### Video Transcript

A body of mass 8.1 kilograms was resting on a smooth plane inclined at an angle of πΌ to the horizontal, where tangent of πΌ equals four-thirds. The body was connected to the end of a string, passing through a pulley fixed at the top of the plane. At the other end of the string a body weighing 26.9 kilograms was hanging freely. The system was released from rest and started moving. And two seconds later, the string snapped. Find the distance the body on the plane ascended after the string broke and before it momentarily came to rest. Take π to equal 9.8 meters per second squared.

Letβs start out by writing down the information weβve been given. The mass on the incline, which we can call π sub one, is equal to 8.1 kilograms. The mass on the other end of the string, which we can call π sub two, is equal to 26.9 kilograms. π sub one sits on an inclined plane inclined at an angle πΌ, where the tangent of the angle πΌ is equal to four-thirds. Weβre also told there is a time interval of two seconds, which weβll call π‘, during which the two masses are connected by a string and moving with mass one up the plane. Weβre also told to take the acceleration due to gravity π as 9.8 meters per second squared.

Here is a diagram of our setup. We have two masses, π one and π two, connected by a string that runs over a frictionless pulley. Weβre told that the plane that π one is on is smooth, meaning that there is no friction between this plane and the mass π one. Under these conditions then, if we release this system from rest, π two will descend and π one will move up the plane. Weβre told that this goes on for a time of two seconds, after which the string that connects the two masses is snapped.

After that happens, π two is no longer able to pull π one up the incline. π one is simply moving under its own inertia. After the string has snapped, because of its inertia, π one will continue moving up the incline some distance we can call π. Itβs this distance π that we want to solve for. As we start, letβs put a pair of coordinate axes on our diagram, with the positive π₯ motion being motion up the plane and positive π¦ motion being motion perpendicular to it.

Now itβs π one whose motion we are primarily concerned with. So letβs start out by drawing the forces that are acting on π one. In the first case, there is the force of gravity on π one. That force is equal to π one times π. There is also a normal force pushing π one away from the plane. And finally, there is a force on π one due to the tension in the string. We can call that force capital π. Recalling Newtonβs second law, that the net force on an object is equal to its mass times its acceleration, we can apply this law to the forces on π one acting in the π₯-direction.

The two forces acting on π one with components in that direction are the tension force π and the weight force π one times π. Considering the weight force, we can break that force up into π¦- and π₯-components, forming a right triangle where the angle in the upper corner of that triangle is equal to πΌ. So the π₯-component of the gravitational force is equal to π one times π times the sine of the angle πΌ. So by Newtonβs second law, we can write that the tension force π minus π one π times the sine of πΌ is equal to the mass of our system times its acceleration π. But what is that tension? And what is the mass of our system?

Maybe itβs just π sub one. Maybe itβs just π sub two. As we look back at our diagram, we see that our system essentially does include the second mass π sub two. Itβs that mass that creates tension in the string, which then supplies the tension force weβve called π. Specifically, itβs the weight force of mass two that provides that tension. So we can replace π in our equation with π two times π. And since the two masses π two and π one are in motion together as long as the string is intact, the mass of our system is equal to the sum of π one and π two.

If we algebraically rearrange this entire expression in order to solve for the acceleration π, we find that is equal to π times the quantity π two minus π one sine πΌ divided by the sum of our two masses. This acceleration π is the acceleration of π one while it moves up the plane under the influence of the tension of the string. But we know that that string tension is not in place forever. After a time of two seconds, that tension goes away as the string is snapped.

So really, what we love to be able to solve for is the speed of π one, weβll call it π£, when time is equal to two seconds; that is, at the instant that the string is cut. Now as we look back at the acceleration of our mass π sub one under the influence of the stringβs tension, we see that that acceleration consists of constant terms; that is, terms that do not change over time. This implies that acceleration itself is constant, meaning that we could use the fact that acceleration is equal to the change in velocity over the change in time.

Since our system of masses started at rest, that means the speed of mass one at time π‘ equals two seconds is equal to Ξπ£ and that Ξπ‘ is equal itself to two seconds. This means we can take the expression π£ when π‘ equals two seconds, divide it by π‘, and substitute it in for π in our equation. When we do that and then multiply both sides of the equation by the time π‘, we find this expression for the speed of π one at the instant when the string is snapped. At that instant, the conditions in our scenario change significantly. There is no longer attention force acting on π one.

In fact, we can call this speed of π one at that instant in time π£ sub π, the speed of π one at the instant that the string is cut. Under these new conditions, when the string is no longer connecting the two masses, we can redraw our diagram to reflect that new information. Mass two is no longer in the picture because the two masses are not connected. Likewise, there is no longer a tension force π because the string is not intact. This means that now, from a force perspective, the only force acting on π one in the π₯-direction is the weight force component down the plane.

Recall that component of the weight force is π one π times this sine of πΌ. Since thereβs no longer a tension force acting on π one, we can reapply Newtonβs second law to π one for the forces in the π₯-direction. Now we can write that negative π one π times the sine of πΌ is equal to the mass of our system, now just π one, times its acceleration, which weβve called π sub π to symbolize that is its acceleration after the string has snapped.

π sub one is common to both sides and so cancels out. So we see that the acceleration of mass one after the string is cut is equal to negative π times the sine of πΌ. Now all the terms in this side of the equation are constant. π doesnβt change and either does the angle πΌ, which means that this acceleration is constant too. That means weβre able to draw on a set of equations called the kinematic equations, which apply whenever acceleration is constant.

One of these four equations holds that an objectβs final velocity squared is equal to its initial velocity squared plus two times its acceleration times its displacement. When we apply this relationship to our scenario, because we wanna solve for the maximum distance π one slides up the plane, its final speed therefore will be zero. π£ sub π, its initial speed; that is, its speed at the moment the string snaps, is something weβve solved for in terms of our variables. Its acceleration π is π sub π, the acceleration after the string is snapped. And the distance π is what we want to solve for.

When we rearrange this equation algebraically to solve for π, we find it equals negative π£ sub π squared over two π sub π. If we plugin for π£ sub π and π sub π, then we can see in this expression that a factor of the gravitational acceleration π cancels out in the numerator and the denominator and so does the minus sign leading in the numerator and denominator. Now that weβve written our equation for π, we can prepare to plug in for each of the values in this expression.

As we walk through these terms one by one, we see that the acceleration due to gravity π is given to us in the problem statement as is the time value π‘. π two and π one are also given in values of kilograms, which means the only remaining unknown is the sine of πΌ. Now we donβt know the sine of πΌ, but we do know that the tangent of that angle is equal to four-thirds. Now if we were to draw out a right triangle and label the hypotenuse β and the two legs π₯ and π¦, then if π was the angle in the lower left-hand corner, that would mean that the tangent of that angle π is equal to π¦ over π₯.

Now in our triangle with the angle πΌ, weβre told the tangent of that angle is equal to four divided by three. Since this is a right triangle, we can tell that the hypotenuse β will have a value of five. Itβs a 3:4:5 triangle. This means weβre now able to figure out what the sine of πΌ is. Itβs equal to π¦ over β or four over five. With that, weβre ready to plug in and solve for π. With our values for π, π‘, π two, π one, and the sine of πΌ plugged in, weβre ready to enter all these terms on our calculator.

When we do, we find that π is 8.34 meters. Thatβs how far π one would slide up the plane after the string has snapped before it comes to a stop.