Physics Help Forum Two Impulse + Momentum related questions

 Apr 12th 2009, 03:18 PM #1 Junior Member   Join Date: Apr 2009 Posts: 4 Two Impulse + Momentum related questions First Question A 2.5 kg ball and a 5.0 kg ball have an elastic collision. The 2.5 kg ball was at rest and the other ball had a speed of 3.5 m/s. What is the kinetic energy of the 2.5 kg ball after the collision? I did 5 x 3.5 = 17.5 so 2.5(v) = 17.5 v = 7 m/s kinetic energy which is 1/2 m v^2 .5 x 2.5 x 7^2 = 61.25 J I got 61.25 J, however the answers have to be.. a) 1.7 J b) 8.1 J c) 3.4 J d) 14 J e) 27 J What is the answer? and how do you come to it? 2nd Question A 5.0 kg boy runs at a speed of 10.0m/s and jumps onto a cart. The cart is initially at rest. If the speed, with the boy on the cart, is 2.50 m/s, what is the mass of the cart? mv + m2v2 = m'v' (5)(10) + m2(0) = (M2 + 5) (2.5) 50 = 2.5M + 12.5 M = 15 kg is what I did, but the answers have to be either.... a) 150J b) 210J c) 175J d) 260 J e) 300J what is the answer and what am I doing wrong? can someone help?? thanks
 Apr 12th 2009, 07:53 PM #2 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 Edit: post deleted as I realised it was incorrect. Correct solution posted separately Last edited by Parvez; Apr 14th 2009 at 11:43 AM.
 Apr 13th 2009, 09:22 AM #3 Junior Member   Join Date: Apr 2009 Posts: 4 I'd love to think so too man, but those are the only options and I don't know what i'm doing wrong
 Apr 13th 2009, 09:55 AM #4 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 I again ask how the mass can be in Joul. Either the question is wrong or the answer.
 Apr 13th 2009, 12:57 PM #5 Junior Member   Join Date: Apr 2009 Posts: 4 yea sorry its a mistake by me its supposed to be 150 kg, 210 kg .... etc. so that was my fault for that
 Apr 14th 2009, 11:41 AM #6 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 First Question M = mass of the larger block m = mass of smaller block u = velocity of larger block before collision V = velocity of larger block after collision v = velocity of smaller block after collision. As the collison is elastic, kinetic energy and the momentum will remain conserved. Hence- (1/2) Mu^2 = (1/2)MV^2 + (1/2)mv^2 (conservation of energy) Mu^2 - MV^2 = mv^2 M(u+V)(u-V) = mv^2 ............(1) Next Mu = MV + mv (conservation of momentum) M(u-V) = mv .................(2) Divide eq (1) by eq (2) (u+V) = v V=v-u put this value of V in eq (2) and you get v= 2Mu/(M+m) = 2*5*3.5/(5+2.5) Kinetic nergy of this (smaller) ball = (1/2)mv^2 = 27.2 J
 Apr 14th 2009, 11:46 AM #7 Junior Member   Join Date: Apr 2009 Posts: 4 thank you so much for solving the first question. any ideas on the secound? my initial thought would be a) only because i keep getting 15 and it seems relevant but hey maybe you got better ideas
 Apr 14th 2009, 12:00 PM #8 Physics Team   Join Date: Feb 2009 Location: India Posts: 365 I tried my best. Seems no way to get the answer. Please check the answers once again. you might have read kg as 0j ( 150J --> 15kg)

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