Question

Two objects with masses 4.20kg and 2.20kg hang 0.500m above the floor from the ends of a cord 7.30m long passing over a frictionless pulley. Both objects start from rest.

a) Find the maximum height reached by the 2.20kg object.

Answer 1

The 2.20 kg object will accelerate upward until the 4.20 kg object strikes the floor, let's look at this segment of the motion of the object first

4.20*a=4.20*g-T where a is the acceleration downward and T is the tension in the cord

or

T=4.20*(g-a)

and
2.20*a=T-2.20*g where T is the tension again and a is the acceleration upward towards the pulley

or

T=2.20*(a+g)

Solve for a

4.20*(g-a)=2.20*(a+g)

a=(4.20-2.20)*9.81/(2.20+4.20)

a=3.06m/s^2

The motion of the object under this acceleration will be constant for y=0.500 m from the starting y of 0.500m lets look at what the velocity will be when the object reaches 0.1000 m

first
0.1000=0.500+.5*3.06*t^2 (v0=0 because they start from rest, solve for t)

t=sqrt(0.1000/3.06) = 0.1807 s
v=3.01*0.1807 = 0.544 m/s

Now, the motion of the 2.20 kg object will have a new dynamic because it will no longer be subject to a tension in the cord.
Therefore the new acceleration will be -g, and the new equation of displacement for the object will be
y=0.1000+.544*t-.5*g*t^2
and
vy=.544-g*t

hen vy=0, the object reaches apogee
t=0.0554/g
and
y(apogee)=0.1000+.544^2/g-.5*g*.544^2/g^2
or
y(apogee)=0.1000+.5*.544^2/9.81 = 0.115 m above the floor