Problem involving SHM and a collision

[palaphys]

Homework Statement

Two particles, 1 and 2, each of mass 𝑚, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at 𝑥0, are oscillating with amplitude 𝑎 and angular frequency 𝜔. Thus, their positions at time 𝑡 are given by
𝑥1(𝑡) = (𝑥0 + 𝑑) + 𝑎 sin 𝜔𝑡
and
𝑥2(𝑡) = (𝑥0 − 𝑑) − 𝑎 sin 𝜔𝑡, respectively, where 𝑑 > 2𝑎.
Particle 3 of mass 𝑚 moves towards this system with speed
𝑢0 = 𝑎𝜔/2, and undergoes instantaneous elastic collision with particle 2, at time 𝑡0. Finally, particles 1 and 2 acquire a center of mass speed 𝑣 cm and oscillate with amplitude 𝑏 and the same angular frequency 𝜔.

Relevant Equations

p=mv, coefficient of restitution, SHM equations

please view the first reply, I was unable to post my attempt here for some reason.

 

[palaphys]

this is the problem statement,
My attempt:-
I considered 1 and 2 as a system, now differentiating the respective positions,
$$ v_1 (t)= aw cos (wt)$$
and $$ v_2(t)= -awcos (wt) $$
however, I'm not sure how this would help.
if the collision occurs at t=0, then v1 reduces to Aw and v2 reduces to -Aw, indicating that the balls are in equilibrium position (as far as I have been learnt in SHM)
Also, I am unsure whether momentum conservation applies due to a spring force on ball 2, if the collision between 2 and 3 is considered. Also, I tried to use impulse momentum theorem (most probably incorrect, but can't see where and why) :

I have attempted by considering 1+2 as a system, and the only external force on it during collision, which causes a change of momentum of the system, which I think is the change in momentum of the center of mass (maybe wrong). clearly from the velocity equations above, the center of mass has an inital V_cm=0. so the inital momentum is 0.

similarly here is ball 3 during the collision. I have assumed the velocity right after the collision as V_1f.

coefficient of restitution equation.
solving all of these, I end up with a wrong answer,
$$ v_{cm} = aw/3$$
I am worried that I have learnt some concept in a really wrong manner, and all help is much appreciated.

 

[kuruman]

I would not consider 1+2 as a system. I would consider finding the speed of particle 2 immediately after the collision with 3. I would assume that the velocity of 1 does not change appreciably while the collision takes place. Thus, I know the velocities of 1 and 2 after the collision.

 

[palaphys]

I would not consider 1+2 as a system. I would consider finding the speed of particle 2 immediately after the collision with 3. I would assume that the velocity of 1 does not change appreciably while the collision takes place. Thus, I know the velocities of 1 and 2 after the collision.

why not consider 1+2 as a system? I have tried what you have suggested and I got v final as 3/4 aw which indeed is the right answer. But why is what i did incorrect?

 

[kuruman]

But why is what i did incorrect?

I don't see equations expressing energy and momentum conservation anywhere. Also how do you conclude that

if the collision occurs at t=0, then v1 reduces to A and v2 reduces to -A, indicating that the balls are in extreme positions.

If the balls are at extreme positions then their velocities must be zero, not $-a\omega$ and $+a\omega.$

The key idea here is that the collision is over before the information that it has taken place is transmitted, through the spring, to mass 1. So the new constant velocity of the CM is determined by the post collision velocity of mass 2 and the pre-collision velocity of mass 1.

 

[palaphys]

I don't see equations expressing energy and momentum conservation anywhere. Also how do you conclude that

If the balls are at extreme positions then their velocities must be zero, not $-a\omega$ and $+a\omega.$

The key idea here is that the collision is over before the information that it has taken place is transmitted, through the spring, to mass 1. So the new constant velocity of the CM is determined by the post collision velocity of mass 2 and the pre-collision velocity of mass 1.

I've mentioned the equations I wrote down on the image. my bad, i wanted to say it was the equilibirum position

 

[palaphys]

The key idea here is that the collision is over before the information that it has taken place is transmitted, through the spring, to mass 1. So the new constant velocity of the CM is determined by the post collision velocity of mass 2 and the pre-collision velocity of mass 1.

sorry but i dont understand what you mean by information transmission here. nor do i understand the last line.

 

[kuruman]

sorry but i dont understand what you mean by information transmission here. nor do i understand the last line.

I will explain in a few hours. I have to sign off now.

 

[haruspex]

sorry but i dont understand what you mean by information transmission here. nor do i understand the last line.

Another way of saying it:
Say the collision occurs over some small time interval $\Delta t$.
During that time, there is a large, varying, but unknown force, $F(t)$, between the particles 2 and 3. The net effect of that is to transfer momentum $\int_{t=0}^{\Delta t} F(t).dt$ between those particles. The momentum transferred can be computed from conservation laws. The shorter $\Delta t$, the larger the average of $F(t)$ must be.
Meanwhile, there is some more modest force, $F'(t)$, acting between particles 1 and 2, mediated by the spring. The magnitude of that force depends on the spring length and will change very little during the collision. The effect of that force is to transfer momentum $\int_{t=0}^{\Delta t} F'(t).dt$ between particles 1 and 2.
As we let $\Delta t\rightarrow 0$, $\int_{t=0}^{\Delta t} F(t).dt$ is constant but $\int_{t=0}^{\Delta t} F'(t).dt \rightarrow 0$.

In the present case, the question has been structured so that $F'$ is negligible during the collision anyway, but the above result does not depend on that. It would still have applied even if the spring had been at maximum compression at the time.

 

[kuruman]

To @palaphys: I think that @haruspex has said more or less what I was going to say. If you need additonal clarifications, please ask.

 

[palaphys]

Another way of saying it:
Say the collision occurs over some small time interval $\Delta t$.
During that time, there is a large, varying, but unknown force, $F(t)$, between the particles 2 and 3. The net effect of that is to transfer momentum $\int_{t=0}^{\Delta t} F(t).dt$ between those particles. The momentum transferred can be computed from conservation laws. The shorter $\Delta t$, the larger the average of $F(t)$ must be.
Meanwhile, there is some more modest force, $F'(t)$, acting between particles 1 and 2, mediated by the spring. The magnitude of that force depends on the spring length and will change very little during the collision. The effect of that force is to transfer momentum $\int_{t=0}^{\Delta t} F'(t).dt$ between particles 1 and 2.
As we let $\Delta t\rightarrow 0$, $\int_{t=0}^{\Delta t} F(t).dt$ is constant but $\int_{t=0}^{\Delta t} F'(t).dt \rightarrow 0$.

here is what I understand from this-some large unknown force acts on the two masses 2 and 3 during collision. And you say that there is some other force acting between particles 1 and 2. (spring force??) But how is this implying that what I have done is incorrect?? still can't figure it out.

 

[haruspex]

But how is this implying that what I have done is incorrect?

You treated 1 and 2 as though connected by a rigid rod. This led to too much momentum being transferred to those two. As @kuruman and I have shown, you can treat the collision as only involving 2 and 3. Its effect on 1 only comes later.
Bear in mind that particle 1 doesn’t "know" about 2 and 3. All it experiences directly is the force from the spring, and that depends entirely on the length of the spring. The collision doesn't suddenly change the length of the spring, only the velocities of particles 1 and 2, so not even the acceleration of particle 1 changes instantly, let alone its velocity.

 

[palaphys]

You treated 1 and 2 as though connected by a rigid rod. This led to too much momentum being transferred to those two. As @kuruman and I have shown, you can treat the collision as only involving 2 and 3. Its effect on 1 only comes later.
Bear in mind that particle 1 doesn’t "know" about 2 and 3. All it experiences directly is the force from the spring, and that depends entirely on the length of the spring. The collision doesn't suddenly change the length of the spring, only the velocities of particles 1 and 2, so not even the acceleration of particle 1 changes instantly, let alone its velocity.

SO I can't just treat the spring block system as a rigid body? correct?

 

[palaphys]

Bear in mind that particle 1 doesn’t "know" about 2 and 3. All it experiences directly is the force from the spring, and that depends entirely on the length of the spring. The collision doesn't suddenly change the length of the spring, only the velocities of particles 1 and 2, so not even the acceleration of particle 1 changes instantly, let alone its velocity.

but how do we find the length of the spring at t=0? i thought Fnet/total mass= acceleration of system or cm

 

[kuruman]

Let me add that in your expression for the coefficient of restitution you are considering the collision between mass 1 and the CM. Mass 1 collides with mass 2, not the CM. That's what we've been trying to tell you.

 

[kuruman]

but how do we find the length of the spring at t=0? i thought Fnet/total mass= acceleration of system or cm

You have the position of each mass at t = 0. What is the length of the string?

 

[palaphys]

Let me add that in your expression for the coefficient of restitution you are considering the collision between mass 1 and the CM. Mass 1 collides with mass 2, not the CM. That's what we've been trying to tell you.

okay but as per the force diagram it seems as if the system masses 1+2 is experiencing a force N, which should lead to a change in momentum of the velocity of the CM. I have been told that external forces acting on a body are external forces on the system.

 

[haruspex]

Let me add that in your expression for the coefficient of restitution you are considering the collision between mass 1 and the CM. Mass 1 collides with mass 2, not the CM. That's what we've been trying to tell you.

The numbering is right to left: 3 collides with 2.

 

[haruspex]

okay but as per the force diagram it seems as if the system masses 1+2 is experiencing a force N, which should lead to a change in momentum of the velocity of the CM. I have been told that external forces acting on a body are external forces on the system.

It does indeed lead to a change in velocity of the CM, but not as great as if masses 1 and 2 were connected by a rigid rod.
Forget the spring for a moment. There is nothing to stop you considering masses 1 and 2 as a system even though they're entirely disconnected. If, initially, 1 and 2 are at rest then the collision will leave 3 at rest and 2 heading towards 1 at velocity $u_0$. The CM of 1+2 is now moving at velocity $u_0/2$.
Alternatively, if we replace the spring by a rigid rod, the collision will leave 3 rebounding at speed $u_0/3$ (velocity $-u_0/3$) and the 1+2 CM moving off at velocity $2u_0/3$.
Work and momentum are conserved in both cases.

 

[palaphys]

. If, initially, 1 and 2 are at rest then the collision will leave 3 at rest and 2 heading towards 1 at velocity $u_0$. The CM of 1+2 is now moving at velocity $u_0/2$.
Alternatively, if we replace the spring by a rigid rod, the collision will leave 3 rebounding at speed $u_0/3$ (velocity $-u_0/3$) and the 1+2 CM moving off at velocity $2u_0/3$.
Work and momentum are conserved in both cases.

sorry but I still cannot understand where I am going wrong. so first of all
1) the spring- mass system is not a rigid body?
2)please point out where specifically i have gone wrong in my work

 

[haruspex]

1) the spring- mass system is not a rigid body?

Clearly it is not rigid.

2)please point out where specifically i have gone wrong in my work

Your equation beginning “e = " is Newton’s Experimental Law. It assumes the two masses are rigid bodies, and that they are not rotating.
For the e=1 case, it can be derived from momentum and work both being conserved, work being taken to be the sum of the KEs of the two bodies. But the KE of the 1+2 system is more than $\frac 12(m+m)v_{cm}^2$.

 

[palaphys]

Your equation beginning “e = " is Newton’s Experimental Law. It assumes the two masses are rigid bodies, and that they are not rotating.

so the mistake I am making is misuse of newtons experimental law?
also how is the KE of the system more than 1/2 2m vcm 2?

 

[palaphys]

so @haruspex the final momentum in my equations i.e 2mvcm,f is incorrect?

 

[haruspex]

so the mistake I am making is misuse of newtons experimental law?

Yes.

also how is the KE of the system more than 1/2 2m vcm 2?

You can think of the motion of the two particles as the sum of the motion of their CM plus their motions relative to it. There is KE in each:
$v_{cm}(m_1+m_2)=m_1v_1+m_2v_2$

$v_{cm}^2(m_1+m_2)+m_1(v_1-v_{cm})^2+m_2(v_2-v_{cm})^2$
$=2v_{cm}^2(m_1+m_2)-2v_{cm}(m_1v_1+m_2v_2)+m_1v_1^2+m_2v_2^2$
$=2v_{cm}(v_{cm}(m_1+m_2)-(m_1v_1+m_2v_2))+m_1v_1^2+m_2v_2^2$
$=m_1v_1^2+m_2v_2^2$

 

[palaphys]

So I've thought about this for a while and here's what I think about where I went wrong. Please correct me here:
So the block from the left collides with the mass 2 connected with the spring. As the mass is in equilibrium position in SHM, no force acts on it. so momentum conservation is valid. Also during the collision, the spring does not transmit the change instantly so as to change the velocity of the block on the far end. I assume there is some sort of delay (not sure how this works). Hence I got to use energy conservation and momentum, as coefficent of restitution is valid for collision of rigid masses.?

 

[palaphys]

Yes.

You can think of the motion of the two particles as the sum of the motion of their CM plus their motions relative to it. There is KE in each:
$v_{cm}(m_1+m_2)=m_1v_1+m_2v_2$

$v_{cm}^2(m_1+m_2)+m_1(v_1-v_{cm})^2+m_2(v_2-v_{cm})^2$
$=2v_{cm}^2(m_1+m_2)-2v_{cm}(m_1v_1+m_2v_2)+m_1v_1^2+m_2v_2^2$
##=2v_{cm}(v_{cm}(m_1+m_2)-(m_1v_1+m_2v_2))+m_

The key idea here is that the collision is over before the information that it has taken place is transmitted, through the spring, to mass 1. So the new constant velocity of the CM is determined by the post collision velocity of mass 2 and the pre-collision velocity of mass 1.

OHHH i get it now, so the collision between the forward two masses is not going to affect the velocity of the right end mass just after collision, and it would only further complicate things if the right two masses are considered as a system as for one, coefficient or restitution would not apply and the kinetic energy of the system would just become more complicated..?

 

[haruspex]

So I've thought about this for a while and here's what I think about where I went wrong. Please correct me here:
So the block from the left collides with the mass 2 connected with the spring. As the mass is in equilibrium position in SHM, no force acts on it. so momentum conservation is valid.

As I wrote, it would still be valid even if the spring were exerting a force. This is because we assume, as an idealisation, that the collision takes negligible time, so the momentum imparted by the spring (force x time) during that time is negligible. The force exerted by particle 3 during the collision is arbitrarily great, so the momentum imparted by that is not negligible.
Something similar applies when a block hits another horizontally on a rough horizontal surface. The friction doesn’t have time to exert an impulse, so can be ignored during the impact.

Also during the collision, the spring does not transmit the change instantly so as to change the velocity of the block on the far end. I assume there is some sort of delay (not sure how this works).

In reality, yes, there is a delay. Einstein explained that. But we do not need to appeal to relativity.
If you accelerate particle 2 at rate $1m/s^2$ for 0.1s then at the end of that time it has only moved 5mm. If the spring is 1m long, that won't have changed the force in the spring much, so the acceleration of particle 1 is much smaller so far.

 

[palaphys]

ok