Main Menu | Contents | Grades | Chat | Syllabus [ Course Contents > Homework Set 6 (due October 13) > 7.38 " Timer Notes Evaluate Feedback Print An m = 6.60-kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22.0 m/s and shatters into three pieces, which all fly backward, as shown in the figure. The wall exerts a force on the ball of 2620 N for 0.110 s. One piece of mass m, = 2.60 kg travels backward at a velocity of 10.9 m/s and an angle of = 32.0 above the horizontal. A second piece of mass my = 1.60 kg travels at a velocity of 8.40 m/s and an angle of 28.0 below the horizontal. m a) What is the speed of the third piece? Submit Answer Incorrect. Tries 4/99 Previous Tries 22 m/s 320 28.T m b) What is the direction of the third piece? (Use positive values for an angle above the horizontal, and negative for below the horizontal.) m2 Submit Answer | Tries 0/99m Vi = 22 m/s 32 28" m m2 Here we will use the Change of momentum of the ball the to collision with the wall along al and I directions . Let the mass of the 382 ball is my then, my = Im-( mitmz) ) - 6+60 - (2-60 + 1.60 )] 19. Jm = 2- 40 kg And also let after collision the division my makes an angle o with the horizontal - Ve re direction as shown in the figure .Change of momentum of ball due to the force exerted by the wall along a direction is AP = FE = ( 26 20 N . XO . 11 0 s. ) = 288. 2 So for change of momentum of ball along the a direction we can write, AP = mv- (- miv,cos 320 - m2V 2 cos 28 - M3 V3 Cos 0 ) where, V. , V2 & V3 are the velocities of the makes my, my & my respectively. So, 288.2 - (6- 60 x 22 ) = 2. 60X 10.9 LOD 320 1 . 60X 8.4 Cos 280 + 2. A V 2 GoSD 143 = + 24 + 11 .9 + 2 . A V3 cose + 2.4 V3 Lose = 143 - 24-11.9 + 2 . 4 V/3 COSO : 2 = 107.1V2 Case = 107 . 2. 4 - V3 Cogo = 44.625 Along y direction . momentum of ball remains Same, so, 0 = Miv , Sin 320 - mcV z sin 280 + myV 3 sine 30 = 2.60 x 10.9 Sin 32- 1.60x8. 40 Sin28" + 2:4 V3 sing = 15 - 6-3 + 204 93 sie - 2.A V3 sing = 6:3-15 - 2. 4 V/ 3 sing = - 8.7 * V3 Sino - 8.7 2. 4 -7 V/3 Sino = - 3.6By Of Q) we get, V3 = /V/3 costo + v, sinto V ( 41. 625)2 +(-3-12 m/s . V 3 - 14. IF m/s. and V3 sing - 3.6 tang = V3 ZeSo 44- 625 tang = - 0.08 o = tan'( - 0-08 ) 0 = - 4.60