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:
A1.8 N s
B2.4 N s
C3.6 N s
D4.8 N s
Answer & Solution
Correct answer: C. 3.6 N s
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: NCERT Class 11 Physics Ch 4 "Laws of Motion", Example 4.4, p. 56_
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