347_Dynamics 11ed Manual

# 347_Dynamics 11ed Manual - = restitution ev b v B v A − =...

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Engineering Mechanics - Dynamics Chapter 15 Guesses e 0.8 = R 10 ft = v A2n 1 ft s = v A2x 1 ft s = v A2y 1 ft s = Given v A2n ev A1n = v A2x v A2n sin θ () v A2t cos () + = v A2y v A2n cos () v A2t sin () = R cos () v A2x t = R sin () g 2 t 2 v A2y t + = e R v A2n v A2x v A2y Find eR , v A2n , v A2x , v A2y , () = R 7.23 ft = e 0.502 = Problem 15-67 The ball of mass m b is thrown at the suspended block of mass m B with velocity v b . If the coefficient of restitution between the ball and the block is e , determine the maximum height h to which the block will swing before it momentarily stops. Given: m b 2kg = m B 20 kg = e 0.8 = v b 4 m s = g 9.81 m s 2 = Solution: Guesses v A 1 m s = v B 1 m s = h 1m = Given momentum m b v b m b v A m B v B +
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Unformatted text preview: = restitution ev b v B v A − = energy 1 2 m B v B 2 m B gh = v A v B h ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ Find v A v B , h , ( ) = v A v B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.55 − 0.65 ⎛ ⎝ ⎞ ⎠ m s = h 21.84 mm = * Problem 15-68 The ball of mass m b is thrown at the suspended block of mass m B with a velocity of v b . If the time of impact between the ball and the block is Δ t , determine the average normal force exerted on the block 345...
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