A target glider, whose mass m2 is 390 g , is at rest on an air track, a distance

Question

A target glider, whose mass m2 is 390 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 590 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

by conservation of momentum

initial momentum = final momentum

0.59 * 0.75 = 0.39 * v1 + 0.59 * v2

by conservation of energy

initial energy = final energy

0.5 * 0.59 * 0.75^2 = 0.5 * 0.39 * v1^2 + 0.5 * 0.59 * v2^2

v1 = 0.903061 m/s

v2 = 0.153061 m/s

time it'll take glider to reach end of the track = 0.903061 / 0.53

time it'll take glider to reach end of the track = 1.7 sec

distance covered by second mass in this time = 0.153061 * 1.7

distance covered by second mass in this time = 0.26 m

they both meet after time t at distance x from the end of the track

t = 0.903061 / x

also,

t = 0.153061 / (0.53 - 0.26 - x)

so,

0.903061 / x =  0.153061 / (0.53 - 0.26 - x)

x = 0.231 m

second collision occur at distance 23.1 cm from end of track

time elapsed before second collision = 1.7 + 0.903061 / 0.231

time elapsed before second collision = 5.61 sec

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