# Question Video: Analysis of the Motion of Two Bodies Hanging Vertically and Connected by a String Passing through a Pulley Mathematics

Two bodies A and B of equal mass π grams were connected to one another by means of a light inextensible string which passed over a smooth pulley. A mass of 44 g was added to body A and the system was released from rest. Body A hit the ground after moving 64 cm, whereas body B continued its motion upward until it momentarily came to rest 80 cm above its starting point. Find the value of π, given that the acceleration due to gravity π = 9.8 m/sΒ².

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### Video Transcript

Two bodies A and B of equal mass π grams were connected to one another by means of a light inextensible string which passed over a smooth pulley. A mass of 44 grams was added to body A and the system was released from rest. Body A hit the ground after moving 64 centimeters, whereas body B continued its motion upward until it momentarily came to rest 80 centimeters above its starting point. Find the value of π given that the acceleration due to gravity π equals 9.8 meters per second squared.

We will begin by sketching a diagram to model the situation. We have two bodies A and B connected by a light inextensible string that passes over a smooth pulley. As the pulley is smooth, the tension in the string will be equal throughout. The bodies have a mass of π grams and will therefore exert a downward force of this mass π multiplied by gravity π. We are told that this gravity π is equal to 9.8 meters per second squared, which is the same as 980 centimeters per second squared. A mass of 44 grams was added to body A. This means that it now exerts a downward force of π plus 44 multiplied by π. And distributing the parentheses, this is equal to ππ plus 44π.

We are then told that the system was released from rest. At this point, body A will accelerate downwards, and body B will accelerate upwards. As the string is inextensible, this acceleration π will be constant for the whole system. We can now use Newtonβs second law, πΉ equals ππ, for body A and body B. The sum of the forces acting on the body is equal to the mass multiplied by acceleration. As body A is accelerating downwards, we will assume this is the positive direction. And the sum of the forces is therefore equal to ππ plus 44π minus π. This is equal to the mass of the body π plus 44 multiplied by π. We will call this equation one.

Repeating this for body B, where the positive direction is vertically upwards, we have π minus ππ is equal to ππ. We now have a pair of simultaneous equations. By adding the two equations, we can eliminate the tension force π and find an expression for the acceleration π in terms of π and π. The left-hand side of our equation becomes ππ plus 44π minus ππ. And after distributing the parentheses, the right-hand side becomes ππ plus 44π plus ππ. This simplifies so 44π is equal to two ππ plus 44π.

Since π is 980, the left-hand side is equal to 43,120. And factoring π on the right-hand side gives us π multiplied by two π plus 44. We can then divide through by this such that π is equal to 43,120 over two π plus 44. We might wonder how this can help us solve this question. However, we will now consider what happens after the system was released. Body A fell a distance of 64 centimeters before it hit the ground. We can therefore use the equations of motion or SUVAT equations to find the velocity of this body when it hits the ground. Taking the positive direction to be vertically downwards, the displacement is 64 centimeters and the initial speed is zero centimeters per second.

We are trying to calculate the final velocity π£. And the acceleration is 43,120 over two π plus 44 centimeters per second squared. We will use the equation π£ squared is equal to π’ squared plus two ππ . Since π’ equals zero, we have π£ squared is equal to two multiplied by 43,120 over two π plus 44 multiplied by 64. This simplifies to 2,759,680 over π plus 22. We could try and square root this. However, we will first consider what happens to body B after the system is released.

We know that body B is accelerating upwards with the same acceleration. This means that after it has traveled a distance of 64 centimeters, it will have the same speed as body A. At this point, body A hits the ground. This means that the string will become slack, and body B will continue to move under the force of gravity. We are told that body B comes momentarily to rest 80 centimeters above its starting point. This means that it travels a further 16 centimeters vertically upwards after body A has hit the ground.

If we take vertically upwards to be positive, the displacement here is 16 centimeters. The final velocity π£ is zero centimeters per second as the body comes to rest. The acceleration due to gravity is negative 980 centimeters per second squared as gravity is acting against the direction of travel. The initial velocity during this part of the journey is the square root of 2,759,680 over π plus 22.

We can now clear some space and use the equation π£ squared is equal to π’ squared plus two ππ  once again. Substituting in our values and squaring π’, we have the following equation. Two multiplied by negative 980 multiplied by 16 is negative 31,360. We can then add 31,360 to both sides such that this is equal to 2,759,680 over π plus 22. This can be rearranged so that we have π plus 22 on the left-hand side. And the right-hand side simplifies to 88. We can then subtract 22 from both sides of this equation such that π is equal to 66. The initial mass of bodies A and B is 66 grams.