PCAPS
- 8
- 5
- Homework Statement
- statement and diagram are in the attached pic
- Relevant Equations
- .
haruspex said:Per forum rules, please quote any standard equations you suppose to be relevant and, more importantly, some attempt.
tension?haruspex said:During the brief period of impact, what horizontal force does the ball exert on its support?
During the (assumed very brief) impact, the string will still be nearly vertical.PCAPS said:tension?
Im still not able to get the solutionharuspex said:During the (assumed very brief) impact, the string will still be nearly vertical.
A few things to think about...PCAPS said:Homework Statement:: statement and diagram are in the attached pic
Relevant Equations:: .
View attachment 277188
I think I got the answer, can u validate my method?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.]
Looks good. Well done!PCAPS said:I think I got the answer, can u validate my method?
thankyouSteve4Physics said:Looks good. Well done!
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.haruspex said:During the (assumed very brief) impact, the string will still be nearly vertical.
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.PCAPS said:thankyou![]()
thanks for pointing that out, is the latter part correct?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.
Yes!PCAPS said:thanks for pointing that out, is the latter part correct?