Question

A bullet (0.1 kg) initially moving at speed u is embedded inside block 1 (4 kg) initially at rest at A which travels 8 m in a rough region (uk = 0.2) to B at speed of 12 m/s, then travels to a frictionless region from B to C, then to a rough region (Uk = 0.2) after point C. Then bulletblock 1 collides with block 2 (15 kg) which is initially at rest and connected with a spring of force constant k = 1500 N/m. Block 2 then travels a distance d and stops at E. 12 m/s ICI --- B - A 8m 6 m 2 mm 50° EDHE 4 m d Figure 5 (a) Find the speed u of the bullet before colliding with block 1. (b) Find the speed ui of the bullet-block 1 just before colliding with block 2 at D. (c) If the collision between bullet-block 1 and block 2 is completely elastic, find the speed of block 2 just after collision. (d) Find the distance d.

Answer 1

a work energy theorm after collision between A and B w (friction ) - DK.E - M (mtimu)g.d, = {(mm) (v +-1,2) -0-2 x 3-81x8 = (12*-Vi%) : Wis = 175.392 Vi= 13:244 mis mu, a (m+M) V, uz (0.1+4) * 13.244 6.1 . Tu= 543 mls | Initial speed of bullet Јок <hеrg> Момы bahovatv B and D. wil granity) tw(triction) - Doleage mtaugh - M (mm), a50-4,(M+71)gd3 5 Lov( x2) ( 4129•PX12.) – 0274-1800 ton set 4) =12x1944 (35=2") 482.16 - 72.6 = 2.05 (vz?-144 )

Na - 18.54 mis colliding at D Velocity of bullet-block just before = 18:54 mis: 7 At this point m = 4.1 kg m2 = 15 kg U; = 18.54 mis Y = 0 mig V = (mm₂) u t2m₂ u2 m. tM2 (M, AM2) vc (4.1-15 ) x 18.54 to (4.1H5) vie -10.58 min am, u of marm, ) u (m, fM₂) (Mitma) (2x4.1 ) x 18:54 (4.14.15) 7.96 mis

after collision speed of VE block 2 7.96 mis as tkd = 1 mv² - ME Mz gd 15000 + 2x0.2 x 1549.81* d - 15x44.96) = 0 15000? + 58.86d – 950.3 -0 da - 58.86 1/ (54-8692+ 4x980-3x1500 2x1500 do -58.86 € 23886 3000 do 0.776 m