### Video Transcript

A mass π one rests on a smooth horizontal table. It is connected by a light inextensible string passing over a smooth pulley fixed at the edge of the table to another mass π two hanging freely vertically below the pulley. A mass of 6.69 kilograms was added to π one. When this system was released from rest, it accelerated at seven twentieths π. Another mass of 6.75 kilograms was added to π one. As a result, the acceleration of the system slowed to thirteen fiftieths π. Determine π one and π two. Take π to equal 9.8 meters per second squared.

We can start off by drawing a diagram of this situation. We start off with two masses, π one and π two, connected by a string over a smooth pulley. π one sits on a surface that offers no frictional resistance. With the system at rest, we add a mass to the top of π one. We can call it π sub π, where π sub π is given as 6.69 kilograms. After adding π sub π, the system is released and allowed to move with π two descending and pulling π one and ππ along with it. Under these conditions, the acceleration of the system, which we can call π sub π, is equal to seven twentieths times the acceleration due to gravity, where we count π as exactly 9.8 meters per second squared. While the system is in motion, we add a third mass to the stack. We can call it π sub π which has a mass of 6.75 kilograms. Under these conditions, the systemβs acceleration slows to a value we can call a sub π, thirteen fiftieths times π.

Based on this information, we want to solve for the mass values π one and π two. To do that, we can begin by considering Newtonβs second law which tells us that the net force on a system is equal to the mass of the system multiplied by its acceleration. If we consider the forces in our scenario, we know that the table top is smooth and provides no frictional resistance, and also that the pulley turns without resistance itself. Each of the masses ππ, ππ and π sub π have weight forces. But all of those act in the vertical direction while motion is in the horizontal. So the weight forces of those three masses do not affect the motion of our system. The only force affecting the motion of our system is the weight force of mass π two which equals π two times π.

If we define motion in that direction as positive motion and we rewind our scenario a bit so that weβve only added π sub π to the stack of masses, meaning that the system is accelerating at π sub π, then we can write that the weight force of mass π two, π two π, is equal to the mass of our system multiplied by π sub π, the acceleration under these conditions. Our system, at this time, consists of our three masses π two, π one and ππ. So we substitute their sum in for π. Rearranging then to solve for π sub π, we find itβs equal to π two times π divided by the sum of the three masses involved. And we also know that itβs equal to seven twentieths π. Now that we have this equation, we can store it off to the side and let it help us later on to calculate π one and π two. What we have with this equality is essentially an equation of motion for our system when it consists of π one, π two, and ππ.

Weβll now add ππ to the system and develop another equation of motion. With two equations of motion and two unknowns, π one and π two, weβll be able to solve for the information we want. So considering the system with all four masses involved, still the only force causing motion in this system is the weight force of π two. So now, when we apply Newtonβs second law to our scenario, we write π two π equals the sum, now, of all four of our masses multiplied by the acceleration of our system under this condition, which is π sub π or thirteen fiftieths π.

We can rewrite this expression to solve for π sub π. Itβs π two times π divided by the sum of the four masses which equals thirteen fiftieths times the acceleration due to gravity. This is our second equation of motion. With these two separate equations and two unknowns we want to solve for, we can begin by solving for π one in terms of π two, using the first equation we derived.

Starting with that equation, weβll rearrange it algebraically to solve for π sub one. We see in this equation that the factor of π cancels from both sides. And if we multiply both sides of the equation by the denominator of the left side and rearrange, we find that π sub one is equal to thirteen-sevenths π sub two minus π sub π. Weβll now take this result and substitute it in for π sub one in our second derived equation. When we do, we see the factors of π once again cancel out, as does the mass π, π sub π, in the denominator of our left side.

We now have an expression that involves only π sub two, our unknown, and otherwise known values. Rearranging, we find that π sub two equals ππ over fifty thirteenths minus twenty-sevenths. Plugging in for the given value of ππ of 6.75 kilograms, when we enter this value on our calculator, we find a result of 6.825 kilograms. Thatβs the value of mass π two.

We can now use our equation for π one, in terms of ππ and π two, to solve for π one. π two, we now know is 6.825 kilograms. And ππ is given as 6.69 kilograms. Entering these values on our calculator, we find 5.985 kilograms. Thatβs the value of the mass π one.