| 4 comments ]

In this system, assume gravitational force on the block is included in the external force. So the block's potential of gravity is not included in calculating total energy of the system.

From previous solution, we get force exerted by the ball to the block is


So it needs an external force with the same value to make the block in equilibrium.


If the block is slowly moved upward, we can still assume the block is in equilibrium, and the external force still works on the block.

Suppose the block is moved upward with a very small distance dh. Total works done by the external force is



The work will result in change of ball's speed only (remember, no potential of gravity included).



Combining these two equations,



Integrating the equations and simplifying it,


Hence,





That's my solution.
If you have any question, correction, or comment, please leave a comment below.
Thank you.

Solution - Multiple Collisions (2)SocialTwist Tell-a-Friend

4 comments

Anonymous said... @ February 17, 2009 at 11:57 PM

I still don't understand why external force must be equal to the force that exerted by the ball. If it's equilibrium, the total force must be equal to the weight of the block.. but, that doesn't mean the two forces magnitudes are same..
can u explain me?

Copycat91 said... @ February 18, 2009 at 4:22 PM

I assume the weight of the block as external force, so we don't need to include gravitational potential energy in calculating system's total energy.

Anonymous said... @ April 29, 2009 at 1:10 AM

In the first problem, you put N in (N>>1), which means you considered a long period. But here I think we cannot use N again. The key for this problem is that the block moves very slowly, or the speed of the ball is very large compared with the speed of the block. This is not only the reason for that F_ext = F, but also for many things else:

1. Because V_block is very small, we can assume that in the period of 2 successive collisions with the block, the height is still h.
Actually there's a little change of h, say Dh (not dh in the solution above). The existence of Dh leads to the change of v (Dh is not zero, so the force of the block acting on the ball does work).
But Dh is very small, so the change of v is very small too (or we can say after the collision, the ball's speed is approximately v), which helps us to deduce the formula of F.

2. Because v >> the speed of the block, we can assume that the time t for the ball to move between 2 successive collisions is very small, which leads to the idea that v and h change continuously, so that we can use dv and dh.

In my opinion, the important difference between the two problems is that while in the first problem, the external force is obviously Mg, in this problem, it is a general external force.

Anonymous said... @ June 19, 2010 at 11:36 AM

I think the condition for this problem for moving up is impossible, because when the block M moves at distance dh, and the change of v is very small, we can see that the force that exerts to the block will decrease because F = m v^2 / h , where v will decrease in a small amounts and h will increase in a small amounts too, hence, F will decrease, and F < Mg, and the block must move downward

Post a Comment