October 29th, 2016, 02:23 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  system of particles
Two particles of masses 3 kg and 5 kg are connected by a light inextensible string, of length 4 m, passing over a light smooth peg of negligible radius. The 5 kg mass rests on a smooth horizontal table. The peg is 2.5 m directly above the 5 kg mass. The 3 kg mass is held next to the peg and is allowed to fall vertically a distance 1.5 m before the string becomes taut. does any one know what this question looks like i cant even imagine it

October 29th, 2016, 07:18 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387  Quote:
 
October 30th, 2016, 12:51 AM  #3 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
Show that when the string becomes taut the speed of each particle is 3root3g all over 8 m/s i tried working this out by looking at the 3kg weight and using the equation v=u+2as u=0 a=g s=1.5 however working this out gives me an answer of root3g this is not the desired answer 
October 30th, 2016, 06:25 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387  Quote:
final velocity of the 5 kg mass = $v_f$ initial velocity of the 5 kg mass = $0$ final velocity of the 3 kg mass = $v_f$ because its motion is opposite in direction to the 5 kg mass initial velocity of the 3 kg mass = $\sqrt{3g}$ $Mv_f = m(v_f  v_0)$ solving for $v_f$ ... $v_f = \dfrac{mv_0}{M+m} = \dfrac{3\sqrt{3g}}{8}$  

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