A ball of mass m is tied to a cart of mass 4m by a string that passes over a pulley. The cart is on wheels, which allows us to neglect friction...

Hello!

Denote the downward speed of a ball as `V(t).` A cart and a point on an outer edge of a pulley have the same speed. I'll use a downward y-axis, starting from the initial position of a ball.

It is known that the kinetic energy of a rotating disk is `1/4 m_d V^2,` and for linearly moving ball and cart it is `1/2 m_b V^2` and `1/2 m_c V^2.` The potential energy with respect to the y-axis I use is `-m_b gh,` and the total energy of the system is conserved. This way we obtain an equation

`(1/2 m_b + 1/2 m_c + 1/4 m_d) V^2 = m_b gh.`

Recall `m_b=m,` `m_c=4m` and `m_d=2m` and obtain

`(1/2 + 2 + 1/2)m V^2=mgh,` or `3V^2(t)=gh(t).`

This is a differential equation because `V(t)=h'(t).` It is simple to solve it and the solution is `h(t)=g/12 t^2.` So the starting acceleration is `V''(0)=g/6` and the answer for A is f.

For B use Newton's Second law, `F=ma.` Net force `F` here is equal to `mg-T,` where `T` is the tension force, so `T=m(g-a)=m(g-g/6)=5/6 mg,` the answer is g.

For C is the same idea but applied to a cart, the answer is `T=m_c*a=4m*g/6=2/3 mg,` the answer is d.

D (about kinetic energy). We saw that the kinetic energies have the ratio 1/2:2:1/2, i.e. 1:4:1 or 1/6:2/3:1/6, so the answer is c.