Is the latter part of the momentum conservation question correct?

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SUMMARY

The discussion centers on the application of momentum conservation in a collision scenario involving a ball and two blocks. Participants emphasize the importance of Newton's 1st law and the relative velocities before and after the collision. The equation $$m_1u + m_0u = (m_1 + m_2)v + m_0u$$ is highlighted as crucial for understanding the momentum transfer, where the mass of the ball can be ignored due to its negligible impact compared to the blocks. The conversation concludes with validation of the user's method and understanding of the problem.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with conservation of momentum principles
  • Knowledge of basic kinematic equations, specifically $$v_{min} = \sqrt{5gh}$$
  • Ability to analyze collision scenarios in physics
NEXT STEPS
  • Study the implications of Newton's 1st law in collision problems
  • Explore advanced applications of conservation of momentum in multi-body systems
  • Learn about different frames of reference in physics, particularly in collision analysis
  • Investigate the derivation and applications of the kinematic equation $$v_{min} = \sqrt{5gh}$$
USEFUL FOR

Physics students, educators, and anyone interested in understanding collision dynamics and momentum conservation principles in mechanics.

PCAPS
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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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