IIT JEE , AIEEE Forum - Physics , Mechanics

ishan

2 identical particles each of mass 4 kg are placed together at the ge of a tall cliff and joined by a string of negligible mass one of the particles is projected with a speed of 10*3^1/2m/s at 60 degree above the horizontal.the string becomes taut after 2 seconds assume g=10m/s^2.whatis the length of the string and what is the speed with which the second particle starts moving and what is the tention when the second particle starts moving?also find the time when the string becomes horizontal for the first time after projection?(answers 20m,2.5m/s,7.5N,3.2s)

krishna.gopal

Ishan here I am giving solution of first two parts. For other two parts even I am not very clear so i don't want to say anything right now.

Let us call the particle which is projected as A and the other particle as B.
Initially uxA (initial velocity of A along x axis)=ucos(60)=10sqrt(3)*cos(60)=5sqrt(3)
Initially uyA (initial velocity of A along y axis)=usin(60)=10sqrt(3)*sin(60)=15 m/s.

Displacement os A in 2 sec
xA=uxA*t=10*sqrt(3)
yA=uyA*t-(1/2)gt^2=10 m
At this stage string will becomes taut. So length of string =sqrt( 10^2 +(10*sqrt(3)^2)
=20m

At this point the string makes an angle of taninverse(10/(10*sqrt(3))=+30 degree with horizontal

Velocity of A just before this is
vxA(2) = uxA = 5sqrt(3) m/s
vyA(2) = uyA-gt = 15-10*2=-5m/s
So velocity of A is 10 m/s making an angle of -30degree with horizontal,

So velocity makes an angle of 60 degree with direction of string. It will have two componets. 5 m/s along string and 5sqrt(3) m/s perpendicular to string. A will continue to move with 5sqrt(3) m/s perpendicular to string as string cannot exert force perpendicular to it. But velocity along string will decrease with coresponding increase in velocity of B.
Let after string becomes taut velocity of A and B along string is v.
Momentum along string just before t=2sec = 4*5=20kgm/s
Momentum along string justsftere t=2sec =(4+ 4)*v=20kgm/s
v = 2.5 m/s

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