So for this, I got v_A = sqrt(2gd/3) by using the energy balance equation between the initial state and the state just before the collision. I then used the conservation of momentum equation mgv_A - 2mgv_A = mgv_C + 2mgv_C - 2mgv_C, giving -mgv_A = mgv_C.
Mathematically, this seems right, but intuitively, it doesn't make sense to me. Shouldn't the velocity of C be lower after the collision?
If I'm reading your equations correctly, I think your issue here is that A & C should be moving at the same speed before the collision, and A,B, & C should all be moving at the same speed after the collision.
This is due to the inextensible cable. With that observation, you should be able to alter both your kinematics and momentum equation.
You may, but such an assumption is not necessary. Since potential energy appears as a CHANGE in potential energy in the work/energy equation, you do not care where the two blocks are located initially. You only need to realize that for this problem the two blocks always travel through the same distance.
I want to be sure that I have the correct fundamentals because I am getting an unexpected value in my answers. I defined four times of interest, 1 - at release, 2 - before hit, 3- after hit, and 4 - at the max height of B. Then, from 1 to 2 there is conservation of energy, which allows you to calculate the the velocity of A, under the assumption that Va2 = -Vc2 because of the cable. Then, from 2 to 3, you can use conservation of momentum because the system itself (A, B, C, pulleys, cable) is not accelerating. With the assumptions that Va3 = Vb3 = -Vc3, this results in downward velocity for A and B. Do I have the correct process/assumptions set up? Did I make an error in my calculation? Thanks!
The change in velocity equals the velocity of A before it collides with block B minus the velocity of blocks a and b right after perfectly inelastic collision.
Professor Krousgrill, I watched the solution video and I'm just wondering why we were able to neglect the fact that gravity exists when solving for the average tension in the cable. I understand from state 2 to 3 how we consider the Favg to consider tension and gravity, but I cannot understand where it is within part b. Thank you.
As we have discussed in class, the force of impact is generally several orders of magnitude larger than "non-impulsive" forces such as weight. With that in mind, during impact one can typically ignore the influence of weight since the impulse of the weight is calculated over a very short time.
So for this, I got v_A = sqrt(2gd/3) by using the energy balance equation between the initial state and the state just before the collision. I then used the conservation of momentum equation mgv_A - 2mgv_A = mgv_C + 2mgv_C - 2mgv_C, giving -mgv_A = mgv_C.
Mathematically, this seems right, but intuitively, it doesn't make sense to me. Shouldn't the velocity of C be lower after the collision?
If I'm reading your equations correctly, I think your issue here is that A & C should be moving at the same speed before the collision, and A,B, & C should all be moving at the same speed after the collision.
This is due to the inextensible cable. With that observation, you should be able to alter both your kinematics and momentum equation.
In finding the force via impulse do we use only the mass of A or is it the combined mass (for the equation mv_i +int(F*dt)=mv_f)
Should the Tavg be extremely high since it is for such a short duration?
Short duration impacts can experience large impact forces.
Since kinetic friction is not defined, are we able to assume all surfaces are smooth?
Yes, friction does not come into play on this problem.
Should we assume that blocks C and B have the same height when the system is at rest?
You may, but such an assumption is not necessary. Since potential energy appears as a CHANGE in potential energy in the work/energy equation, you do not care where the two blocks are located initially. You only need to realize that for this problem the two blocks always travel through the same distance.
I want to be sure that I have the correct fundamentals because I am getting an unexpected value in my answers. I defined four times of interest, 1 - at release, 2 - before hit, 3- after hit, and 4 - at the max height of B. Then, from 1 to 2 there is conservation of energy, which allows you to calculate the the velocity of A, under the assumption that Va2 = -Vc2 because of the cable. Then, from 2 to 3, you can use conservation of momentum because the system itself (A, B, C, pulleys, cable) is not accelerating. With the assumptions that Va3 = Vb3 = -Vc3, this results in downward velocity for A and B. Do I have the correct process/assumptions set up? Did I make an error in my calculation? Thanks!
when i am using the conservation of energy should i consider the entire system or individual objects
When solving for average tension force, should I be including block C's momentum?
The change in velocity equals the velocity of A before it collides with block B minus the velocity of blocks a and b right after perfectly inelastic collision.
Professor Krousgrill, I watched the solution video and I'm just wondering why we were able to neglect the fact that gravity exists when solving for the average tension in the cable. I understand from state 2 to 3 how we consider the Favg to consider tension and gravity, but I cannot understand where it is within part b. Thank you.
As we have discussed in class, the force of impact is generally several orders of magnitude larger than "non-impulsive" forces such as weight. With that in mind, during impact one can typically ignore the influence of weight since the impulse of the weight is calculated over a very short time.
Does this make sense?
Yes that does make. Thank you.