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A target glider, whose mass m2 is 320 g, is at rest on an air track, a distance

ID: 1483090 • Letter: A

Question

A target glider, whose mass m2 is 320 g, is at rest on an air track, a distance d = 53 cm from the end of the track. A projectile glider, whose mass ml is 560 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)? Assume that the air track has infinite mass so that the target glider conserves its energy when colliding with the end of the air track. Thus its velocity v2 becomes -v2. Also assume that the gliders have no size and are thus at the same position when they collide. 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

m1 = 0.55 kg                   m2 = 0.32 kg


speeds before collision


u1 = -75i                     u2 = 0i m/s

speeds after collision


v1 = ?                         v2 = ?


initial momentum before collision


Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from momentum conservation


total momentum is conserved

Pf = Pi


m1*u1 + m2*u2 = m1*v1 + m2*v2 .....(1)


from energy conservation


total kinetic energy before collision = total kinetic energy after collision


KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2


KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2


KEi = KEf


0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2 .....(2)

solving 1&2


we get

v1 = [ ((m1-m2)*u1) + (2*m2*u2) ] /(m1+m2)

a)


v1 = ( -((0.56-0.32)*75) + (2*0.32*0) ) /(0.56+0.32)

v1 = -20.45 cm/s

v2 = ( ((m2-m1)*u2) + (2*m1*u1) ) /(m1+m2)

v2 = ( ((0.32-0.56)*0) + (-2*0.56*75) ) /(0.56+0.32)


v2 = -94.83 i cm/s


t1 = d1/v2 = 53/95.45


t1 = 5.55*10^-1 s


x2 = 53-v1*t1 = 53-(20.45*5.55*10^-1) = 41.65


after t2 time they will meet


v1*t2 + v2*t2 = x2


(20.45+95.45)*t2=41.65


t2 = 3.59*10^-1s

distance tavelled = x3 = v2*t2 = 95.45*3.59*10^-1 = 34.266 cm <<------answer

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