Question

Block A (M = 10 kg) is released from rest and slides down an initially frictionless ramp (θ= 30°, L = 10 m). It then passes over a rough patch of ground (µ_k= 0.633 and d = 5m) before impacting block B. Block B (m = 20 kg) is initially at rest and suspended by a rigid bar of negligible mass and length R = 5m. The coefficient of restitution during the impact is e = 0.75. Determine the maximum deflection angle reached by Block B after impact.

R A B 10

Answer 1

= 28.357° max. deflection angle From the formula of co-efficient of restitation. Now, Velocity of Block A just before impact (un)3 = 6 m/s Velocity on Block B just before impact (UB), - omys let velocity of Block A and B just after impact be (VA) and (v.) 28 ea VB-VÁ VB-VA (Ung (UB), 6-0 4.50 = Vg-VA equation (1) Now; oupply conservation of linea momentum principle to above problem; MACUAg + mg (UB)3 = manat molvo') 10 (6) +2010) - 10(V)t 20 (vb) IVÁ +2 ve equation (2) Solve the above 2 equations simultaneously; whet get = -1m's ( Vé = 3.5ms (+)) Now; apply principle of.conservation of energy principle to block B : After reaching the highest point its velocity becomes o Therefore; {(kieta {{P.5)436 -☺ žmová)= yoghe Rooso here; ho = R-RCOSØ R(1-cos) zein equ ③ hel (3.5) - 9.81x RC1-Cosy) F.B.D Note R/ substituting a B 2 cosø = (1-0-12) = 0.88

1 ma mgsing ex ing Rom 0 = 30° e 10104570 Indo National L 101049 e L = 10m d=5m e-0-75 Hot 14 uk 93 Consider Free Body diagsam of block A : Consider algebrauc sum of forces down theslope for block A: EFx = maa mg sino = rma a = 4.9 M/S2)(down the stope) Now; from kinematic equation of motion. NA = 12 + 2aAL : He block A was released from hest Up=0. (at posihon 1) VA = 12x 4.9x10 9.90 m/s (at the bottom of slope) UA, (position 2 of blockA) Now, consider the motion of block from position 2 to 3 : frictional surface EFgc = ma (de to flichen Her block will decelerate f Invoice Margin / TOT 47016505 DNPR/15- Also; (VA) on a NO - FA = MARA - - MK NA = MARA -MK Mag= MARA .: Jan = –6.2014. deceleration at end of positions) Again applying kinematic equation of motion between position 2 & 3 (VA) = Ua)+ 2as :- (VA)X = (9.9) * - 216.20)(5) )3 = 736 6 m/s .. UA);