A 0.2 kg ball moving at 10 m per second is brought to rest in 0.05 s. The average force on the ball is:
A20 N
B100 N
C40 N
D4 N
Answer & Solution
Correct answer: C. 40 N
1. Initial momentum p_i = m·u = 0.2 × 10 = 2 kg m per second.
2. Final momentum p_f = 0 (ball at rest).
3. Change in momentum Δp = p_f − p_i = −2 kg m per second.
4. Impulse F·t = Δp, so F = Δp / t.
5. F = (−2) / 0.05 = −40 N (the sign means decelerating force).
6. The magnitude of the average force is 40 N.
_Source: Selina Concise Physics ICSE Class 9, Ch 3 'Laws of Motion' (aplustopper.com extract)_
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