Question

> $ d 8:12 PM 7./29/2020 X X + 0 10 Module 3: Planar Kinematics of A PowerPoint Presentation Mastering Engineering Mastering X CA Stop Block, S. Prevents A Crat x https://session.masteringengineering.com/myct/itemView?assignmentProblemID=11103181&offset=prev UTRGV Register your busin... Get federal and stat. My Drive - Google.. C Engineering Mecha... A My Giving - The We... NP 9 VED Other favorites Homework for MO2.5 Problem 15.64 19 of 2 > Review Bal A has a mass of 3 kg and is moving with a velocity of (UA): 9 m/s when makes a direct collision with ball B. which has a mass of 25 kg wis moving with a velocity of (0) = 5 m/s Suppose that c = 0.7. Neglect the size of the balls. Figure 1) Part A Determine the velocity of Ajust after the collision measured to the night Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is directed to the right and negative value if the velocity is directed to the batt HA + BI? Value Units Submit Request Am. Part 8 Determine the veloty or just after the collision measured to the right Express your answer to three significant figures and include the appropriate units. Enter positive value of the velocity is directed to the right and negative value of the velocity directed to the left th Figure < 1 of 1 > Value Units Submit Request Answer Next > Provide Feedback

I

Answer 1

Given data Ball A: Ball Bi MA = 3kg Befde collision (VA), = 9 m/s after collision (VA), = ? MB = 2.5kg Before collision (VB), = -5m/s Negative sign indicates that bau B moving towards left direction. After collision ball is said to be moving towards right. so take (VB), as positive. : 0.7 coeficient o restitution : (MB), Olet . Before collision (1972 ("B12 Ofter collision from laws of conservation of linear momentum before equal to momentum collision is to collisio momentum after collision ñ MA (VA), + MB (VB), = MA(VA)+ Mg(VB) 3(9) + 2-5(-5) = 3 (Val2 + L05( 48), Ja? + 2.5 (V) 14.5. restitution , ea Voly-V), (M), Coefficient of

0.7 (VB2-VA), 9-(-5) (2-) : 9.8 (ven - (V2 +9.8 Substitute ® in equ o 3. VAL + 2.5 (VB)2 = 14.5 : -10 = 3(VA), + 2.5 {(+ 9.8] - 14.5 5.5W), (VA), -1.818 m/s CV = (VA), +9.8 =-1.818 +9.8 NB) = 7.981 m/s 2 il after collision , balls move balls move away from each other 7,981 m/s 1.888 m/s B A