| Problem statement Solution video |
DISCUSSION THREAD
Ask and answer questions here. You learn both ways.
DISCUSSION and HINTS
Initially Block A slides to the right along Block B which is traveling to the right. However, with friction acting between A and B, both A and B slow down. At some point, A instantaneously comes to rest, and the starts to move to the left. Once the speed of A to the left matches that of the speed of B to the left, the two stick and move together. You can see this in the animation that follows.
Recall the following four-step plan outline in the lecture book and discussed in lecture:
Step 1: FBDs
Draw single free body diagram (FBD) for the entire system (A+B). Do NOT consider A and B in separate FBDs because you will need to deal with the friction force acting between A and B (which you do not know).
Step 2: Kinetics (linear impulse/momentum)
Consider all of the external forces that you included in your FBD above. If there are no external forces acting in the horizontal direction (x-direction) on your system, the linear momentum in the x-direction is conserved.
Step 3: Kinematics
As described above, A comes to rest with respect to B when vA = vB.
Step 4: Solve
Combine your kinetics equation from Step 2 with your kinematics that you found in Step 3, and solve for the velocity of B.
QUESTION: Are you surprised that your answer for the final speed of B (and A) does not depend on the coefficient of friction acting between A and B? I was the first time that I worked the problem. š
For this problem I was wondering if there is a way to solve using the frictional force? I know it's not needed but I am not sure if it is still possible to use.
You would need some additional information, such as the time that it took for the two blocks to stick together.
For our free body diagram should we show two separate normal forces acting on block B or is one fine?
I doubt that matters since there is no movement in the y direction.
Agreed. Although there are two normal forces, they will not be used in your analysis since the horizontal surface is smooth.
I agree, separating the normal forces would make no difference on the problem either way as well. The forces in the y-direction do not really matter for this problem since the bottom surface is smooth, and the x-direction is what is needed to be analyzed.
As the hints mention, if you analyze the system as a whole the friction force between blocks A and B becomes an internal force which allows for an easier solving process. However, if there was friction between B and the floor it becomes more complicated and more information would be necessary.
Is there a way to find how long block B would have to be to give room for block A to come to a stop relative to block B?
To determine the minimum length of B that would allow block A to come to rest fully, one would need the initial speed of A and the coefficient of kinetic friction. From this one can follow similar steps taken in H.4.L to calculate the length needed
To determine the minimum length of B that would allow block A to come to rest fully, one would need the initial speed of A and the coefficient of kinetic friction. From this one can follow similar steps taken in H.4.L to calculate the length needed
An important note is that for T_2, our kinetic energy at position 2, the mass of the system is all added together, so make sure you account for the mass of A AND B in the equation.
Could this problem also be solved using the work-energy equation, since there would be no potential energy change or nonconservative forces to consider?
The use of the work/energy equation would require you to determine the work done by the friction force acting between A and B. You do not know the coefficient of kinetic friction, so that would not be possible. The advantage of the momentum equation is that you can treat A and B as a single system, leaving the friction force as internal (and not contributing to the external forces).
Thank you. As a followup, since we are only dealing with velocity in the X direction in this problem, do we assume a direction for final velocity, and attach a sign to it accordingly? In general, do we keep the sign of the initial speed, or do we assume it to be positive and let the math tell us the direction if we were to plug in some numbers?
You need to define your positive direction (e.g., either left or right). Stick with that sign convention throughout, including the signs given to the initial velocities for A and B.
If the positive x is defined to the right, is it necessary to assume that vb2 is negative in our calculations, or should the signs work out on their own?
I would assume that since we have already defined x to be positive to the right, we should let the signs work out since we don't know if the final vb is going to be to the left or right (as this depends of va1 and vb1)
According to the simulation, V_2 moves as a system between both block A and B. That being said, the mass moving with the ending velocity is the mass of the blocks combined.
It is important to note that when block A has come to rest relative to block B, blocks A and B can essentially be viewed as one body of mass 3m. The friction between the two blocks causes them to move together.
This can be done with basic momentum equation from high school.
Is that momentum equation different from the one that we use in this class?
Are kinematics required for this problem? It seems like the problem can be solved with only the linear impulse-momentum equation.
Sometimes that happens. When it does, simply include:
Step 3 - N/A
Does Va1 take into account the speed of b? Is vA1 relative to b or is it the true velocity?
I guess the question Iām asking is, what is the expression for initial momentum of block A? Is it m(vb1-vA1), or just -mva1
The answer is in terms of va1 and vb1, right?
Yes
If we make the system to include both blocks A and B, we can neglect friction (the only horizontal force) and see a conservation in momentum. Doing this makes this question easier to solve since we aren't given anything to solve for friction. Since we're given v_A1 and v_B1, all we need to do is determine the speeds of v_A2 and v_B2 when the speed of block A relative to B is 0 (this is mentioned in the hints).
This a really simple problem and requires minimal work if you do it correctly. This question uses the linear impulse and momentum equation and your answers should be in terms of Va1 and Vb1. You don't have to worry about the vertical forces but you will also learn that the only horizontal force is friction in the same but opposite directions from the 2 blocks. This means it can be canceled out and reduced down to a simple momentum equation. Also, note that if you already defined which way is positive and negative (x), you shouldn't assume the sign of the final Vb because the signs that you give for Va1 and Vb1 should give you the proper signs for Vb2.