Motion of an Object in a SemiCircle

[Student149]

1. The Setup

Given a Schema as in the Image attached. Each of the 2 balls have the velocity = V0 and Mass = M0. The system in the lower part of the image has Velocity = V1 and Mass = M1. The arcs are tubes that are hollow and are 1/4th (or a quadrant) of a circle that are perfectly aligned with the 2 balls.

Let the Radius of each of the external arc (that forms the outer circle) of the tube be R1.

Lets ignore all the complicating factors (gravity, air resistance, friction, heat loss and so on) and assume 100% efficiency. When each of the balls leaves the system simultaneously after traveling through the tubes:

1. What is the resultant velocity of the balls as they exit the Schema (they exit exactly horizontally)?
2. What is the new (vertical) velocity of the Schema when the balls exit the Schema?
3. How does the velocity depend on the radius R1?

2. Approach

If the balls and the schema have a direct head collision all the momentum of the system (balls and schema) would be completely utilized in the vertical direction. But since the balls follow a circular arc a part of momentum is used in vertical direction and the rest is used in horizontal direction:

1. The momentum available in the vertical direction is given by: M1*V1 + 2* M0*V0* sin(45)
2. The momentum available in the horizontal direction is given by: 2* M0*V0*cos(45)

 

[BvU]

Nice exercise !
Please use the template, don't erase it.

Momentum is a vector quantity and it is conserved in collisions. [edit[ In the horizontal direction, initial momentum is zero, so final momentum is zero too.

In the vertical direction the momentum is conserved. Simplify the situation (e.g. M₁=2M₀ and V₁=0 to get some ideas.

I do not understand where your formulas come from. What is the 45$^\circ$ ?

 

[Student149]

Nice exercise !
Please use the template, don't erase it.

Momentum is a vector quantity and it is conserved in collisions. [edit[ In the horizontal direction, initial momentum is zero, so final momentum is zero too.

In the vertical direction the momentum is conserved. Simplify the situation (e.g. M₁=2M₀ and V₁=0 to get some ideas.

I do not understand where your formulas come from. What is the 45$^\circ$ ?

My mistake. Will take care now onward.

I assumed part of the 'energy'/'velocity' would be used by the balls as they try to push the system in the horizontal direction as they follow the arc. Thus, only remaining part of the energy would be used to push the system in the vertical direction. Thus I used sin(45) since the starting and ending part of the arcs form a 45 degree angle.

I understand the generic Idea (conservation of momentum) but I am not sure of what formula for net velocity of the system and the balls is.

 

[BvU]

1. The momentum available in the vertical direction is given by: M₁*V₁ + 2* M₀*V₀ * sin(45)
2. The momentum available in the horizontal direction is given by: 0 2* M₀*V₀*cos(45)

There is no horizontal component in any of the three velocity vectors.

Can you distinguish the symmetry in the scenario ?
The balls push in opposite directions with equal force, so the horizontal component of the total momentum remains zero.

Momentum considerations will only get you so far in this exercise. In the template, under 2: relevant equations", you'll need more. Any idea what ?

By the way:
A full circle is 360$^\circ$ so a quarter circle is 90$^\circ$

 

[Student149]

There is no horizontal component in any of the three velocity vectors.

Can you distinguish the symmetry in the scenario ?
The balls push in opposite directions with equal force, so the horizontal component of the total momentum remains zero.

Momentum considerations will only get you so far in this exercise. In the template, under 2: relevant equations", you'll need more. Any idea what ?

By the way:
A full circle is 360$^\circ$ so a quarter circle is 90$^\circ$

Thank you. As you said I understand the momentum is in horizontal direction. And conserved in the vertical direction. I understand that. But, I am more looking in terms of velocity of each of the components (rather as the resultant directions are known speed). And to get the speed of each component, I am not sure what is missing (but of course as you said something is :) .

I used 45 instead of 90 as the straight line connecting top to bottom of arc would make a 45 degree angle (It was a guesswork, so please ignore it)

 

[BvU]

As you said I understand the momentum is [I think a zero is missing here ?] in horizontal direction. And conserved in the vertical direction. I understand that. But, I am more looking in terms of velocity of each of the components (rather as the resultant directions are known speed).

Good. Remember that a velocity is a vector; for our case there are horizontal and vertical components of the velocities.

I sincererly think there is a mistake in the exercise statement :

1. What is the resultant velocity of the balls as they exit the Schema (they exit exactly horizontally)?

This last part should be something like "with the same vertical speed as the object with mass M₁"

And to get the speed of each component, I am not sure what is missing

The scenario is a description of a collision. There are classes of collisions. What does the instruction

ignore all the complicating factors (gravity, air resistance, friction, heat loss and so on)

hint at for the kind of collision in this scenario ?

.

 

[Chestermiller]

How does the vertical velocity of each of the balls compare with the vertical velocity of the schema when the balls are just about to exit the schema?

 

[BvU]

Hello Chet! Do you agree about this 'horizontally' ?

 

[Chestermiller]

Hello Chet! Do you agree about this 'horizontally' ?

I agree with you. It is not horizontal.

 

[Chestermiller]

To me, the key words in the problem statement are "assume 100% efficiency." I interpret this as "kinetic energy is conserved."

 

[Student149]

I agree with you. It is not horizontal.

I think we can assume that there is a small (as small as necessary) horizontal tail for each of the arcs so the balls exit the system in the horizontal direction. The 100% efficiency refers to the conservation of momentum or kinetic energy as you pointed.

 

[Chestermiller]

I think we can assume that there is a small (as small as necessary) horizontal tail for each of the arcs so the balls exit the system in the horizontal direction. The 100% efficiency refers to the conservation of momentum or kinetic energy as you pointed.

No. They exit with a vertical velocity component equal to that of the schema.

 

[BvU]

conservation of momentum or kinetic energy

Beware! Your 'or' worries me...

There is a big difference between kinetic energy and momentum.

In elastic collisions there is conservation of kinetic energy. (this is what I was fishing for - didn't want to just give it away)
In all collisions there is conservation of momentum.

Now, do you have a battle plan to conclude this exercise ?

 

[Chestermiller]

What is the initial upward momentum of the schema plus the balls (in terms of the masses and the initial velocities)?

 

[Student149]

What is the initial upward momentum of the schema plus the balls (in terms of the masses and the initial velocities)?

The net initial momentum in the vertical plane if I am correct is: M₁*V₁ - 2*M₀*V₀ (I am assuming the downwards direction as negative as convention).

 

[Chestermiller]

The net initial momentum in the vertical plane if I am correct is: M₁*V₁ - 2*M₀*V₀ (I am assuming the downwards direction as negative as convention).

Good. If the final upward velocity components of the balls and the schema are all V, what is the final upward momentum of the balls plus schema?

 

[Student149]

Good. If the final upward velocity components of the balls and the schema are all V, what is the final upward momentum of the balls plus schema?

Then the final upward momentum in that case should be V*(M₁+2*M₀).

 

[Chestermiller]

Then the final upward momentum in that case should be V*(M₁+2*M₀).

Good. So, if momentum is conserved, what is the value of V?

 

[Student149]

Good. So, if momentum is conserved, what is the value of V?

We can equate both the equations to get the V for the complete system. V=(M₁*V₁ - 2*M₀*V₀)/(M₁+2*M₀)

 

[Chestermiller]

We can equate both the equations to get the V for the complete system. V=(M₁*V₁ - 2*M₀*V₀)/(M₁+2*M₀)

Excellent. That finishes the vertical direction. Now for the horizontal direction. If Vx is the exit horizontal velocity component of one of the balls and -Vx is the exit horizontal velocity component of the other, then, in terms of V and Vx, what is the final kinetic energy of the two balls?

 

[Student149]

Excellent. That finishes the vertical direction. Now for the horizontal direction. If Vx is the exit horizontal velocity component of one of the balls and -Vx is the exit horizontal velocity component of the other, then, in terms of V and Vx, what is the final kinetic energy of the two balls?

Thank you Prof. The horizontal I am a bit confused. V is in vertical direction. Vx is in horizontal so not sure how they can be related.

 

[Chestermiller]

Thank you Prof. The horizontal I am a bit confused. V is in vertical direction. Vx is in horizontal so not sure how they can be related.

What is the equation for the kinetic energy of a body of mass M with horizontal and vertical velocity components vx and vy?

 

[Student149]

Apologies. Connectivity issue.

Kinetic Energy of course is = 1/2*M*V².
where Velocity (V) = Sqrt{Vx²+Vy²}

 

[Chestermiller]

Apologies. Connectivity issue.

Kinetic Energy of course is = 1/2*M*V².
where Velocity (V) = Sqrt{Vx²+Vy²}

Good. Now, based on this, please reconsider my question in post #20.

 

[Student149]

Good. Now, based on this, please reconsider my question in post #20.

I think we can use the value Vx (calculated earlier) =(M₁*V₁ - 2*M₀*V₀)/(M₁+2*M₀) in the above equation pair to find Vy.

But, to do that we need the value of V to be substituted in (1/2*M*V²)?

 

[Chestermiller]

I think we can use the value Vx (calculated earlier) =(M₁*V₁ - 2*M₀*V₀)/(M₁+2*M₀) in the above equation pair to find Vy.

But, to do that we need the value of V to be substituted in (1/2*M*V²)?

You're getting your V's and $V_x$'s confused. In post #19, we determined the value of V. Our next objective is to determine $V_x$.

In terms of V and $V_x$, the final total kinetic energy of the two balls exiting the schema is $$m_0(V^2+V_x^2)$$
In terms of V, what is the final kinetic energy of the schema?

 

[Student149]

You're getting your V's and $V_x$'s confused. In post #19, we determined the value of V. Our next objective is to determine $V_x$.

In terms of V and $V_x$, the final total kinetic energy of the two balls exiting the schema is $$m_0(V^2+V_x^2)$$
In terms of V, what is the final kinetic energy of the schema?

I did get V and V_x confused. Thanks.

The Kinetic energy of the Schema is: 1/2*m₁*(V²+0)

(as there is no motion in the horizontal plane for schema, V_x in this case is 0).

 

[Chestermiller]

I did get V and V_x confused. Thanks.

The Kinetic energy of the Schema is: 1/2*m₁*(V²+0)

(as there is no motion in the horizontal plane for schema, V_x in this case is 0).

So then, in terms of V and Vx, what is the final kinetic energy of the balls plus schema?

 

[Student149]

So then, in terms of V and Vx, what is the final kinetic energy of the balls plus schema?

The final Kinetic Energy of the System would be simple addition of both: m₀(V²+V_x²) + 1/2*m₁*V²

I suppose if I am correct we can equate this with the kinetic energy of the system initially and thus calculate the V_x.

 

[Chestermiller]

The final Kinetic Energy of the System would be simple addition of both: m₀(V²+V_x²) + 1/2*m₁*V²

I suppose if I am correct we can equate this with the kinetic energy of the system initially and thus calculate the V_x.

Yes

 

[Student149]

Yes

Thank you Prof. for your immense patience and help.

 

[Chestermiller]

Thank you Prof. for your immense patience and help.

Please thank BVU also. He got you started.

 

[Student149]

Beware! Your 'or' worries me...

There is a big difference between kinetic energy and momentum.

In elastic collisions there is conservation of kinetic energy. (this is what I was fishing for - didn't want to just give it away)
In all collisions there is conservation of momentum.

Now, do you have a battle plan to conclude this exercise ?

thank you Sir for your patience and help.