Video Transcript
Two bodies of masses 374 grams and 102 grams were connected to each other by a light inextensible string passing over a smooth pulley. The two bodies started at rest on the same horizontal level. Then one second after the system was released, the string broke. Determine the vertical distance between the two bodies one second after the string broke. Take acceleration due to gravity 𝑔 equal to 9.8 meters per square second.
We can begin by modeling the situation where we have two masses 𝐴 and 𝐵. The two masses are 374 and 102 grams. We are told that gravity is equal to 9.8 meters per square second. When dealing with grams, we need the acceleration to be in centimeters per square second and the velocities in centimeters per second. As there are 100 centimeters in a meter, 9.8 meters per square second is equal to 980 centimeters per square second. We will use this value when dealing with gravity for the remainder of this question.
The first body exerts a downward force of 374 multiplied by gravity. Likewise, the second mass exerts a downward force of 102 multiplied by gravity. We have a light inextensible string passing over a smooth pulley. This means that the tension forces going vertically upwards will be equal. When the system is released, the 374-gram body will accelerate downwards and the 102-gram body will accelerate upwards. The magnitude of these accelerations will be equal.
We can now use Newton’s second law, force is equal to mass multiplied by acceleration, to create equations for body 𝐴 and body 𝐵. As body 𝐴 is accelerating downwards, the sum of the net forces is equal to 374𝑔 minus 𝑇. This is equal to 374𝑎. Body 𝐵 is moving upwards. So we will consider this to be the positive direction. This means that the sum of the forces is 𝑇 minus 102𝑔. This is equal to 102𝑎.
We now have a pair of simultaneous equations. And we can eliminate 𝑇 by adding equation one and equation two. This gives us 272𝑔 is equal to 476𝑎. We can then divide both sides by 476. And as 𝑔 is equal to 980, 𝑎 is equal to 272 multiplied by 980 divided by 476. This means that the acceleration of the system is 560 centimeters per square second.
Our next step is to use the equations of motion or SUVAT equations to calculate the displacement and velocity of the bodies after one second. This is the point where the string breaks. We know that the initial velocity is zero centimeters per second. The acceleration is 560 centimeters per square second. And the time is one second.
We can use the equations 𝑣 is equal to 𝑢 plus 𝑎𝑡 and 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. This gives us a value of 𝑣 equal to 560 centimeters per second. At the point the string breaks, the bodies are moving with velocity 560 centimeters per second. 𝑠 is equal to zero multiplied by one plus a half multiplied by 560 multiplied by one squared. This is equal to 280. This means that the bodies have moved a distance of 280 centimeters when the string breaks.
At the point the string breaks, body 𝐴 is 280 centimeters below its start point and body 𝐵 is 280 centimeters above its start point. This means that they are currently 560 centimeters apart. We need to find the distance between the bodies one second after this. As the string has now broken, their acceleration will be simply due to gravity. Body 𝐴 is moving downwards with a speed of 560 centimeters per second and is accelerating at 980 centimeters per square second.
Once again, we can use the equation 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. 𝑠 is equal to 560 multiplied by one plus a half multiplied by 980 multiplied by one squared. This is equal to 1050. One second after the string breaks, body 𝐴 has fallen a further 1050 centimeters.
Let’s now consider body 𝐵. This time, the acceleration is negative as gravity is pulling the body downwards. 𝑠 is equal to 560 multiplied by one plus a half multiplied by negative 980 multiplied by one squared. This is equal to 70. One second after the string breaks, body 𝐵 has risen a further 70 centimeters.
We can now calculate the vertical distance between the two bodies one second after the string broke. We need to add 70, 280, 280, and 1050. This gives us an answer of 1680. The distance between the bodies is 1680 centimeters. This is equivalent to 16.8 meters. Had we converted the masses into kilograms, 0.374 and 0.102, as well as keeping the acceleration in meters per square second and velocities in meters per second, we would’ve ended up with an answer of 16.8 meters. Either one of these answers is correct.
