Question Video: Finding the Velocity of a System of Two Vertically Hanging Masses Connected by a String through a Pulley after Its Release Mathematics

Video Transcript

Two masses of 143 grams and 77 grams are connected to the ends of a light inextensible string passing over a smooth fixed pulley. Given that the two masses were hanging freely vertically below the pulley and that the system was released from rest, find its velocity four seconds later. Take 𝑔 equal to 9.8 meters per second squared.

We will begin by drawing a diagram to model the situation. We have two masses of 143 grams and 77 grams that we will call 𝐴 and 𝐵, respectively. These are connected by a light inextensible string which passes over a smooth pulley. As the pulley is smooth, we know that the tension in the string will be constant throughout. The two bodies will exert a downward force equal to their weight. And we know that this weight is equal to the mass of the body multiplied by the acceleration due to gravity. Since there are 1000 grams in a kilogram, the weight of body 𝐴 is equal to 0.143 multiplied by 9.8. This is equal to 1.4014 newtons. The weight of body 𝐵 is equal to 0.077 kilograms multiplied by 9.8 meters per second squared. This is equal to 0.7546 newtons.

When the system is released from rest, body 𝐴 will accelerate downwards and body 𝐵 will accelerate upwards. As the string is inextensible, this acceleration will be constant throughout the whole system. It will therefore also move with the same velocity, and it is this that we are trying to calculate four seconds later. We will begin by using Newton’s second law to calculate the acceleration of the system. This states that 𝐹 equals 𝑚𝑎. The sum of the forces is equal to the mass multiplied by acceleration. If we consider body 𝐴, we know that this is accelerating downwards. Letting this be the positive direction, the sum of our forces is 1.4014 minus 𝑇. This is equal to the mass, 0.143 kilograms, multiplied by the acceleration 𝑎.

We can repeat this process for body 𝐵 which is accelerating upwards. If we let this be the positive direction, we have 𝑇 minus 0.7546. And this is equal to 0.077𝑎. We now have a pair of simultaneous equations that we can solve by elimination. Adding equations one and two will eliminate the tension force 𝑇. This gives us 0.6468 is equal to 0.22𝑎. We can then divide through by 0.22, giving us 𝑎 is equal to 2.94. The acceleration of the system is 2.94 meters per second squared.

As this acceleration is constant, we can use the equations of motion or SUVAT equations. We know that the initial velocity 𝑢 is zero meters per second. The acceleration of the system is 2.94 meters per second squared. And we want to calculate the velocity 𝑣 four seconds after the system is released. We can use the equation 𝑣 is equal to 𝑢 plus 𝑎𝑡. And substituting in our values, we have 𝑣 is equal to zero plus 2.94 multiplied by four. This is equal to 11.76. The velocity of the system four seconds after it is released is 11.76 meters per second.