Determining the accelerations -- two masses connected by a spring

[JoeyBob]

Homework Statement

See attached

Relevant Equations

F=ma, F=kx^2

So first I looked at the forces acting on m1

m1a1=F spring on m2-F spring

Then m2

m2a2=F spring-F spring on m1

using 3rd law,

m2a2=F spring+F spring on m2

m2a2=F spring+m1a1+F spring

m2a2=2F spring+m1a1

Not entirely sure if I've done the above correctly, but I am stuck now because I have two unknowns, a1 and a2.

The answer is suppose to be 20.488 m/s^2 btw.

 

[TSny]

There is only one horizontal force that acts on $m_2$. What is this force?

 

[JoeyBob]

There is only one horizontal force that acts on $m_2$. What is this force?

Its not force from the spring then because kx^2=4.704. Dividing by m2 gives 1.15, an incorrect acceleration.

If its the F spring on m2 then its m1a1+Fspring=m2a2, which is two unknowns still.

 

[TSny]

Its not force from the spring then because kx^2=4.704.

$kx^2$ is not the correct expression for the spring force.

 

[JoeyBob]

$kx^2$ is not the correct expression for the spring force.

Youre right, getting it half mixed up with potential energy (1/2kx^2).

Why isn't the force impacted by the other mass? Like I imagine as the spring decompresses, not all of the decompression pushes m2, isn't some of it used up to push m1?

For instance, if i attached a spring to an unmovable object and compressed it by 1m with a moveable object, won't this movable object experience a greater force than if both object were movable? Force is also related to energy, so shouldn't m1 also be taking some of the energy m2 would otherwise have if it was the only object moving?

 

[haruspex]

Relevant Equations:: F=kx^2

Not so. F is a force, kx² is energy. The energy stored in a spring compressed or extended by x is ½kx².

m2a2=F spring-F spring on m1

m₂ is 'unaware' of m₁. As @TSny asks, what force(s) is m₂ subject to?

 

[JoeyBob]

Not so. F is a force, kx² is energy. The energy stored in a spring compressed or extended by x is ½kx².

m₂ is 'unaware' of m₁. As @TSny asks, what force(s) is m₂ subject to?

But wouldn't m2 move further if m1 didnt move?

 

[TSny]

Why isn't the force impacted by the other mass? Like I imagine as the spring decompresses, not all of the decompression pushes m2, isn't some of it used up to push m1?

For instance, if i attached a spring to an unmovable object and compressed it by 1m with a moveable object, won't this movable object experience a greater force than if both object were movable? Force is also related to energy, so shouldn't m1 also be taking some of the energy m2 would otherwise have if it was the only object moving?

Suppose we consider the case where $m_1$ is held fixed while $m_2$ is allowed to move. At any instant of time during the decompression of the spring, the magnitude of the force that the spring exerts on $m_1$ equals the magnitude of the force that the spring exerts on $m_2$. The force on each mass is $kx$, where $x$ is the instantaneous compression of the spring. But, no work is done on $m_1$ since $m_1$ is fixed. The total work done by the spring is the work done on $m_2$. So, all of the potential energy of the spring is transferred to $m_2$ as kinetic energy.

Suppose both masses are allowed to move. It's still true that at any instant, the force is $kx$ on each mass. But now, work is done by the spring on both masses. So, some of the initial potential energy of the spring is transferred to $m_1$ and the rest to $m_2$. Less work is done on $m_2$ in this case compared to the case where $m_1$ was held fixed because in this case $m_2$ moves less distance during the decompression compared to the first case.

 

[JoeyBob]

Suppose we consider the case where $m_1$ is held fixed while $m_2$ is allowed to move. At any instant of time during the decompression of the spring, the magnitude of the force that the spring exerts on $m_1$ equals the magnitude of the force that the spring exerts on $m_2$. The force on each mass is $kx$, where $x$ is the instantaneous compression of the spring. But, no work is done on $m_1$ since $m_1$ is fixed. The total work done by the spring is the work done on $m_2$. So, all of the potential energy of the spring is transferred to $m_2$ as kinetic energy.

Suppose both masses are allowed to move. It's still true that at any instant, the force is $kx$ on each mass. But now, work is done by the spring on both masses. So, some of the initial potential energy of the spring is transferred to $m_1$ and the rest to $m_2$. Less work is done on $m_2$ in this case compared to the case where $m_1$ was held fixed because in this case $m_2$ moves less distance during the decompression compared to the first case.

But if it moves less distance wouldn't that mean it experienced less force? For instance, If I pushed something with an applied force of 100N, it would move more than if I pushed it with an applied force of 50 N.

You seem to imply that the same amount of force can move the object different amounts, which doesn't make sense to me.

 

[TSny]

You seem to imply that the same amount of force can move the object different amounts, which doesn't make sense to me.

If you apply a constant force of 10N on a 1 kg block, the block will move farther during the push if it is pushed for a longer time interval.

The spring doesn't exert a constant force. But, it's still somewhat similar. If $x_0$ is the initial compression of the spring, then the force on each mass will vary from $k x_0$ to zero. However, it takes more time for the force to vary from $k x_0$ to zero when only $m_2$ is allowed to move compared to when both masses are allowed to move. So, the distance that $m_2$ moves during the decompression is greater in the first case.

 

[JoeyBob]

If you apply a constant force of 10N on a 1 kg block, the block will move farther during the push if it is pushed for a longer time interval.

The spring doesn't exert a constant force. But, it's still somewhat similar. If $x_0$ is the initial compression of the spring, then the force on each mass will vary from $k x_0$ to zero. However, it takes more time for the force to vary from $k x_0$ to zero when only $m_2$ is allowed to move compared to when both masses are allowed to move. So, the distance that $m_2$ moves during the decompression is greater in the first case.

So its because the spring pushes m2 for less time?

How would you go about determining the work on m2 in this problem? I assume you don't just divide the compression by 2 when determining the potential energy to find work.

 

[TSny]

So its because the spring pushes m2 for less time?

That's part of the story. Here are the graphs for the force as a function of time on $m_2$ for the two cases using the data given in the problem. The blue graph is for the case where only $m_2$ is allowed to move. The orange graph is for the case where both masses move.

You can see that the spring exerts a force on $m_2$ for about .082 seconds in the first case and only for about .054 s in the second case. Not only that, but you can also see that for any instant of time, the magnitude of the force in the first case is greater than or equal to the force in the second case.

How would you go about determining the work on m2 in this problem? I assume you don't just divide the compression by 2 when determining the potential energy to find work.

Right. That wouldn't "work".

If both masses are free to move, you could use conservation of momentum and energy to find the final kinetic energy of each mass. The kinetic energy of one of the masses represents the amount of work done on the mass.

 

[JoeyBob]

That's part of the story. Here are the graphs for the force as a function of time on $m_2$ for the two cases using the data given in the problem. The blue graph is for the case where only $m_2$ is allowed to move. The orange graph is for the case where both masses move.

View attachment 274385

You can see that the spring exerts a force on $m_2$ for about .082 seconds in the first case and only for about .054 s in the second case. Not only that, but you can also see that for any instant of time, the magnitude of the force in the first case is greater than or equal to the force in the second case.

Right. That wouldn't "work".

If both masses are free to move, you could use conservation of momentum and energy to find the final kinetic energy of each mass. The kinetic energy of one of the masses represents the amount of work done on the mass.

So you would get the potential energy from the spring, 0.5kx^2, then this would all be converted to kinetic energy. Then would you find the total momentum from the system using the velocity from this kinetic energy?

 

[haruspex]

If both masses are free to move, you could use conservation of momentum and energy to find the final kinetic energy of each mass. The kinetic energy of one of the masses represents the amount of work done on the mass.

Equivalently, because the CoM stays fixed the distances the masses move are in inverse proportion to their mass. Since, at any instant, the spring exerts the same force on each, the work done on each and the velocity of each are also in inverse proportion to their mass.

 

[TSny]

So you would get the potential energy from the spring, 0.5kx^2, then this would all be converted to kinetic energy.

Yes. So, you have $\frac{1}{2} k x^2 = K_1 + K_2$, where $K_1$ and $K_2$ are the final kinetic energies of the two masses.

Then would you find the total momentum from the system using the velocity from this kinetic energy?

When both masses are allowed to move, the total momentum remains constant. Since the total momentum is initially zero, the total momentum is always zero. This tells you something about how the magnitude of the final momentum of $m_1$ compares to the magnitude of the final momentum of $m_2$. The magnitude of the momentum of a block, $p$ , can be expressed in terms of the kinetic energy of the block, $K$: $\,\,\, p = \sqrt{2mK}$. (To see why, substitute $K = \frac{1}{2}mv^2$ and see that the right side reduces to $mv$.) So, conservation of momentum gives another relation between $K_1$ and $K_2$. This relation and the energy relation gives two equation for $K_1$ and $K_2$.

 

[JoeyBob]

Yes. So, you have $\frac{1}{2} k x^2 = K_1 + K_2$, where $K_1$ and $K_2$ are the final kinetic energies of the two masses.

When both masses are allowed to move, the total momentum remains constant. Since the total momentum is initially zero, the total momentum is always zero. This tells you something about how the magnitude of the final momentum of $m_1$ compares to the magnitude of the final momentum of $m_2$. The magnitude of the momentum of a block, $p$ , can be expressed in terms of the kinetic energy of the block, $K$: $\,\,\, p = \sqrt{2mK}$. (To see why, substitute $K = \frac{1}{2}mv^2$ and see that the right side reduces to $mv$.) So, conservation of momentum gives another relation between $K_1$ and $K_2$. This relation and the energy relation gives two equation for $K_1$ and $K_2$.

So m1v1+m2v2=0

p1=sqrt(2m1*Ek). Then v1=p1/m1?

I don't think this is right though.

 

[TSny]

So m1v1+m2v2=0

In order for the total momentum to be zero, the masses must have equal magnitudes of momentum but in opposite directions. So, in terms of magnitudes, $p_1 = p_2$.

p1=sqrt(2m1*Ek).

Yes. Similarly for $p_2$. Then, $p_1 = p_2$ becomes a relation between $K_1$ and $K_2$.

 

[JoeyBob]

In order for the total momentum to be zero, the masses must have equal magnitudes of momentum but in opposite directions. So, in terms of magnitudes, $p_1 = p_2$.

Yes. Similarly for $p_2$. Then, $p_1 = p_2$ becomes a relation between $K_1$ and $K_2$.

Would the kinetic energy be 1/2(m1+m2)v^2? But then velocity from what?

 

[TSny]

Would the kinetic energy be 1/2(m1+m2)v^2?

No. The two blocks will not have the same speed v.

In post #11 you asked, " How would you go about determining the work on m2 in this problem? ". The work done on a block equals the final kinetic energy of the block according to the "work-energy theorem". So, you can find the work done on each block by finding the final kinetic energies $K_1$ and $K_2$.

You have the energy equation $PE_0 = K_1 + K_2$, where $PE_0$ is the initial potential energy in the spring.

You have the momentum equation $p_1 = p_2$. This can be written in terms of $K_1$ and $K_2$.

So, you have two equations for the two unknowns, $K_1$ and $K_2$.