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MHT CET 2026 (19 May, Shift 1) · Impulse & Momentum Conservation

Impulse & Momentum Conservation question from MHT CET 2026 (19 May, Shift 1)

Q1MHT CET 2026 · 19 MAY · SHIFT 1Easy

Impulse & Momentum Conservation

A cricket ball of mass $120$ g is moving with a velocity of $8$ m/s and is hit by a bat so that the ball gets turned back with a velocity of $12$ m/s. If the duration of contact between the ball and bat is $0.02$ s, the impulse (in Ns) given to the ball by the bat is

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Answer: (c)

Impulse $= \Delta p = m(v_f - v_i)$ ; the velocity reverses direction.
$J = 0.12 \times (12 - (-8)) = 0.12 \times 20$
$J = 2.4\text{ N s}$
So the answer is (c) 2.4.

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