A target glider, whose mass m2 is 340 g, is at rest on an air track, a distance

Question

A target glider, whose mass m2 is 340 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 570 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

Apply conservation of linera momentum

570*(-75)= 340*(v2)+570*(v1); ----(A)where v1-velocity of mass1

Apply conservation of energy

570*(-75)2/2 = 340*(v2)2/2+570*(v1)2/2----(B)

Solve (A) and (B)

v2=-93.956 cms-1 ;

v1=-18.956 cms-1 ;

After get bounced form the spring, assume m2 has the same speed.

So now v2= +93.956

relative velocity respect to m1 (v)= v2-v1=93.956+18.956=112.912

time taken to m2 to get to the end of the track(t0)= 53/93.956 =0.56 s

displacement of m1 relative to earth(s0)= t0*18.956=10.693 cm

initial distance betwwen m1 and m2(just after the bounce back of m2)= 53-s0=42.3cm

time taken to collide after bounceback of m2(t1)= v/s0 =112.912/42.3 =2.67s

distance from end to the poin of second collision= t1*v2=2.67*93.956=250.86 cm

time elapsed before the second collision=t1+t0= 2.67+0.56 =3.23 s

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