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 June 12th, 2012, 06:54 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Direct central impact occurs between a 300N body moving to Hi 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]
 June 13th, 2012, 09:27 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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: $v_a=\frac{m_a u_a+m_b u_b+m_bC_R$$u_b-u_a$$}{m_a+m_b}$ $v_b=\frac{m_a u_a+m_bu_b+m_aC_R(u_a-u_b)}{m_a+m_b}$ We are given: $m_a=\frac{300}{9.81}\:\text{kg}=\frac{10000}{327}\ :\text{kg}$ $u_a=6\text{ \frac{m}{s}}$ $m_b=\frac{150}{9.81}\:\text{kg}=\frac{5000}{327}\: \text{kg}$ $u_b=-10\text{ \frac{m}{s}}$ $C_R=0.8$ Hence: $v_a=-3.6\text{ \frac{m}{s}}$ $v_b=9.2\text{ \frac{m}{s}}$
 June 14th, 2012, 05:38 AM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: Direct central impact occurs between a 300N body moving that my answer is correct same which you are got

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