Question

Three blocks of masses m₁=1.00 kg, m₂=2.00 kg, and m₃=3.00 kg are set at rest on a level air track from right to left. Then m₃ is pushed toward m₂ with a speed of 3.00 m/s. Assuming that all collisions are elastic, what are the final speeds of (a) m₁, (b) m₂, and (c) m₃?

Answer 1

Using momentum conservation:

for collision between m3 and m2

Pi = Pf

m3u13 + m2u2 = m3v3 + m2v2

3*3 + 2*0 = 3*v3 + 2*v2

3v3 + 2v2 = 9

Also, for elastic collision

v2 - v3 = u3 - u2

v2 - v3 = 3 - 0

v2 = 3 + v3

3v3 + 2v2 = 9

3v3 + 2(3 + v3) = 9

5v3 = 3

v3 = 3/5 = 0.6 m/sec

v2 = 3 + 0.6 = 3.6 m/sec

Now there will be collision between m2 and m1

m2*v2 + m1*v1 = m2*v2f + m1*v1f

2*3.6 + 1*0 = 2*v2f + 1*v1f

2*v2f + v1f = 7.2

Also for elastic collision

v1f - v2f = v2 - v1

v1f - v2f = 3.6 - 0

v1f = v2f + 3.6

v1f + 2v2f = 7.2

v2f + 3.6 + 2v2f = 7.2

3*v2f = 3.6

v2f = 1.2 m/sec

v1f = 1.2 + 3.6 = 4.8 m/sec