Momentum conservation question

In summary: Now it's correct.In summary, when applying the conservation of momentum equation for a collision between a ball and two blocks, it is important to consider the velocities before and after the collision and to take into account the masses of the objects involved. It is also helpful to use the formula for minimum velocity in a circular motion relative to the observer's frame of reference when determining the velocity of the ball after the collision.
  • #1
PCAPS
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Homework Statement
statement and diagram are in the attached pic
Relevant Equations
.
attached pic.jpeg
 
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  • #2
Per forum rules, please quote any standard equations you suppose to be relevant and, more importantly, some attempt.
 
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  • #3
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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  • #4
During the brief period of impact, what horizontal force does the ball exert on its support?
 
  • #5
haruspex said:
During the brief period of impact, what horizontal force does the ball exert on its support?
tension?
 
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  • #6
PCAPS said:
tension?
During the (assumed very brief) impact, the string will still be nearly vertical.
 
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  • #7
haruspex said:
During the (assumed very brief) impact, the string will still be nearly vertical.
Im still not able to get the solution 😔
 
  • #8
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.]
 
Last edited:
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  • #9
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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Related to Momentum conservation question

1. What is momentum conservation?

Momentum conservation is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time. In simpler terms, it means that the total amount of motion in a system does not change unless an external force acts upon it.

2. Why is momentum conservation important?

Momentum conservation is important because it helps us understand and predict the behavior of objects in motion. It is also a fundamental principle in many areas of physics, such as mechanics, thermodynamics, and electromagnetism.

3. How is momentum conserved in collisions?

In collisions, momentum is conserved through the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision. This means that the combined mass and velocity of the objects involved in the collision will remain constant.

4. Can momentum be lost or gained?

No, momentum cannot be lost or gained in a closed system. This is because of the law of conservation of momentum, which states that the total momentum of a closed system remains constant. However, momentum can be transferred between objects through collisions or other interactions.

5. How does momentum conservation relate to Newton's laws of motion?

Momentum conservation is closely related to Newton's laws of motion, specifically the first and third laws. The first law states that an object in motion will remain in motion unless acted upon by an external force, which is essentially the principle of momentum conservation. The third law states that for every action, there is an equal and opposite reaction, which also plays a role in the conservation of momentum in collisions.

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