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A cricket ball of mass m is struck by a batsman and flies into the air; a fielder catches it near the boundary. The de Broglie wavelength of the ball just after being struck is lambda1 and just before being caught is lambda2. The ball is hit and caught at the same height. Find the total work done by air resistance (assume mechanical energy is lost only due to air resistance).
- h² / (2m) * (1/lambda1² - 1/lambda2²)
- h² / m * (1/lambda1 - 1/lambda2)
- h² / (2m) * (lambda1 * lambda2) / (lambda1 - lambda2)
- h² / (2m) * (1/lambda1 - 1/lambda2)
Correct answer: h² / (2m) * (1/lambda1² - 1/lambda2²)
Solution
De Broglie relation: lambda = h/p = h/(mv), so p = h/lambda and KE = p²/(2m) = h²/(2m*lambda²). Since initial and final heights are the same, delta(PE) = 0. By work-energy theorem, W_air = KE_final - KE_initial = h²/(2m) * (1/lambda2² - 1/lambda1²). Since air resistance slows the ball, lambda2 > lambda1 meaning 1/lambda2² < 1/lambda1², so W_air is negative (energy lost). Total work done BY air resistance = h²/(2m) * (1/lambda2² - 1/lambda1²) = -h²/(2m) * (1/lambda1² - 1/lambda2²). The option asking for total work done by air resistance (which is negative) corresponds to h²/(2m) * (1/lambda1² - 1/lambda2²) in magnitude with a negative sign, but among options the correct form matching standard answer is h²/(2m) * (1/lambda1² - 1/lambda2²).
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