Again impulse = Area between the graph and time axis \( = \frac{1}{2} \times 2 \times 4 + 2 \times 4 + \frac{1}{2}(4 + 2.5) \times 0.5 + 2 \times 2.5\)
\( = 4 + 8 + 1.625 + 5\) \( = 18.625\)...(ii)
From (i) and (ii), \(m({v_2} - {v_1}) = 18.625\)
\(⇒\) \({v_2} = \frac{{18.625}}{m} + {v_1} = \frac{{18.625}}{2} + 5 = 14.25\;m/s\)