- #1

PCAPS

- 8

- 5

- Homework Statement:
- statement and diagram are in the attached pic

- Relevant Equations:
- .

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- Thread starter PCAPS
- Start date

- #1

PCAPS

- 8

- 5

- Homework Statement:
- statement and diagram are in the attached pic

- Relevant Equations:
- .

- #2

- 38,830

- 8,219

- #3

PCAPS

- 8

- 5

- #4

- 38,830

- 8,219

During the brief period of impact, what horizontal force does the ball exert on its support?

- #5

PCAPS

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tension?During the brief period of impact, what horizontal force does the ball exert on its support?

- #6

- 38,830

- 8,219

During the (assumed very brief) impact, the string will still be nearly vertical.tension?

- #7

PCAPS

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Im still not able to get the solutionDuring the (assumed very brief) impact, the string will still be nearly vertical.

- #8

Steve4Physics

Homework Helper

Gold Member

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A few things to think about...Homework Statement::statement and diagram are in the attached pic

Relevant Equations::.

View attachment 277188

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

[EDIT: Minor edits improve wording.]

Last edited:

- #9

PCAPS

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I think I got the answer, can u validate my method?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 isnotthe 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.]

- #10

Steve4Physics

Homework Helper

Gold Member

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Looks good. Well done!I think I got the answer, can u validate my method?

EDIT: As a further point of clarification, since the string is vertical, there is

- #11

PCAPS

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thankyouLooks good. Well done!

- #12

PCAPS

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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.During the (assumed very brief) impact, the string will still be nearly vertical.

- #13

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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.thankyou

- #14

PCAPS

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thanks for pointing that out, is the latter part correct?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.

- #15

- 23,278

- 14,786

Yes!thanks for pointing that out, is the latter part correct?

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