Given :
Mass of ball = m
Initial velocity = u
Final velocity = v
Now,
\(⇒e=\frac{\text{Veloctiy of seperation}}{\text{Velocity of approach}}\)
\(⇒1=\frac{v_2-v_1}{u_2-u_1}\)
\(1=\frac{v_0-v}{u+v}\)
⇒ u + v = v0 - v
⇒ v0 = u + 2v
So, the correct option is (C) : u + 2v.
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)): 
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