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}} …