In the figure, two identical containers of sugar are connected by a cord that passes over a frictionless pulley. The cord and pulley have negligible mass, each container and its sugar together have a mass of 620 g, the centers of the containers are separated by 55 mm, and the containers are held fixed at the same height. What is the horizontal distance between the center of container 1 and the center of mass of the two-container system (a) initially and (b) after 38 g of sugar is transferred from container 1 to container 2? After the transfer and after the containers are released, (c) what is the acceleration of the center of mass, taking up to be the positive direction?**Diagram Description:**The diagram shows two containers labeled 1 and 2, each hanging from the ends of a cord that passes over a pulley. The containers are shown at the same height, and the centers of the containers are separated by 55 mm.**Questions:**- **(a)** Enter the initial horizontal distance.- **(b)** Enter the horizontal distance after the sugar transfer.- **(c)** Enter the acceleration of the center of mass. Answer the questions in the provided fields, including appropriate units.

part(a) What is the horizontal distance between the center of container 1 andthe center of mass of the two-container system (a) initially and Part(b) after 20 g of sugar is transferred from container 1 to container 2? After the transfer and after the containers are released, part(c) what is the acceleration of the center of mass. Given, Distance of separation between the containersd=55 mm=55×10-3 mmass of each container with sugarm=620 g=620×10-3 kgamount of sugar transferred from container 1 to 2, mt=38 g=38×10-3 kg Part(a)∘Taking the origin of the coordinate system at the center of the pulley and considering that the x-axis is horizontal and y-axis is vertically downward.∘So, the center of pulley will be located halfway between the containers on the x-axis at x=0.∘let L is the vertical distance from the center of the pulley to each of the containers.∘Since, the distance of separation between the containersd=55 mm, Therefore, center of mass of the containers will be initilly located at a horizontal distance of 55 mm2=27.50 mm from each of the containers.Part(b)Now, amount of sugar transferred from container 1 to 2, mt=38 gSo, remaining mass of container 1 will be: m1=620 g-38 g=582 g at x1=-27.5 mmalso, mass of container 2 including transferred mass will be:m2=620 g+38 g=658 g at x2=27.5 mmSo, new center of massCOM will be:Xcm=m1x1+m2x2m1+m2substituting values,Xcm=582 g-27.5 mm+658 g27.5 mm582 g+658 g=1.685 mmTherefore, new center of mass will be at: 27.5 mm+1.685 mm=29.185 mm away from the m2 or lighter mass.So, after mass transfer the ceneter of mass of new container system will be 29.185 mm away from the lighter mass.Part(c)∘When the containers are leleased after mass transfer, heavier container will move downward and the lighter container will move upward.∘ Since, the rope is same, the magnitude of acceleration will be same forboth container but in the opposite direction.Let the magnitude of acceleration is a.Then , acceleration of center of massacm will be:acm=m1a+m2-am1+m2=am1-m2m1+m2From the free body diagram of each container.We have,T-m1g=m1a ....(1)and m2g-T=m2a ....(2)adding (1) and (2)T-m1g=m1a m2g-T=m2a m2g-m1g=m1+m2a⇒a=m2g-m1gm1+m2=-gm1-m2m1+m2Now, from above,acm=am1-m2m1+m2substituting value of a,acm=-gm1-m2m1+m2m1-m2m1+m2Or,acm=-gm1-m22m1+m22substituting values,acm=-9.81 m·s-2582-6582582+6582=3.7×10-2 m·s-2Therefore, acceleration of the center of mass is 3.7×10-2 m·s-2.

Answered: In the figure, two identical containers…