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NCERT Exemplar · Q4

Q.In the previous problem (5.3), the magnitude of the momentum transferred during the hit is

(a) Zero
(b) 0.750.75 kg m s−1^{-1}
(c) 1.51.5 kg m s−1^{-1}
(d) 1414 kg m s−1^{-1}.
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The momentum transferred equals the magnitude of the ball's change in momentum, ∣Δp⃗∣=1.5 kg m s−1|\Delta\vec{p}| = 1.5\ \text{kg m s}^{-1}, so option (C) is correct.

This question refers back to problem 5.3, where a cricket ball of mass m=150 g=0.15 kgm = 150\ \text{g} = 0.15\ \text{kg} has initial velocity u⃗=(3i^+4j^) m s−1\vec{u} = (3\hat{i} + 4\hat{j})\ \text{m s}^{-1} and, after being hit straight back by the bat, final velocity v⃗=−(3i^+4j^) m s−1\vec{v} = -(3\hat{i} + 4\hat{j})\ \text{m s}^{-1} — same speed, exactly reversed direction. The "momentum transferred" is the change in the ball's momentum, which by the impulse-momentum theorem equals the impulse delivered by the bat.

  1. Change in momentum. …

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