Video Transcript
A body A of mass 180 grams is resting on a smooth horizontal table. It is connected by a light inelastic string which passes over a smooth pulley, fixed to the edge of the table, to another body B of mass 120 grams hanging freely vertically below the pulley. When body A is 90 centimeters away from the pulley, the system is released from rest. Determine the speed at which body A collides with the pulley. Take 𝑔 to be equal to 9.8 meters per square second.
When dealing with these sorts of problems, it can be really useful to draw a diagram. Here, we have a little sketch of the pulley system. Body A is resting on the table. And body B is hanging freely from the string vertically below the pulley. We’re told that body A has a mass of 180 grams. Now, usually, we would look to work in kilograms, but actually we’re going to continue working in grams in this question. What that will mean is we’re going to need to convert any units for acceleration into centimeters per square second. And then, our units for force are going to be dynes.
Body A exerts a force vertically downward on the table. That force is mass times acceleration due to gravity. So, that’s 180𝑔. And of course, we did say we’re going to work in centimeters per square second to match the mass being in grams. So, we’re going to let 𝑔 be equal to 9.8 times 100. That’s 980 centimeters per square second. Body B has a mass of 120 grams, so the downward force due to gravity is 120 times 𝑔.
There are a number of other forces acting on the bodies. There is tension in the string, which remains consistent throughout that piece of string. There’s the reaction force of the table on the body. And that’s it. We’re told that the table is smooth, so there’s no frictional force. And the string is light, so we don’t need to take into account its mass.
We’re looking to find the speed at which body A collides with the pulley. Now, we’re told that body is 90 centimeters away from the pulley. So, we’re going to work out the speed after it’s been moving for 90 centimeters. And since it starts from rest, we know its initial velocity must be equal to zero. So, in order to solve this problem, we’re going to need to use the equations of constant acceleration, sometimes known as SUVAT equations.
So far, we know that 𝑢 is equal to zero and 𝑠 is equal to 90. We’re looking to find the velocity at the point that it reaches 90 centimeters. And therefore, it would be really useful to know the acceleration of the particle once it’s released from rest. Once we have an acceleration, we can use the formula 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠.
But how do we find the acceleration of the particle? Well, we’re going to resolve forces in our diagram. Remember, 𝐹 equals 𝑚𝑎; the overall force is equal to the mass times the acceleration. We make an assumption about the direction that the particles are going to move in. And then, we form equations for the overall force, or the net force, on each particle. For body A, the only force acting horizontally is tension. So, we can say that 𝐹 is equal to 𝑇, and mass times acceleration is 180𝑎.
There’s a little bit more happening on particle B. We have the downward force due to gravity. That’s 120𝑔. And then, we have the tension acting in the opposite direction. Since the particle is moving downwards, we know that the net force will be 120𝑔 minus 𝑇. And that’s equal to mass times acceleration. We’re going to replace 𝑇 with 180𝑎. And in doing so, we have an equation purely in terms of 𝑎. So, 120𝑔 minus 180𝑎 equals 120𝑎. And we’re now going to add 180𝑎 to both sides. That gives us 120𝑔 equals 300𝑎.
But of course we know that 𝑔 is 980 centimeters per square second. And so, we substitute 𝑔 equals 980 into our equation. And we get 120 times 980 to be 117600. Finally, we divide through by 300. And we find 𝑎 is equal to 392, or 392 centimeters per square second. Of course, we could go back and use this to find the tension in the string in dynes, but we really don’t need to. We’re now going to use the equation 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠. We substitute everything into this equation. We get 𝑣 squared equals zero squared plus two times 392 times 90, which simplifies to 70560.
Our last step is to square-root both sides of this equation. And when we do, we find that 𝑣 is equal to 84 root 10. We’ve worked in centimeters, centimeters per second squared, throughout. So, velocity here must be in centimeters per second. Remember, speed is the magnitude of the velocity. So, here, it’s 84 root 10 centimeters per second.
