| Problem statement Solution video |
DISCUSSION THREAD
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HINTS:
STEP 1 - FBD: Draw a SINGLE free body diagram (FBD) of the system of cart + cannon + cannonball.
STEP 2 - Kinetics: Write down the impulse/momentum equation in the horizontal direction (x-direction) for the the system of cart + cannon + cannonball. Based on the above FBD, is the momentum conserved in the x-direction for that system?
STEP 3 - Kinematics
STEP 4 - Solve. Solve for the velocity of the cart + cannon.
QUESTION: The above analysis allows you to find the answer to the first part of the problem. Unfortunately, it is not useful for the find the answer to the second part of the problem where you want to find the force on the cart/cannon. What do you need to change in the analysis to find this force?
For the initial momentum, should we use the instant before the cannonball is fired to equate with the instant shown in the problem so that both the initial velocity of the cannonball and cart are 0?
Yes, that would be a good plan.
Since the air resistance and friction are all zero for the system, wouldn't that mean that the horizontal forces acting on the system from 0<t<dT would be 0? or is this referring to internal forces? Thank you
The answer depends on what you define as your system. If your system is the same as for the hints for this problem (cannon, ball, cart), then you would be correct and the horizontal forces would all be internal. However, to answer part b you would need to redefine the system so that you can find the average horizontal force.
While completing this problem I found it interesting to consider what the problem might have looked like if the bottom of the cart wasn't smooth. In this scenario we also have to consider the possibility of the cart tipping when the cannon ball rebounds off the wall. This would involve math similar to the slipping vs tipping problem we did in ME270. To solve this problem we might need information such as the launch angle the distance from cannon to wall, or wall thickness. This problem could of course be solved in several different way based on different amounts/types of information given.
Will the answer for average force be simply a magnitude or should it be a vector with sign and direction?
The answer will be a magnitude - the component of force in the horizontal direction.
I am thoroughly confused by part B. By reworking your free body diagram to include the internal force of impact, you can construct two equations for linear momentum in the y and x directions, however, since we are missing one more known, namely the angle at which the ball collides with the wall, how can one move forward?
For the first part, we will need two FBDs. One right after release and one for the instant shown in the given figure right?
You only need one, if you put everything in your system, it'll give you no external forces which makes the calculation very easy for Vc2.
Since we don't need to factor in g for the linear/impulse momentum equation, do we need to divide it out of the given values, if so, what does that make the units of m. Lbs/m/s^2 seems a little confusing.
I believe I have answered my own question, since the given units are Lbs there is no need to divide out g and the units will work themselves out in further calculations in order to get us the correct answer.
The weights of the cart/cannon are given, as well as the weight of the ball. In order to find the mass, divide by "g".
I am still confused by part be in this problem as if we take the system as cart+cannon+ball, then there are no horizontal forces. So do we just need to redefine the system to just the cart and cannon and find the horizontal force on the cart from the ball when it contacts it?
Yes. As we talked about in lecture, making the system "BIG" is convenient when wanting to avoid dealing with forces that can be made internal.
On the other hand, sometimes you want to know such a force. In that case, chose your system (and FBD) that makes the force of interest EXTERNAL. As you suggest here, choose your system to be just the cart + cannon.
So using just the cart as the system and then having that external horizontal force, could we then use the newton formulas and assume that the work from 1 -> 2 is the average force on the system over the time interval and solve for that?
Keep in mind that you are looking for the average force over a period of TIME. The W/E equation will not be very helpful there. The linear/impulse momentum equation has time built right into the equation - use that instead.
Because the masses are given in lbs what does that make of the force as newtons are Kgm/s^s and we would get Lbsm/s^2 so should we just leave it as the Lbsm/s^s?
We are not given masses; instead, we are giving weights. To find mass, divide by "g". No need to convert to SI.
Oh I found my issue for the force the time cancels out thus giving you units that makes sense, I just rushed my calculations on that.
Can we assume that the force of impact is only in the x direction?
Conceptual question C4.9 (page 276 in the lecture book) follows a very similar process for part A of this problem. Fundamentally, the problem is worked out the same way, except this problem has one less mass since the cannon and cart are combined into one mass.
Should we pay attention to potential energy since we can define the line of zero potential energy at the horizontal line on the impact point, and the cart has PE = 0.
In part B, how do you determine the force on the cart/cannon if we do not know the angle that the cannonball hits the cart with? Would you have to assume that the force is only in the x direction?
You can assume the impact is only in the x
While the problem does say that the cart and cannonball start at rest, it is important to note that the force created by the cannon firing the cannonball is not accounted for, so don't worry about that aspect of the physics for your analysis. The only force vector that needs to be considered in our analysis is the cannonball making contact with the wall of the cart.
When solving for part B of this problem, it is very important to be careful how you set up your FBD and realize that you must choose one of the components (the cart or the cannonball) as your focus point. This will allow you to set up the equation with the proper force you are solving for.
The ball bounces off of the cart with both x and y components to its velocity (since theta=20 deg), so will we need to do two separate momentum equations (one for the x direction and one for the y direction) to solve for each component of the cart's final velocity vector?
No, we can only solve for velocity in the x direction since that is the only direction in which momentum is conserved. Also, the cart can only move in the x direction.
Again, this problem is pretty straightforward as long as you use conservation of momentum. Furthermore, for the last step of finding force, remember to divide your answer by 32.17 ft/s2 in order to a number in terms of force only.
It seems the firing of the cannon would cause some initial movement of both the ball and the cart. How is this force accounted for? I am assuming the energy used to do this work would be chemical and thus can not be calculated given what we know about the problem.
Can we assume that when the ball is fired, the cart does not move?
The solution hinges on your system definition. If you align your system with the hints provided for this problem (cannon, ball, cart), your assertion is accurate, and the horizontal forces would be internal. Yet, for Part B, it becomes imperative to redefine the system, enabling the determination of the average horizontal force.
Looking at problem C4.9 from the lecture book provides a great plan of how to approach this problem. From there it is easy to see that setting up two momentum equations will allow the solution to be easily found.