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 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
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Quote:
 Originally Posted by markosheehan 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
sketch attached
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 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
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Quote:
 Show that when the string becomes taut the speed of each particle is 3root3g all over 8 m/s
conservation of momentum ...

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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