Question

A billiard ball rolling across a table at 1.35 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs.
(a) The second ball is initially at rest.

first ball     m/s
second ball     m/s

(b) The second ball is moving toward the first at a speed of 1.10 m/s.

first ball     m/s
second ball     m/s

(c) The second ball is moving away from the first at a speed of 0.90 m/s.

first ball     m/s
second ball     m/s

Answer 1

a) when u2 = 0

By the momentum conservation:-

m1u1+m2u2 = m1v1+m2v2

u1+u2 = v1+v2

1.35+0 = v1+v2 -----------(i)

for elastic collision:-

v1 - v2 = u2 - u1

v1-v2 = 0 - 1.35 -----------(ii)

By (i) + (ii) :-

2v1 = 0

v1 = 0

v2 = 1.35

1st ball = 0 m/s
2nd ball = 1.35 m/s

b) u2 = -1.10 m/s

1.35 - 1.10 = v1+v2

v1+v2 = 0.25 -----------(i)

v1-v2 = -1.10-1.35

v1-v2 = - 2.45 ----------(ii)

By (i) + (ii) :-

v1 = -1.1 m/s

v2 = 1.15 m/s

1st ball = -1.1 m/s

2nd ball = 1.15 m/s

c) when u2 = 0.9 m/s

1.35 + 0.9 = v1+v2

v1+v2 = 2.25 -----------(i)

& v1- v2 = 0.9 -1.35

v1-v2 = - 0.45 ----------(ii)

By eq (i) + (ii)

v1 = 0.9 m/s

v2 = 1.35 m/s

1st ball = 0.9 m/s

2nd ball = 1.35 m/s

u2 =