Question

need help with this two questions, please !!

Two ice skaters, Daniel (mass 60.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 7.00 m/s at an angle of 54.1" from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink. What is the magnitude and direction of Daniel's velocity after the collision? Ans: 7.91 m/s, 32.5" below +x-axis On a very muddy football field, a 100 kg linebacker tackles an 80 kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.5 m/s north and the halfback is sliding with a velocity of 7.6 m/s east. What is the magnitude and direction of the velocity at which the two players move together immediately after the collision? Ans: 5.8 m/s, 54 north of east

Answer 1

RE 7 ms. REX 154.1 Ooft Oos voft mos mass of Daniel MRE Mass of Skaters Rebecca = 45 kg Ri= 13 m/s = initial velocity of Rebecca – Op. 13 m/s UD - omys = initial velocity of Daniel of - final velocity of Daniel vos = final velocity of Danies along x-axix. Oofy - final relocity of Danies along Y-axx Rfx = final velocity of Rebecca along x-axix Offy = final velocity of Rebecca along y-axis Rf- fisas relocity of Rebecca = 5 m/s - 54.1 / URfx & Opf coso - (1 m/s) x cos (54. 1) = 4.10 m/s ORFY = Opp sing, = (17/3) + sin ( 54. 1) = 5.67 / So Order to solve the pooblem , noe need to the conservation of momentem equation along X & Y axix Dospectively 2 mo D2x + MR URix - Mopf & MR UREX » (Cokg)* Os +(45 kg ) *(1.3 m) = (CO KB ) x 8fx + (45 K8)x(4.10m/s) → 585 kg. m/s = 60 x UDsX + 184.5 kg. m/s 2 60 60 ODG Of = - 585 kg. ml – 184.5 kg m/s 400.5 Dfx - co - 6.675 m/ - Uofy = 6.675 m/

Along - axe Mo Upix & MR. UR 2X = MD. ODY & MR. URTY Goxo & 45x0 - 60x oppy + 450 (5.67 /2) 7 0 +0 = 60 X COFY + 255.15 KQ m/s - 255.15 KG. - 60 x Vosy 255-15 GO = -4.252 m/s lory = – 4. 252 m/ 62.639 = 7.91 My Sofy tana of = V(Lox) + (Vory 1 (6.675) + (-4.252)² = [op 7.92 my tanto e = -0,03700 ) - tan' ( -0.637003) = -22.49° >> TO - - 32.490 -32.50° -4.252 6.675 - 0.637003 -ve Sion Endicates below the X-axex in clock Wise direction. re. X-axis 232.50 OOS