Given:-

$ V_A=3m/s \\ e=0.8$

To find:-

Velocity after collision $(V^1_A)$

Solution:

The surface is smooth and rigid $(µ=0)$, the component of velocity parallel to the surface will not change.

$$V^1_{Ax} = V_{Ax}=3 \cos60 $$

$$ \therefore V^1_{Ax}=1.5 m/s $$

Now,

$e=\dfrac { V^1_{Ay} - V^1_{\text {Surface}}}{V_{\text {surface}}-V_{Ay}} =-\dfrac {V^1_{Ay}}{V_{Ay}} [ \because \text {Surface is rigid}] \\ \therefore V^1_{Ay} =-eV_{Ay} \\ =0.8 × 3 \sin60=-0.2078 \\ \therefore V^1_{Ay}=2.078 m/s (\downarrow ) $

The resultant velocity,

$V^1_A=2.563 m/s (54.182⁰ )$