Is the latter part of the momentum conservation question correct?

AI Thread Summary
The discussion focuses on the application of momentum conservation in a collision scenario involving a ball and heavier blocks. Participants emphasize the importance of Newton's first law and the horizontal velocity of the ball before and after impact. It is noted that during the brief impact, the string remains nearly vertical, resulting in no horizontal force acting on the ball, thus maintaining its horizontal velocity. The relevance of the equation v_min = √(5gh) is debated, with clarification provided on its application in different frames of reference. Overall, the discussion leads to a validation of the participant's method and solution.
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Homework Statement
statement and diagram are in the attached pic
Relevant Equations
.
attached pic.jpeg
 
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Per forum rules, please quote any standard equations you suppose to be relevant and, more importantly, some attempt.
 
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haruspex said:
Per forum rules, please quote any standard equations you suppose to be relevant and, more importantly, some attempt.
WhatsApp Image 2021-01-31 at 12.22.15 PM.jpeg
 
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During the brief period of impact, what horizontal force does the ball exert on its support?
 
haruspex said:
During the brief period of impact, what horizontal force does the ball exert on its support?
tension?
 
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PCAPS said:
tension?
During the (assumed very brief) impact, the string will still be nearly vertical.
 
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haruspex said:
During the (assumed very brief) impact, the string will still be nearly vertical.
Im still not able to get the solution 😔
 
PCAPS said:
Homework Statement:: statement and diagram are in the attached pic
Relevant Equations:: .

View attachment 277188
A few things to think about...

You have ignored Newton’s 1st law! Just before the collision, the ball has (horizontal) velocity u. What is its velocity just after the collision (and why)? (Consider @haruspex's replies.)

You are told the blocks are much heavier than the ball. When calculating the velocity of [A and B stuck together] using conservation of momentum, you can simply ignore the mass of the ball.

You have used ##v_{min} = \sqrt{5gh}##. This is not usually a ‘standard formula’ but, if you are allowed to use it without derivation, OK.

The above formula gives ##v_{min}## in a frame of reference in which the centre of the circular rotation is stationary. This is not the lab’ frame. It may help if you imagine you are an observer sitting on block A. Note that ##v_{min}## is a velocity relative to you (just after impact).

[EDIT: Minor edits improve wording.]
 
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Steve4Physics said:
A few things to think about...

You have ignored Newton’s 1st law! Just before the collision, the ball has (horizontal) velocity u. What is its velocity just after the collision (and why)? (Consider @haruspex's replies.)

You are told the blocks are much heavier than the ball. When calculating the velocity of [A and B stuck together] using conservation of momentum, you can simply ignore the mass of the ball.

You have used ##v_{min} = \sqrt{5gh}##. This is not usually a ‘standard formula’ but, if you are allowed to use it without derivation, OK.

The above formula gives ##v_{min}## in a frame of reference in which the centre of the circular rotation is stationary. This is not the lab’ frame. It may help if you imagine you are an observer sitting on block A. Note that ##v_{min}## is a velocity relative to you (just after impact).

[EDIT: Minor edits improve wording.]
I think I got the answer, can u validate my method?
 

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  • #10
PCAPS said:
I think I got the answer, can u validate my method?
Looks good. Well done!

EDIT: As a further point of clarification, since the string is vertical, there is no horizontal force acting on the ball . Therefore there is no horizontal acceleration and the ball's velocity remains u (until the ball starts moving upwards after the collision).
 
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  • #11
Steve4Physics said:
Looks good. Well done!
thankyou :smile:
 
  • #12
haruspex said:
During the (assumed very brief) impact, the string will still be nearly vertical.
thankyou too for providing hints instead of plain answer, I am happy that I have done it myself .It was bothering me from the morning.
 
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  • #13
PCAPS said:
thankyou :smile:
The first equation you wrote is not quite right. You should have: $$m_1u + m_0u = (m_1 + m_2)v + m_0u$$ The mass ##m_0## disappears from the equation at this point. You use ##m_0 << m_1 + m_2## subsequently, by ignoring any change in the motion of ##m_1 +m_2## as ##m_0## executes its circuit.
 
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  • #14
PeroK said:
The first equation you wrote is not quite right. You should have: $$m_1u + m_0u = (m_1 + m_2)v + m_0u$$ The mass ##m_0## disappears from the equation at this point. You use ##m_0 << m_1 + m_2## subsequently, by ignoring any change in the motion of ##m_1 +m_2## as ##m_0## executes its circuit.
thanks for pointing that out, is the latter part correct?
 
  • #15
PCAPS said:
thanks for pointing that out, is the latter part correct?
Yes!
 
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