A batsman hits back a ball of mass $$0.4$$ kg straight in the direction of the bowler without changing its initial speed of $$15$$ m s$$^{-1}$$. The impulse imparted to the ball is ______ N s.

We need to find the impulse imparted to the ball when the batsman hits it back without changing its speed.

Since the mass of the ball is $$m = 0.4$$ kg and its speed is $$v = 15$$ m/s, and the ball is hit back in the opposite direction with the same speed, we define a direction convention to set up our calculations.

Let the initial direction of the ball towards the batsman be negative, so $$v_i = -15$$ m/s. After being hit back, the ball moves in the opposite direction, which we take as positive: $$v_f = +15$$ m/s.

Now the impulse can be calculated as the change in momentum: Impulse = Change in momentum = $$m(v_f - v_i)$$, which gives $$0.4 \times (15 - (-15))$$, or equivalently $$0.4 \times 30$$, resulting in $$12$$ N s.

The impulse imparted to the ball is 12 N s.