31st March 2007 - 09:34 PM
Forgive the probable incorrect use of words of physics, I'm not used to writing them in English
Hope you will understand anyway.
Object number 1 is moving horizontally (left to right) with Uniform rectilinear motion at the speed of 10.5 m/s. Its mass is 7.5 Kg. The coefficient of dynamic friction is 0.15. A drawing force drags object number one from left to right (with an angle of 45) and its vertical component(projection) has the same value of its horizontal component, which is 9,6N.
Evidently the frictional force is -9,6N, otherwise the motion would not be uniform rectilinear.
Now Object number 2 (same mass of number 1, 7.5 Kg) falls on object number 1 vertically, causing a perfectly inelastic collision and it stays attached to object number 1. The drawing force stays constant (it doesn't change from 9.6N vertically and 9.6N horizontally).
what is the speed right after the collision?
what distance does the object number 1+2 cover before it stops?
Thanks for the help.
1st April 2007 - 04:12 AM
Here is the simple version of the solution (the one that textbooks and teachers usually want):
Because the collision takes very little time to complete, the drawing force and frictional force do not have time to transfer a significant amount of momentum to the system. Therefore, momentum is conserved during the moment of the collision, and so the final momentum = (2m)v_final = mv_initial = initial momentum. The velocity just after the collision is therefore exactly 1/2 of the velocity before the collision, 5.25 m/s.
Here is the more careful solution (which is actually true):
At the moment of impact, there will be a very large downward force on the bottom mass, as the top mass transfers all of its vertical momentum to the bottom mass. Because of this, the normal force between the bottom block and the ground will temporarily be very large, as the downward momentum of the top block passes through the bottom block and is absorbed by the ground. Therefore, the friction will also be very large during this brief time. Therefore, you CANNOT ignore the change of momentum due to friction during the collision.
The total momentum removed from the system by friction will be the average value of the friction force multiplied by the time of the collision. Since the original friction force is canceled by the horizontal part of the drawing force, the loss of momentum will be due only to the excess friction, which is caused by the impact. This is easy to calculate: If the second mass has initially a downward velocity v_y, it must lose all of its vertical momentum mv_y during the collision time t. Therefore, the average net upward force on the second block is mv_y/t. Since F_net = N - mg, it follows that N = mv_y/t + mg. The "mg" term is not really part of the collision, but is part of the new steady normal force which the first mass will feel after the collision. Therefore, the collision itself exerts, by Newton's third law, an average force mv_y/t driving the first mass downward during the collision, and so it generates an average friction of 0.15m(v_y/t).
During the time of collision, this friction will remove a quantity of momentum which is 0.15m(v_y/t)*t = 0.15mv_y. Therefore the momentum of the system immediately after the collision will be the initial momentum minus this change, i.e. (2m)v_final = mv_initial - 0.15mv_y. Thus the velocity just after the collision will be v_final = 0.5v_initial - 0.075v_y.
Since the problem did not specify v_y, this more careful solution cannot lead to a numerical answer. This is an example of poor teaching in my opinion, because the problem is constructed so that a student who does NOT see deeply into the problem is rewarded with a correct mark, while a student who DOES see deeply into the problem becomes lost because the necessary information is not present. The important thing to see is that the simple answer (which your textbook will say is correct) is ACTUALLY WRONG. To make the simple answer be the correct one, it is necessary to specify that the second mass is dropped onto the first mass from a height very close to zero, so that it does not have a vertical velocity when it strikes the first mass. Only if v_y = 0 is the simple solution correct, and the problem fails to state this assumption.
It may or may not be wise to inform your teacher of this fact. Some teachers are glad to see that students are really thinking, while other teachers hate to be corrected by a student. I do not know your teacher, so I cannot predict the reaction.
Now for the second part of the problem, assuming that the simple solution is correct, you must find the net force which is slowing the blocks. This is fairly easy. The normal force on the bottom of the first block is now 2mg - 9.6, and so the friction force is now 0.15*(2mg - 9.6) = 0.15*(2*7.5*9.81 - 9.6) = 20.6325 Newtons. Since the drawing force is still pulling forward with 9.6 Newtons, the net force slowing the blocks is 9.6 - 20.6325 = -11.0325 Newtons. Therefore, its acceleration is -11.0325/(2m) = -11.0325/15 = -0.7355 m/s^2.
Since its initial velocity is 5.25 m/s, its final velocity is 0, and its acceleration is -0.7355 m/s^2, you can use the kinematic equation (v_final)^2 - (v_initial)^2 = 2*a*(x_final - x_initial). Solving this gives x_final - x_initial = (0 - 5.25^2)/(2*-0.7355) = 18.737 meters. You can check this by doing the kinematic calculation another way: The time to stop must be (v_final - v_initial)/a = 7.138 seconds. Since the average speed during constant accleration is always (v_initial + v_final)/2, you have in this problem an average speed of (0 + 5.25)/2 = 2.625 m/s. Therefore, the distance traveled should be average speed * time = 2.625*7.138 = 18.737 meters, the same answer as you get by the other method.
Hope this helps!
1st April 2007 - 07:47 AM
Thank you Stuart, I really appreciated the detailed explanation. It seemed to me a bit banal that the speed after collision was just half the initial speed.
I'll take advantage or your help again
Have a good day,