# Question Video: Finding the Vertical Distance between Two Bodies Hanging Freely by a String through a Smooth Pulley after Moving for a Certain Time Mathematics

Two bodies of masses 8.7 and 11.6 grams hung vertically from the ends of a light inextensible string passing over a smooth pulley. When the bodies were released from rest, they were on the same horizontal level. Determine the vertical distance between them one second after they started moving. Take the acceleration due to gravity 𝑔 to be 9.8 m/s².

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### Video Transcript

Two bodies of masses 8.7 and 11.6 grams hung vertically from the ends of a light inextensible string passing over a smooth pulley. When the bodies were released from rest, they were on the same horizontal level. Determine the vertical distance between them one second after they started moving. Take the acceleration due to gravity 𝑔 to be 9.8 meters per second squared.

We will begin by sketching a diagram of the two bodies labeled A and B, where body A has a mass of 8.7 grams and body B has a mass of 11.6 grams. These bodies will exert a downward force equal to their weight. And we can calculate this for each body by multiplying the mass by the acceleration due to gravity. We are told this is equal to 9.8 meters per second squared, which is the same as 980 centimeters per second squared. Multiplying 8.7 by 980 gives us 8526. The weight of body A is 8526 dynes. Multiplying 11.6 by 980 gives us 11368. This means that body B exerts a downward force of 11368 dynes.

We are told that the bodies 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 the same throughout. And as the string is inextensible, when the bodies are released from rest, the system will move with the same acceleration. We are asked to calculate the vertical distance between the bodies one second after they started moving. In order to do this, we will first calculate the acceleration of the system. We can do this using Newton’s second law, 𝐹 equals 𝑚𝑎. The sum of the forces is equal to the mass multiplied by the acceleration.

We know that after the system is released, body A will accelerate vertically upwards. Taking this to be the positive direction, the sum of our forces is 𝑇 minus 8526. And this is equal to the mass of 8.7 multiplied by 𝑎. Body B will accelerate downwards. So if we take this to be the positive direction, the sum of its forces will be 11368 minus 𝑇. This is equal to 11.6𝑎. We now have a pair of simultaneous equations that we can solve by elimination. If we add equation one and equation two, the tension forces 𝑇 will cancel. Negative 8526 plus 11368 is equal to 2842. This is equal to 20.3𝑎. Dividing through by 20.3 gives us 𝑎 is equal to 140. The acceleration of the system after release is 140 centimeters per second squared. This is equal to 1.4 meters per second squared.

We are told in the question that the bodies start at the same horizontal level. This means that at any time, the displacement of body A vertically upwards will be equal to the displacement of body B vertically downwards. The distance that each body will be away from the initial position at any time interval will be the same. This means that we can calculate the vertical distance between them by calculating the displacement of one body and then multiplying it by two. We will do this using the equations of motion or SUVAT equations. The initial speed of body A is zero meters per second, and it accelerates at 1.4 meters per second squared. We want to calculate its displacement after one second.

We can use the equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. Since 𝑢 is equal to zero, 𝑠 is equal to a half multiplied by 1.4 multiplied by one squared. This is equal to 0.7. After one second, body A has moved a distance of 0.7 meters vertically upwards. This means that body B will have moved a distance of 0.7 meters vertically downwards. Recalling that the body started on the same horizontal level, the distance between them is equal to two multiplied by 0.7. This is equal to 1.4. After one second, the distance between the two bodies is 1.4 meters.