# A block of mass 2

Mass of the first block is m = 2kg

Initial speed before collision is u= 2 m/s

Mass of the second block is m’ = 2 kg

Final speed after collision is v

From law of conservation of momentum

{a} Loss in kinetic energy in elastic collision will be

{b} Actual loss = (maximum loss)/2 =2/2 =1J

Let final velocities of the blocks of masses 2kg are v1 and v2.

Where v1+v2=u and v1 - v2= eu.

Then, actual loss will be

Since, v1+v2=u and v1 - v2= eu

So, the coefficient of restitution (e) is

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