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Solution : - Plus-1 maximum height = boom So, from conservation of energy velity of ( Ball + Block ) just after collision So , from conservation of momentum mid, + maz = (mitm z ) v 0.0 1 0 XU +0 = ( 6 0 0 10 + 5 ) * 1 . 25 2 2 627 . 35 mis. Dues - 2 Solution Given Mass cy Casts My = 1.2 KQ M,clock Speed of cart may 19 2 6 m/s Apply momentum conservation offer collision 1 - 2 x 6 = 1 , 2 19 , + 12 8 19 2 7 12 = 1 : 2 26 + 1 . 810 2 Spiral -Now using conservation of energy 4 3 . 2 = 1. 2 ( 6 - 1 1509 2 ) 2 + 1.810, 2 ( Using can 1 ) 43 : 2 = 1 12 ( 36 + 2 . 2 5 1 1 2 - 1 8x 2) + 1:802 43. 2 = 43. 2 + 2.7 1- - 21:6 V , + 1. 82 32 4:5 1075 - 21.62 2 20 2 2 2 4 . B mis Velocity 1 = 6 - ( 1 , 5 x 4 . 8 ] 1 = - 1 .2 mis I Partl'of Using energy conservation K X ( 0 . 02 ) - = 1,8 x ( 4 : 8 ) K = 103680 N / m Final Answers (a, velocity of carts 2 = 1. 2 mis( W (D) Spring constants * = 1036 90 Dim SpiralIwas- 3 Given Two magnetic bucks A and B collide each other elastically by means of repelling magnets . Find up = final velocity ( velocity after collision) of puck A 8 12 = final velocity ( velocity after collision ) of magnetic puck B? Let direction to right is treated as positing direction to left is treated as negative 1X , = Some parameter = mixy, + me x 42 U. and 42 = initial velocity of puck, mand my ~ masses of puck A andB Ky = ( 6 . 5 x 4 ) + 0 . 9 8 ( - 3 . 0 ) = - 0 . 7 K gm K1 2 - 0.7 kym/S K2 = 6 . 9 x ( 4 - ( - 3 ) = 6. 30 kgm/s K2 = 6: 30 Kam/ s V. = final velocity of magnetic buck A K 1 - * 2 (- 7) - (6 30)- - 5 m/s mitm z 50 5 + 61 9 SpiralNegative Sign of MJ shows that it is directed to left Let Us - final velocity of cant Be U z = 4 + 1 - 5 ) - ( - 3 ) = 2 mis 12. is positive hence it is directed to Answers ( G ) Velocity of magnetic puckA after Collision U, = - 5 MLS (6 ) Velocity of magnetic puck B( my =0.90 KG ) outer collision Vz z zm/s