# Question Video: Analyzing the Motion of Two Bodies Hanging Freely Connected by a String Passing through a Pulley Mathematics

Two bodies of masses 644 g and 156 g were connected to the ends of a light inextensible string passing over a smooth pulley. The system was released from rest and, 2 seconds later, the larger mass hit the ground. Find the maximum height the smaller mass reached above its initial position. Take the acceleration due to gravity π = 9.8 m/sΒ².

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### Video Transcript

Two bodies of masses 644 grams and 156 grams were connected to the ends of a light inextensible string passing over a smooth pulley. The system was released from rest, and two seconds later the larger mass hit the ground. Find the maximum height the smaller mass reached above its initial position. Take the acceleration due to gravity π equal to 9.8 meters per second squared.

We will begin by sketching a diagram to model the situation. We have two bodies A and B of masses 644 grams and 156 grams. They are connected by a light inextensible string that passes over a smooth pulley. As the pulley is smooth, we know that the tension in the string will be constant throughout. The bodies will exert a downward force equal to their mass multiplied by the acceleration due to gravity. We are told that this is equal to 9.8 meters per second squared, which is the same as 980 centimeters per second squared. Body A exerts a downward force of 944 grams multiplied by 980 centimeters per second squared. This is equal to 631,120 dynes. And body B exerts a downward force of 156 grams multiplied by 980 centimeters per second squared, which is equal to 152,880 dynes.

When the system is released from rest, body A will accelerate downwards, and body B will accelerate upwards. As the string is inextensible, the acceleration of the system will be constant. We are told that the larger body hits the ground after two seconds. During this period, both bodies will have traveled the same distance. And we are asked to calculate the maximum height that the smaller mass reaches above its initial position. Once body A hits the ground, the string will become slack and body B will continue to move under the action of gravity.

We begin to solve this problem using Newtonβs second law, which states that πΉ equals ππ. The vector sum of the forces is equal to the mass multiplied by the acceleration. There are two forces acting on body A. And if we let the positive direction be vertically downwards, they have a sum of 631,120 minus π. This is equal to the mass of 644 grams multiplied by π. We will call this equation one.

Repeating this process for body B, where we take the positive direction to be vertically upwards, we have π minus 152,880 is equal to 156π. We will call this equation two. We now have a pair of simultaneous equations. And we can eliminate the tension force π by adding equation one to equation two. This gives us 478,240 is equal to 800π. We can then divide through by 800, giving us π is equal to 597.8. This is the acceleration of the system in centimeters per second squared.

We can now use the equations of motion or SUVAT equations to find the displacement of body B. As already mentioned, there are two parts to this, firstly, the displacement of the whole system in the first two seconds. We know that the initial speed π’ is zero centimeters per second. We have just calculated the acceleration π is 597.8 centimeters per second squared. As π‘ is equal to two, we can use the equation π  is equal to π’π‘ plus a half ππ‘ squared. Since π’ equals zero, this simplifies to π  is equal to a half multiplied by 597.8 multiplied by two squared. π  is therefore equal to 1,195.6. In the first two seconds, both bodies travel a distance of 1,195.6 centimeters.

We can calculate the velocity of the bodies at this point using the equation π£ is equal to π’ plus ππ‘. Since π’ is zero, π£ is equal to 597.8 multiplied by two. This is also equal to 1,195.6. At the point at which body A hits the ground, the system is moving with velocity 1,195.6 centimeters per second. This is the initial velocity at the point at which the string becomes slack. We know that when body B reaches its maximum height, its velocity will be equal to zero. The acceleration of the body doing this part of the motion is negative 980 centimeters per second squared, as gravity is acting against the direction of motion. We can use the equation π£ squared is equal to π’ squared plus two ππ  to calculate the displacement in this part of the journey.

Substituting in our values, we have zero is equal to 1,195.6 squared plus two multiplied by negative 980 multiplied by π . This can be simplified to 1,960π  is equal to 1,429,459.36. Dividing through by 1,960, we have π  is equal to 729.316. Body B travels a further distance of 729.316 centimeters after body A hits the ground. We can now calculate the maximum height that the smaller mass reaches above its initial position by adding 1,195.6 to 729.316. This is equal to 1,924.916. The smaller mass reaches a height of 1,924.916 centimeters above its initial position. Alternatively, we could give our answer in meters by dividing by 100, giving us 19.24916 meters.