# 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 between the cart and the horizontal table the cart is on. The pulley is a uniform solid disk of mass 2m that rotates without friction about an axle passing through its center. A. Accounting for the mass of the pulley, what is the ball's acceleration, after the system is released from rest? a._ g  b._ g/2  c._ g/3  d_g/4 e._ g/5 f._ g/6 g._ g/7 B. Accounting for the mass of the pulley, what is the tension in the vertical part of the string after the system is released from rest? a._ mg b._ mg/3 c._ mg/2 d._ 2mg/3 e._ 3mg/4 f._ 4mg/5 g._ 5mg/6 h._ 6mg/7 C. Accounting for the mass of the pulley, what is the tension in the horizontal part of the string after the system is released from rest? a._ mg b._ mg/3 c._ mg/2 d._ 2mg/3 e._ 3mg/4 f._ 4mg/5 g._ 5mg/6 h._6mg/7 D. Accounting for the mass of the pulley, after the ball has accelerated through some distance, how is the system's kinetic energy split between the cart, the pulley, and the ball? a._ the cart has 4/7 of the kinetic energy, the pulley has 2/7, and the ball has 1/7b._ the can has 80% of the kinetic energy and the ball has 20%; the pulley has none because it is not experiencing straight-line motionc._ the cart has 2/3 of the kinetic energy and the pulley and the ball each have 1/6d._ the cart has 60% of the kinetic energy and the pulley and the ball each have 20%e._ the cart has 60% of the kinetic energy and the ball has 40%; the pulley has none because it is not experiencing straight-line motionf._ each has 1/3 of the kinetic energyg._ the cart and the ball each have 50% of the kinetic energy; the pulley has none because it is not experiencing straight-line motion

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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...

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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.

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