Impulse And Momentum

Question

A cricket ball of mass 0.15 kg approaches a bat at 20 m/s and rebounds straight back along the same line at 20 m/s after contact. What is the magnitude of the impulse delivered by the bat to the ball?

RankGuru · Accepted Answer

Impulse equals the change in linear momentum of a body, expressed as $J = \Delta p = m(v_f - v_i)$, and it acts in the direction of that momentum change. Taking the initial direction of approach as positive, the incoming velocity is $+20	ext{ m/s}$ and the rebound velocity is $-20	ext{ m/s}$ because the ball reverses direction. The change in velocity is therefore $-20 - 20 = -40	ext{ m/s}$ in magnitude 40 m/s. Hence $J = 0.15 	imes 40 = 6	ext{ N s}$. The value 3 N s wrongly uses a velocity change of only 20 m/s, ignoring the reversal. The 4.5 N s value comes from an incorrect 30 m/s difference. The 0 N s choice mistakenly assumes equal speeds mean zero change, but direction reversal makes the momentum change largest. This follows the NCERT impulse-momentum theorem central to the Laws of Motion. As a sign check, the impulse points opposite to the incoming ball, exactly the direction in which the bat strikes.

Impulse And Momentum

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