Collision involving blocks and pulley

In summary, the conversation is discussing a problem involving a particle of mass M/2 colliding with a pan and the resulting tension and acceleration in the string connecting the pan to another block. Multiple possible solutions are proposed, including treating the two blocks as a system and considering the centripetal force on block A, but the final answer is not agreed upon. The conversation ends with a question about how the leftwards impulse on block A can result in a downwards acceleration.
  • #71
theodoros.mihos said:
You are right. Sorry for my English. A body can have acceleration while its velocity is steel zero.
But this problem is more complicate because the "radius" of A movement is not constant.
Thank you for finally acknowledging that mass A can be accelerating upward even though its initial upward velocity is zero.
All works with assumption about "circular" movement with radius l probably are basically wrong.
Excuse me, but my analysis does not assume that the movement of mass A is circular with radius l, and it does take into account the fact that the radial distance of mass A from the pulley is not constant. The equations that I presented in post #62 give the actual trajectory of mass A at short times, where x is the distance mass A moves to the left, and y is the distance mass A moves upward. If we eliminate time t from these equations, we obtain:
$$y=\frac{x^2}{4l}$$
This is consistent with a radius of curvature of the circular arc of mass A of 2l, rather than l. So mass A does execute a circular arc at short times, but, as a result of the increasing radial distance of mass A from the pulley, it is initially traveling along an arc of radius 2l rather than l.

Chet
 
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  • #72
Theodoros inspired me to do the Lagrangian method. I have to admit that TSny and Chet were right, the upward acceleration of B is v02/(2L) at t=0, just after the collision .

@theodoros.mihos : I see, you considered upward positive for y, so the sign is OK in the constraint.
But why did you write that the potential energy of A is ##\frac{gy_A^2}{\sqrt{x_A^2+y_A^2}}## instead of gyA?
 
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  • #73
@ehild you are right. This mistake comes for a previous look about the direct force.
@chet. I see match more works except yours. Nothing personal.
Now all differs.
 
  • #74
theodoros.mihos said:
@ehild you are right. This mistake comes for a previous look about the direct force.
@chet. I see match more works except yours. Nothing personal.
Now all differs.
I have no idea what you are saying. But I can tell you that I am totally thrilled that ehild, TSny, and I are now in total agreement.

Now that full consensus has been reached, I am closing this thread.

Chet
 

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