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Question: *A big block of mass $2m$ is placed near a wall in such way that only its left side touches the vertical wall. Now a ball of mass $m$ is released from top (near to the side of the wall) of a smooth cylindrical groove of radius $R$ made inside the bigger block of mass $2m$, placed on a smooth frictionless floor.During the motion, the ball crosses the bottom most and reaches a point which makes an angle theta with the vertical. Now find the velocity of the bigger mass right after the block crosses the bottom most point.

(An image is added at the end of the question for better visualization )

My attempt using kinematics and momentum conservation:

The ball imparts a normal force on the surface of the cylindrical groove, (which tends to impart velocity to the bigger mass towards left due to the horizontal component of the normal force but is prevented by the reaction forces of the wall on the left) during it's motion from top to the bottom-most point.

Since the only motion that exists now is that of the ball & that too the surface is smooth it is equivalent to a free fall thus the velocity gained by the ball on the bottom most point is $v_{bottom}=\sqrt{2gR}$

Applying the momentum conservation principle,

The velocity gained by the bigger mass towards the right as the ball just crossed the bottommost point, $m*\sqrt{2gR}$ = $(3m)_{system}*V_{system}$

Thus $V_{system}$ =$\frac{1}{3}*\sqrt{2gR}$

MY DOUBT:

Why do I get a different answer when I try to apply the Work Energy theorem from the initial point where the ball is released to the bottom-most point:

This is how I applied the Work energy theorem wrt ground:

$TE_{release point}$=$TE_{bottom}$ where TE represents total energy

$KE_{intial}$+$U_{initial}$=$KE_{final}$+$U_{final}$

0 + 0 = $\frac{1}{2} * 3m * v_{bottom}^2$ + $mgR$

$v_{bot}$ = $\sqrt{2gR/3}$

Question Image (I drew a diagram for better visualization):

enter image description here

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  • $\begingroup$ All my previous questions were criticized and closed without any answers due to faults from myside which I humbly accept. This time I learnt the latex and tried to present my doubt as well as possible. I spent about 45 min just trying write this one. I hope you will understand my efforts. Thanks $\endgroup$
    – Elizabeth Huffman
    Commented Mar 31, 2023 at 3:57
  • $\begingroup$ I also read various other answers especially this one which is similar: physics.stackexchange.com/questions/496740/… But none of them helped in my understanding. $\endgroup$
    – Elizabeth Huffman
    Commented Mar 31, 2023 at 4:02
  • $\begingroup$ For one thing, I think the question is poorly worded. "Velocity of the entire system" is an ambiguous concept. I can only assume it means "velocity of the center of mass" of the two objects $\endgroup$
    – RC_23
    Commented Mar 31, 2023 at 4:14
  • $\begingroup$ @RC_23 By the term 'Velocity of the entire system' I meant the velocity of the bigger mass. Ill edit rn $\endgroup$
    – Elizabeth Huffman
    Commented Mar 31, 2023 at 4:16
  • $\begingroup$ @RC_23 Yes Edited...You can now check the question $\endgroup$
    – Elizabeth Huffman
    Commented Mar 31, 2023 at 4:19

1 Answer 1

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Here is the complete analysis of the situation as per my understanding.

1. Until m reaches the bottom most point...

Looking at the forces, we know the interaction forces between m and M will be balanced by normal contact from the left wall and the surface below.

While the above statement holds true until it reaches the bottom most point, the directions switch just after it crosses the bottommost point causing the vertical component of the forces to be balanced by normal reaction from the ground and the horizontal component to cause acceleration of M towards the right since its movement is not constrained anymore.

2. Velocity just after m crosses the bottomost point O'

It's exactly zero at the instant m is at O' since there hasn't and slowly gains velocity as m moves ahead of the point O.

The velocity just after crossing is virtually 0 although we can very well determine the velocity of M, when m makes an angle $\theta$ from, let's say the horizontal.

3. Why it went wrong

Everything puk has mentioned about your assumptions is correct. You seem to have considered them to stick and move with the same velocities while in reality they have individual velocities and that too in different directions.

4. What next ?

Why don't we try to build an expression for velocity of M when m makes an angle $\theta$ from the horizontal.

Hints: Conserve the mechanical energy of the system M + m between the instant where m is at O' and its final position determined by $\theta$.

Momentum conservation between these two points is also completely valid in the horizontal direction.

The expression will be lengthy but you need not do the algebra. It's only two equations try getting them right.

Here's what I got...

Here's what I got

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  • $\begingroup$ Thank you so much. Just 1 last doubt. So Work Energy theorem isnt valid in the period of the motion where ball reaches from release to bottom as momentum conservation aint valid there right? $\endgroup$
    – Elizabeth Huffman
    Commented Mar 31, 2023 at 6:49
  • $\begingroup$ Work energy theorem is applicable everywhere just that it isnt required in this particular problem. I think you're getting confused between conservation of mechanical energy and work-energy theorem in general. Maybe go over their specific statements and derivations once again. Work energy theorem would've involved writing the work done by forces that has been done at the expense of the energy of the system. $\endgroup$
    – mewtonscat
    Commented Mar 31, 2023 at 8:40

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