Any help would be great thanks
A light inextensible string passes over a fixed smooth pulley. Particles A and B, of o.o3kg and o.o5kg respectively, are attached to the ends of the string. The system is held at rest with A and B at the same horizontal level and the string taut. The two parts of the string not in contact with the pulley are vertical. The system is released at time t=0, where t is measured in seconds, the particle B moves downwards for 2s before it hits the floor. The string then becomes slack and B remains a rest. Neglecting air resistance, show that the string becomes taut again when t=3.
When released, there is a difference in forces, so the system moves down.
Originally Posted by gracey
This is F = 0.05*9.8 -0.03*9.8 = 0.196 newton
The system will accelerate downward
F = m*a
0.196 = (0.05 +0.03)a
a = 0.196 /0.08 = 2.45 m/sec/sec
When B hit the floor, A continued moving up because of inertia.
A will stop moving after somer time, t1 second
V = Vo -g(t1) -------------(i)
V = final velocity of A after t1......which is zero
Vo = velocity of A at t = 2 sec
Vo = (initial velocity) +a*t
Vo = 0 +2.45(2) = 4.9 m/sec
So, substitute those into (i),
0 = 4.9 -9.8(t1)
t1 = 4.9 /9.8 = 1/2 sec
Total time now is 2 +1/2 = 2.5 sec after the release.
In that 0.5 sec it moved upward some more before stopping, A traveled:
h1 = Vo*t -(1/2)g(t^2)
h1 = 4.9(0.5) -(1/2)(9.8)(0.5)^2
h1 = 1.225 meters
Then, from rest, A will fall down.
When it falls down, it will travel 1.225 m and then the string will become taut again because by then, A will start pulling up B.
h1 = (initial velocity)*t2 +(1/2)g(t2)^2 ....."+" because going down is positive here
1.225 = (0)(t2) +(1/2)(9.8)(t2)^2
(t2)^2 = 1.225 / 4.9 = 0.25
t2 = 0.5 sec
So, total time now after release is 2.5 +0.5 = 3 seconds.