Momentum conservation question

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  • #1
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Homework Statement:
statement and diagram are in the attached pic
Relevant Equations:
.
attached pic.jpeg
 

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  • #2
haruspex
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Per forum rules, please quote any standard equations you suppose to be relevant and, more importantly, some attempt.
 
  • #3
PCAPS
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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
 
  • #4
haruspex
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During the brief period of impact, what horizontal force does the ball exert on its support?
 
  • #5
PCAPS
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During the brief period of impact, what horizontal force does the ball exert on its support?
tension?
 
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  • #6
haruspex
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tension?
During the (assumed very brief) impact, the string will still be nearly vertical.
 
  • #7
PCAPS
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During the (assumed very brief) impact, the string will still be nearly vertical.
Im still not able to get the solution 😔
 
  • #8
Steve4Physics
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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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  • #9
PCAPS
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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
Steve4Physics
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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).
 
  • #12
PCAPS
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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.
 
  • #13
PeroK
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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.
 
  • #14
PCAPS
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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
PeroK
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thanks for pointing that out, is the latter part correct?
Yes!
 

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