest on an air track, a dista

Question

A target glider, whose mass m2 is 310 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 640 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.

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

b)How long does the target glider take to reach the end of the track?Â

c)How much more time elapses before the gliders collide again?Â

Explanation / Answer

By inserting correct numbers in the formula mentioned previously

x=53*(0.31+0.64) / (3*64-0.31) = 31.27cm

V1f=(0.64-0.31) / (0.64+0.31)*-75 = -26.05 m/s

V2f=2*0.64 / (0.64+0.31)*-75 = -101.05 m/s

After the second collision solving momentum and energy equations give

V2f = -70.2 m/s

V1f = 56.90

Time taken for second collision after the first collision = (53 + 31.27) / 101.05 = 0.83 s

Time to reach end after the second collision = (31.27 + 53) / 70.2 = 1.2 s

Total time taken from first collision to reach track end = 2.03 s

Third collision occurs at distance 31.27 + y cm therefore

(31.27 + y) / 70.2 = y / 56.9

y = 133.78

Therefore third collision occurs after time = 133.78 / 56.9 = 2.35 s after second collision

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