This question involves the concepts of the law of conservation of momentum.
The magnitude of the final momentum of the eight ball is "0.22 N.s".
According to the law of conservation of momentum:

where,
= initial momentum of the cue ball = 0.23 N.s
= initial momentum of the eight ball = 0 N.s (since ball is initially at rest)
= final momentum of the cue ball = 0.01 N.s
= final momentum of the eight ball = ?
Therefore,

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Answer:
-0.7 m/sec
Explanation:
Mass of first block = m1 =3.0 kg
Mass of second block = m2= 5.0 kg
Velocity of first block = V1= 1.2 m/s
Velocity of second block = V2 = ?
Momentum of Center of mass MVcom is sum of both blocks momentum and is given by
MVcom= m1v1+m2v2
Where
M= mass of center of mass
Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)
Putting values, we get;
0= 3×1.2+5v2
==> v2= 3.6/5= - 0.7 m/s
-ve sign indicates that block 2 is moving in opposite direction of block 1
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