"Solving Masses Accelerating Homework Statement

[Karol]

Homework Statement

All surfaces are smooth. mass m₁ is attached to m₃ with a smooth rail, it can't detach.
What are the accelerations of the masses relative to the floor.

Homework Equations

Conservation of momentum: $m_1v_1+m_2v_2=0$

The Attempt at a Solution

a₃ and a₂are the acceleration of masses m₃ and m₂ relative to the floor. $a_2'$ is m₂'s acceleration relative to m₃.
In the accelerating frame of m₃:
$$\left\{ \begin{array} {l} m_2a_3+T=m_2a'_2 \\ m_1g-T=m_1a'_2 \end{array} \right.$$
Accelerations relative to the floor: $a_2=-a_3+a'_2$
These equations yield a relation between a₃ and a₂ that includes g.
If i consider relative to the floor, from conservation of momentum:
$$(m_1+m_3)v_3+m_2v_2=0\Rightarrow (m_1+m_3)a_3+m_2a_2=0$$
From this we get relation without g.

 

[Titan97]

Are you sure you can conserve momentum of the system? The centre of mass of the system has an acceleration.

 

[Qwertywerty]

Post edited .

 

[Qwertywerty]

Are you sure you can conserve momentum of the system? The centre of mass of the system has an acceleration.

Yes , this is correct ( as in , your comment ) .

If i consider relative to the floor, from conservation of momentum:

(m1+m3)v3+m2v2=0⇒(m1+m3)a3+m2a2=0​

Also , another hint : Is there any force acting on m₃ ? So what is it's acceleration ?
Does this show how you have gone wrong again ?

 

[Karol]

Are you sure you can conserve momentum of the system? The centre of mass of the system has an acceleration.

I don't think that it matters that the masses accelerate, there is still conservation of momentum since there aren't any external forces in the x direction, so the whole of the masses can't move. the C.O.M remains in place, and from that i also got the same answer.
The only force i see that acts on m₃ is the tension T. relative to the floor:
$$\left\{ \begin{array}{l} m_1g-T=m_1a_2 \\ T=m_2a_2 \end{array}\right.\Rightarrow a_2=\frac{m_1}{m_1+m_2}g$$
The tension T acts on m₂ and on m₃:
$$F=ma\rightarrow m_2a_2=(m_1+m_3)a_3$$
This is the same as momentum conservation, same as before.

 

[Qwertywerty]

I don't think that it matters that the masses accelerate, there is still conservation of momentum since there aren't any external forces in the x direction, so the whole of the masses can't move. the C.O.M remains in place, and from that i also got the same answer.
The only force i see that acts on m3 is the tension T.

No , you can't conserve momentum . Firstly , there is a net force in the x - direction , from the tension acting on m₂ .

No force acts on m₃ . It remains stationary .

 

[haruspex]

No , you can't conserve momentum . Firstly , there is a net force in the x - direction , from the tension acting on m₂ .

No force acts on m₃ . It remains stationary .

Wrong on both counts.
No masses are given for the wheels, so we should take them as massless. The consequence is that it will be just as though m3 is sliding on the ground without friction.
As Karol posted, there are no external horizontal forces on the m1+m2+m3 system, so its mass centre stays put, and conservation of momentum can be used. However, the trap is that m3 will lose contact with m1.

Karol, I'm not sure what question you are asking with your post. You ended up with two equations and two unknowns, which is fine if the equations are correct. What stopped you at that point?

 

[Qwertywerty]

Wrong on both counts.
No masses are given for the wheels, so we should take them as massless. The consequence is that it will be just as though m3 is sliding on the ground without friction.

Why would m₃ start moving ? There is no force acting on it .

As Karol posted, there are no external horizontal forces on the m1+m2+m3 system, so its mass centre stays put, and conservation of momentum can be used. However, the trap is that m3 will lose contact with m1.

The tension acting on m₂ ?

 

[haruspex]

Why would m3 start moving ? There is no force acting on it .

The string exerts a normal force around the curve of the pulley. The net of this will be a force acting down and left at 45 degrees.

The tension acting on m2 ?

I don't understand your question. are you suggesting that this is a force external to the m1+m2+m3 system?

 

[Karol]

However, the trap is that m3 will lose contact with m1.

I am very sorry i didn't draw it more clearly and didn't explain the conditions.
Mass m₁ can't detach from m₃ because it slides on a rail that is attached to m₃!

 

[haruspex]

I am very sorry i didn't draw it more clearly and didn't explain the conditions.
Mass m₁ can't detach from m₃ because it slides on a rail that is attached to m₃!

No, my mistake - you did indeed write that in the OP.
So, what's your issue?

 

[Karol]

The issue is that when i solve in the non inertial frame attached to m₃:
$$\left\{ \begin{array} {l} m_2a_3+T=m_2a'_2 \\ m_1g-T=m_1a'_2 \end{array} \right.$$
Accelerations relative to the floor: $a_2=-a_3+a'_2\rightarrow a'_2=a_2+a_3$
With conservation of momentum: $(m_1+m_3)a_3+m_2a_2=0$ i get:
$$a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)-m_1m_2}$$
Relative to the floor:
$$\left\{ \begin{array}{l} m_1g-T=m_1a_2 \\ T=m_2a_2 \end{array}\right.\Rightarrow a_2=\frac{m_1}{m_1+m_2}g$$
With conservation of momentum: $(m_1+m_3)a_3+m_2a_2=0$ i get:
$$a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)}$$
The denominator is different

 

[haruspex]

Relative to the floor:
$$ m_1g-T=m_1a_2 $$

That does not look right to me.

 

[Karol]

Relative to the floor:
$$\left\{ \begin{array}{l} m_1g-T=m_1a'_2 \\ T=m_2a_2 \\ a_2=a_3+a'_2 \end{array}\right.\Rightarrow a_2=\frac{m_1g-m_1a_3}{m_1+m_2}g$$
Conservation of momentum:
$$(m_1+m_3)a_3+m_2a_2=0$$
$$\Rightarrow a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)-m_1m_2}$$

 

[haruspex]

Relative to the floor:
$$\left\{ \begin{array}{l} m_1g-T=m_1a'_2 \\ T=m_2a_2 \\ a_2=a_3+a'_2 \end{array}\right.\Rightarrow a_2=\frac{m_1g-m_1a_3}{m_1+m_2}g$$
Conservation of momentum:
$$(m_1+m_3)a_3+m_2a_2=0$$
$$\Rightarrow a_3=-\frac{m_1m_2g}{(m_1+m_2)(m_1+m_3)-m_1m_2}$$

All good now?

 

[Karol]

Thank you very much Haruspex and the others