A 24 kg mass and a 13 kg mass?

Question

are suspended by a pulley with a radius of 4.9 cm and mass of 6 kg. Cord is negligible in mass, pulley rotates witout slipping or friction. The masses start from rest 4 m apart (on is higher than the other), the pulley is a uniform disk. Acceleration of gravity = 9.8 m/s^2.

Calculate the speeds of the two masses as they pass each other in units of m/s.

Answer 1

F=Ma.
Let T= tension in the cord.

For 24kg mass:

F=24g-T
24g-T=24a
T=24g-24a

For 13kg mass:

F=T-13g
T-13g=13a

Substitute value of T for 24kg mass:

24g-24a-13g=13a
37a=11g
a=11g/37
=2.91m/s^2

Compute time t:

s=ut+1/2at^2

For 24kg mass, s=x

x=0+1/2*2.91t^2

For 13kg mass, s=4-x

4-x=0+1/2*2.91t^2

Substitute value of s for 24kg mass:

4-1/2*2.91t^2=1/2*2.91t^2
2.91t^2=4
t^2=4/2.91
t=1.17s

Speed v of 24kg mass is equal to speed of 13kg mass

v=u+at

v=0+2.91*1.17
v=3.4m/s

The assumptions here are that tension T in cord for 24kg mass equals that for 13kg mass, acceleration of 24kg mass equals that of 13kg mass, and v of 24kg mass equals v of 13kg mass. The 24kg mass and 13kg mass meet at point x from 24kg mass, and at 4-x from 13kg mass.

Answer 2

the diameter and mass of the pully are irrelevant. The way it is stated you could just take the easy way out and say the 13 kg body is the one tha is higher to start, then the object never pass each other. You can treat this like a problem of an 11kg mass dropping 2 m, what is it's velocity. The other object will have the same velocity but in the opposite direction.

Answer 3

3.4 m/s

Conservation of energy method is easiest:

Reference level is midway between weights at 2m.

So initial energy is all potential energy:

total energy = m1gh1 + m2gh2 = 24g(2) + 13g(-2) = 22g

as the weights travel past each other at the midway point:

m1v^2/2 + m2v^2/2 = 22g
37v^2 = 44g
v = sqrt(44g/37) = 3.4 m/s