Question

What is the magnitude of the acceleration of the masses?

Two masses are connected by a weightless cord, which passes over a frictionless pulley. The masses are held stationary and then released. The acceleration due to gravity is g.

Answer 1

`g/4`

Explanation:

For mass `m_1`

`T-m_1 g = m_1 a_1`

for mass #m_ 23

`T-m_2 g = m_2 a_2`

assuming the weightless cord is inextensible

`a_1 = -a_2 = a` then

`a = ((m_2-m_1)/(m_1+m_2))g = 1/4 g`

Answer 2

g/4`(m/s^2)` for 5 kg it's downwards and for 3 kg it's upwards

Explanation:

Clearly the 5 kg will go downwards and as a result the cord will try to pull the 3 kg above.
Let tension in cord is T
So,for 5 kg we can write,
`5g-T = 5a` ..1( a is the acceleration of the system)
For, 3 kg we can write,
`T- 3g = 3a` ..2
Eliminating T from both we get, a= `(5g-3g)/(5+3)` m/`(s^2)`
I.e `g/4` m/`(s^2)`