A batsman hits a 0.15 kg ball back along the line of approach without changing its speed of 12 m s⁻¹. The magnitude of the impulse imparted to the ball is:

1. The ball's velocity reverses, so Δv = 12 − (−12) = 24 m s⁻¹ in magnitude. 2. Impulse J = m Δv = 0.15 × 24 = 3.6 N s. 3. Forgetting the sign flip gives 0.15 × 12 = 1.8 N s — the common trap. _Source: syllabus Physics Ch 4 "Laws of Motion", Example 4.4, p. 56_