Two identical spheres A and B, each of m mass, attached toinextensible inelastic
ID: 1815243 • Letter: T
Question
Two identical spheres A and B, each of m mass, attached toinextensible inelastic cord length L. they rest at a distance afrom one another on horizontal surface (frictionless). SphereB has velocity v0 in direction perpendicular to line AB and moveswith out friction until hits B' when cord is taut. Find : a) magnitude of velocity of each sphere after becoming taut(immediately). b)energy lost as cord becomes taut diagram:http://img522.imageshack.us/img522/4781/ss20090429043101.pngPlease if you can solve this as soon as possible Find : a) magnitude of velocity of each sphere after becoming taut(immediately). b)energy lost as cord becomes taut diagram:http://img522.imageshack.us/img522/4781/ss20090429043101.png
Please if you can solve this as soon as possible Two identical spheres A and B, each of m mass, attached toinextensible inelastic cord length L. they rest at a distance afrom one another on horizontal surface (frictionless). SphereB has velocity v0 in direction perpendicular to line AB and moveswith out friction until hits B' when cord is taut. Find : a) magnitude of velocity of each sphere after becoming taut(immediately). b)energy lost as cord becomes taut diagram:http://img522.imageshack.us/img522/4781/ss20090429043101.png
Please if you can solve this as soon as possible
Explanation / Answer
let the angle = angle B'AB, so cos = a/L. after taut, the velocity of A is vA, that of B isvA + u (vector sum), where vA is along AB'and u is perpendicular to vA momentum conservation: mvo = mvA + m(vA + u), or vo = 2vA + u, (vector sum) the direction to the right: 0 = 2vAcos -usin, so 2vA =usin/cos the up direction: vo = 2vAsin +ucos, so vo = usin2/cos +ucos = u/cos u = vocos then vA = vosin/2 = vo/2 *(1 - a2/L2) the magnitude of the vector sum vA + u is(vA2 + u2) =vo(sin2/4 +cos2) = vo/2 * (1 +3cos2) = vo/2 * (1 +3a2/L2) a) magnitude of velocity of A = vA = vo/2 *(1 - a2/L2) magnitude of velocity of B = vB = vo/2 *(1 + 3a2/L2) b) energy lost = mvo2/2 - mvA2/2 -mvB2/2 = mvo2/2 * [1 - (1/4) (1 -a2/L2) - (1/4) (1 +3a2/L2)] = mvo2(1 - a2/L2)/4
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