(hrwc10p77_6e) A target glider, whose mass m2 is 310 g , is at rest on an air tr

Question

(hrwc10p77_6e) 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. 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

m2 = 310/1000 = 0.31 Kg
m1 = 640/1000 = 0.64 Kg

Initial Velocity of m1, v1 = 75/100 m/s = 0.75 m/s
Initial Velocity of m2 = 0

Using Momnetum Conservation
m1 * v1 + m2 * v2 = m1 * v1f + m2 * v2f
0.64 * 0.75 + 0 = - 0.64 * v1f + 0.31 * v2f ----------1


As the Collision is Elastic , Kinetic Energy Remains Conserved
1/2 *m1v1^2 + 1/2 * m2v2^2 = 1/2 * m1v1f^2 + 1/2 * m2v2f^2
1/2 * 0.64 * 0.75^2 + 0 = 1/2 * 0.64 * v1f^2 + 1/2 *0.31 * v2f^2 ----------2


Using eq 1 & 2
v1f = - 0.26 m/s
v2f = - 1.01 m/s


Now Distance = 53 cm = 0.53 m

time = Distance / Speed
t = 0.53/1.01
t = 0.52 s

Let the Distance from the end of the track = x m

Now we can say time taken by both is same therefore -
(0.53 + x)/1.01 = (0.53 - x) / 0.26
x = 0.313 m
x = 31.3 cm

time = (0.53 - 0.313 )/0.26
t = 0.835 s

How far from the end of the track does this second collision occur ( cm)
x = 18.4 cm

How long does the target glider take to reach the end of the track?
t = 0.52 s

How much more time elapses before the gliders collide again?
t = 0.835 s

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