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Direct central impact occurs between a 300N body moving toHi all 1 ) Direct central impact occurs between a 300N body moving to the right with velocity of 6 m/s and 150 N body moving to the left with velocity of 10 m/s . Find the velocity of each body after impact if the coefficient of restitution is 0.8 . my answer : [color=#BF4000]Given [/color]:: m1 = 300 /9.81 = 30.58 kg u1 = 6 m/s m2 = 150/9.81 = 15.2 kg u2 = -10 m/s e = 0.8 V 1 ? v2 ? m1u1+m2u2 = m1v1+m2v2 30.58X 6 + 15.2 X (-10) = 30.58v1+ 15.2 v2 183.48 -152 = 30.58v1+ 15.2 v2 31.48 = 30.58v1+ 15.2 v2 ===== [color=#FF0000]this is first equation (1)[/color] e(u1 - u2) = (v2 - v1) 0.8(6 + 10) = v2 - v1 4.8 + 8 = v2 - v1 12.8 = v2 - v1 v2 = 12.8 + v1 ===== [color=#FF0000]this is first equation (2) [/color] make v2 in fist equation 31.48 = 30.58v1+ 15.2 (12.8 + v1) 31.48 = 30.58v1+ 194.56 + 15.2 v1) 31.48 - 194.56= 30.58v1 + 15.2 v1 -163.08 = 45.78 [color=#0040FF]v1 = -3.6 m/s [/color] [color=#0040BF]to find v2 = 12.8 + (-3.6) = 9.2 m/s[/color] |

Re: Direct central impact occurs between a 300N body moving Let the body initially moving to the right be body A and the other be body B, hence: We are given: Hence: |

Re: Direct central impact occurs between a 300N body moving that my answer is correct same which you are got |

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