A ball of mass m is moving towards a batsman at a speed v. The batsman strikes the ball and deflects it by an angle θ without changing its speed. The impulse imparted to the ball is

See fig. The resolved components of velocity v of the ball before and after striking the bat are shown in the figure. From figure, AF = EB and. AE = FB. The  velocity components EB and BD along the bat are equal in magnitude and are along the same direction.∴Change in momentum along the bat =mvsin⁡(θ/2)−mvsin⁡(θ/2)=0Now impulse = change in momentum along BF                    =mvcos⁡θ/2−{−mvcos⁡θ/2}=2mvcos⁡(θ/2)