Repeated Collision: Find Minimum Mass M

[peripatein]

Hi,

Homework Statement

I'd like to find the minimum mass M which will render possible a second collision between M and the two masses m and the spring in the attachment, and also the time between the two collisions. It is stated that the initial velocity of M is v0 as indicated and that the surface is frictionless. It is also stated that this is an elastic collision, whose time is very short. M doesn't get attached to the masses.

Homework Equations

The Attempt at a Solution

I am not quite sure how to formulate a condition on M, which will enable a second collision. I am not even sure how a second collision would be possible unless there were let's say a wall on the other side hitting M back towards the masses and the spring. The kinetic energy of M is turned into an elastic energy of the spring and kinetic energy of both M and the reduced mass = m/2, right (due to conservation of energy)?
Furthermore, there is conservation of linear momentum, isn't there?
Must I work in the CM reference frame?

 

[Fightfish]

It is also stated that this is an elastic collision, whose time is very short

Physically, this is what happens:

(1) Mass M strikes the left mass m elastically. Speed of mass M is reduced, while mass m acquires a finite speed. At this point in time, the right mass m is unaffected and the spring is still uncompressed.

(2) As the left mass m moves leftward, the spring compresses, and the right mass m begins to move as well. The system then reaches equilibrium when both masses are moving at the same speed. This speed is obviously less than the original speed of the left mass just after the collision. This is why the mass M can catch up and collide with the system again if conditions are met.

 

[peripatein]

Alright, so a fraction of a second after the collision:
(1/2)Mv0^2 = (1/2)mv^2 (conservation of energy)
Mv0 = -mv (conservation of momentum)
Are these correct?
Now, as the left mass m begins to move, could I write:
(1/2)mv^2=(1/2)kx^2 + (1/2)*mred*u^2
where mred = m/2?

 

[Fightfish]

(1/2)Mv0^2 = (1/2)mv^2 (conservation of energy)
Mv0 = -mv (conservation of momentum)

No. Mass M does not stop after the collision.

Now, as the left mass m begins to move, could I write:
(1/2)mv^2=(1/2)kx^2 + (1/2)*mred*u^2
where mred = m/2?

No either. Why do you want to use reduced mass?
For the second part, conservation of momentum would suffice.

 

[peripatein]

Let me then try this:
(1) (1/2)M(v0)^2 = (1/2)mv^2 + (1/2)M(v1)^2 (conservation of energy)
(2) (1/2)mv^2 + (1/2)M(v1)^2 = (1/2)kx^2 + (1/2)*2m*u^2
(3) (M/2m)v0 = [2m*u + M(v1)]/[2m+M]
Are these somewhat better?

 

[Fightfish]

No, you should treat the two collisions separately.

Collision 1
$$\frac{1}{2}Mv_{0}^{2} = \frac{1}{2}Mv_{1}^{2} + \frac{1}{2}mu_{0}^{2}\\
Mv_{0} = Mv_{1} + mu_{0}$$

Collision 2
$$\frac{1}{2}mu_{0}^{2} = \frac{1}{2}k x^{2} + mu_{1}^{2}\\
mu_{0} = 2 mu_{1}$$

For collision 2, the COE equation is not useful because you don't need the spring compression and you cannot directly compute it without already knowing the initial and final speeds.

 

[peripatein]

But mass M is not at rest after collision 1 takes place, as you've pointed out. How come it is not included in any of the equations referring to collision 2?

 

[Fightfish]

But mass M is not at rest after collision 1 takes place, as you've pointed out. How come it is not included in any of the equations referring to collision 2?

Sorry if I haven't been clear, but collision 2 refers to the 'collision' between the two masses m, and not the repeated collision between mass M and the spring-mass system.

 

[peripatein]

Right, but whilst writing your COE equations, aren't you expected to refer to the system as a whole?

 

[Fightfish]

You could, but nothing happens to the mass M in the collision 2 that I am referring to. So the terms before and after would just cancel each other out.

 

[peripatein]

Fair enough. How could I now set a condition on mass M to make a second collision possible? You wrote "if conditions are met". Which are the conditions?

 

[Fightfish]

Fair enough. How could I now set a condition on mass M to make a second collision possible? You wrote "if conditions are met". Which are the conditions?

Well, to solve for the conditions is the goal of the question isn't it?
Well, you need the mass M to be able to catch up and hit the spring-mass system. So that means that $v_{1}$ must be greater than $u_{1}$.

 

[peripatein]

Did you not mean that v1 must be less than u1? If v1 were greater, M would "flee" away before the spring-mass system could hit it again.

 

[Chestermiller]

Why don't you just solve for the motion of each of the three masses as a function of time using Newton's second law on each of them (and applying the appropriate initial conditions for conservation of energy and momentum at t = 0)? Then there will be no guessing involved. You already have written the equations for establishing the initial conditions on M and the first mass m.

 

[Fightfish]

Did you not mean that v1 must be less than u1? If v1 were greater, M would "flee" away before the spring-mass system could hit it again.

Mass M must be moving leftwards even after the first collision in order for the repeated collision to occur. This is because the spring-mass system will always be moving leftwards.

 

[peripatein]

Alright, I have managed to find the condition: M_min = 2m.
However, I am not quite sure how to determine the time elapsed between the two collisions.
Need I to transfer to the center of mass frame of reference, determine x_CM(t) before and after the first collision?

 

[peripatein]

Rereading the previous replies, should I instead find x(t) for M and x(t) for the reduced mass with the spring k and find out the times at which they are equal? Or, even more simply, determine when x(t) for the reduced mass is zero?

 

[peripatein]

Shouldn't the time between the two collisions simply be pi/omega, where omega=sqrt(k/reduced mass)?

 

[Chestermiller]

Let the subscript M refer to the mass M, and let the subscripts 1 and 2 refer to the first mass and the second mass. From the equations you've already derived, the initial velocities of the masses after the initial collision are v_M=v₀(M-m)/(M+m), v₁=2mv₀/(M+m), and v₂=0. The initial locations of the masses at the initial collision are x_M=x₁=0, and x₂=L, where L is the undeformed length of the spring. The velocity of the mass M does not change after this. Therefore, x_M=v₀(M-m)t/(M+m).

The force balance equations for two small masses are:
m\frac{d^2x_1}{dt^2}=k(x_2-x_1-L)
and
m\frac{d^2x_2}{dt^2}=k(x_1-x_2+L)
If we add these two equations, we get
2m\frac{d^2x_1}{dt^2}=0
The solution to this equation, subject to the initial conditions is :
x_1+x_2=L+2mv_0t/(M+m)
If we subtract the two force balance equations, we obtain:
m\frac{d^2δ}{dt^2}=-2kδ
where δ=(x_1-x_2+L).
Solve this equation for δ subject to the initial conditions δ=0 and dδ/dt=2mv₀/(M+m) at t = 0.

 

[peripatein]

Okay, I got:
δ = (2mv₀/[ω(M+m)]) * sin(ωt)
where ω = √(2k/m)
Finding x₁ and equating x₁(t) and x_M(t) I obtained:
sin(ωt)=(1/2)ωt
which can be solved graphically to give t≈1.9/ω
Would you agree?

 

[haruspex]

Okay, I got:
δ = (2mv₀/[ω(M+m)]) * sin(ωt)
where ω = √(2k/m)
Finding x₁ and equating x₁(t) and x_M(t) I obtained:
sin(ωt)=(1/2)ωt
which can be solved graphically to give t≈1.9/ω
Would you agree?

You can do better than resorting to a graphical solution.
You have reduced the problem to solving an equation pair like
y = at, y = sin(ωt)
for some t > 0.
But you are specifically looking for the lowest (most negative) a for which there is a solution.
What, geometrically, is the relationship between the straight line and the sine curve in that case?

 

[Chestermiller]

Okay, I got:
δ = (2mv₀/[ω(M+m)]) * sin(ωt)
where ω = √(2k/m)
Finding x₁ and equating x₁(t) and x_M(t) I obtained:
sin(ωt)=(1/2)ωt
which can be solved graphically to give t≈1.9/ω
Would you agree?

Your solution for δ looks right, but what happened to all the M's and m's in the solution? You're trying to find the value of M that barely allows two hits. Please show us your work. Also, as Haruspex was alluding to, not only must the displacements match, but also something else must match.

Chet

 

[peripatein]

Mind you, your expressions for v₁ and v_M would not be accurate unless M were 2m. I have substituted them in all the equations and an equality was only attained upon equating M to 2m. This is what I ended up substituting. In any case, I did make a mistake earlier. Redoing some of the work, I obtained:
x₁ = (mv₀t)/(M+m) + δ/2 = x_M = (v₀(M-m)t)/(M+m)
which yielded sin(ωt)=0, hence t = π/ω.

 

[haruspex]

Mind you, your expressions for v₁ and v_M would not be accurate unless M were 2m.

Is that in response to Chester's analysis? You certainly should not be making such an approximation. It isn't necessary.

t = π/ω.

That is not the right answer.

 

[peripatein]

If I do not make such assumption/approximation, I obtain:
t(M-2m) = (m/w)sin(wt)

 

[peripatein]

Also, the general expression for v1 = Mv0/(M+m). How then Chester obtained what he wrote without equating M to 2m is beyond me.

 

[haruspex]

Also, the general expression for v1 = Mv0/(M+m). How then Chester obtained what he wrote without equating M to 2m is beyond me.

Typo. He meant v₁=2Mv0/(M+m)

 

[peripatein]

Okay, so let's go back to the reduced problem:
y=at, y=sin(wt)
How should I solve this if not graphically? What condition should I enforce?

 

[haruspex]

Okay, so let's go back to the reduced problem:
y=at, y=sin(wt)
How should I solve this if not graphically? What condition should I enforce?

In general, those two equations will have a solution at t = 0 and maybe no other solutions, or maybe lots of solutions. You want the minimum value of M (which will correspond to some extreme relationship between a and w) such that there is at least one solution for t > 0. Sketch that.

 

[peripatein]

But the minimum value of M is 2m (so that a second collision is possible), and so sin(wt) equals 0. Isn't it?
Did you not indicate that that was not how it ought to be approached?
And what do you mean by "sketch that"? I thought you insisted I should refrain from solving this graphically.
Please clarify.

 

[Chestermiller]

Typo. He meant v₁=2Mv0/(M+m)

Thanks haruspex. That was a mistake in algebra, not a typo.

With this correction, the equation for δ becomes:
\delta=\frac{2Mv_0}{ω(M+m)}sin(ωt)
and
x_1(t)=\frac{Mv_0t}{(M+m)}+\frac{Mv_0}{ω(M+m)}sin(ωt)=\frac{Mv_0}{ω(M+m)}(ωt+sin(ωt))
So, if x_M=\frac{v_0(M-m)t}{(M+m)}, we get
\frac{v_0(M-m)t}{(M+m)}=\frac{Mv_0t}{(M+m)}+\frac{Mv_0}{ω(M+m)}sin(ωt)
From this, it follows that:
-mωt=Msin(ωt)
An additional relationship is required if M barely catches up with m for a second time. This is that the velocities must match.
\frac{v_0(M-m)}{(M+m)}=\frac{Mv_0}{(M+m)}+\frac{Mv_0}{(M+m)}cos(ωt)
From this, it follows that
-m=Mcos(ωt)
The angle we're looking for is in the third quadrant where cosine and sine are both negative. So, sin(ωt)=-\sqrt{1-(\frac{m}{M})^2}
This is sufficient to solve for the value of M/m that makes good on both these equations.

Chet

 

[peripatein]

Actually, I didn't want to bring this up, but you're still wrong as far as I am concerned. Recheck your algebra. I have checked my calculations numerous times and v1 = Mv0/(m+M) not 2Mv0/(m+M).

 

[peripatein]

Adding that latter condition, I finally obtained:
tg(x)=x
where x=wt
That gives t~4.49/w

 

[Chestermiller]

Actually, I didn't want to bring this up, but you're still wrong as far as I am concerned. Recheck your algebra. I have checked my calculations numerous times and v1 = Mv0/(m+M) not 2Mv0/(m+M).

Uh Oh. I check it numerous times too, and got 2Mv0/(m+M). Here is the sequence of steps:

I started with the following:
v_M=v_0\frac{M-m}{M+m}
v_1=\frac{M(v_0-v_M)}{m}

Substituting the first equation into the second equation, I get:
v_1=\frac{Mv_0}{m}\left(1-\frac{(M-m)}{(M+m)}\right)=\frac{Mv_0}{m}\frac{(M+m)-(M-m)}{(M+m)}=\frac{Mv_0}{m}\frac{2m}{(M+m)}=\frac{2Mv_0}{M+m}
I can't see where I made a mistake.

 

[Chestermiller]

From the previous equations I presented, I get:

ωt=\sqrt{\left(\frac{M}{m}\right)^2-1}
so,
\cos\left(\sqrt{\left(\frac{M}{m}\right)^2-1}\right)=-\frac{m}{M}
This needs to be solved for m/M.

 

[peripatein]

I have checked mine several times as well and obtained the following:
v_M = (Mv₀ - 2mv₁)/M
v₁ = Mv₀/(M+m)
Hence,
v_M=v₀(M-m)/(M+m)

 

[haruspex]

Uh Oh. I check it numerous times too, and got 2Mv0/(m+M). Here is the sequence of steps:

I started with the following:
v_M=v_0\frac{M-m}{M+m}
v_1=\frac{M(v_0-v_M)}{m}

Substituting the first equation into the second equation, I get:
v_1=\frac{Mv_0}{m}\left(1-\frac{(M-m)}{(M+m)}\right)=\frac{Mv_0}{m}\frac{(M+m)-(M-m)}{(M+m)}=\frac{Mv_0}{m}\frac{2m}{(M+m)}=\frac{2Mv_0}{M+m}
I can't see where I made a mistake.

I get the same. peripatein, please post your calculation.

what do you mean by "sketch that"?

Sketching is not the same as solving graphically (which requires an accurate drawing).
I was trying to get you to see that the straight line representing the subsequent motion of M had to be tangential to the curve for the motion of m. This is the same as Chester's observation that the velocities must match as well as the positions, and therefore produces the same equation.

 

[peripatein]

I am going to leave it the way it is, so no need to repost. In any case, it doesn't matter much. The physics is the essence.
But thank you very much for your help; I am sincerely grateful! :)

 

[Chestermiller]

I have checked mine several times as well and obtained the following:
v_M = (Mv₀ - 2mv₁)/M
v₁ = Mv₀/(M+m)
Hence,
v_M=v₀(M-m)/(M+m)

The first of these equations is not consistent with the momentum balance. Where did the 2 come from? See post #6 by fightfish (which is correct).

 

[Chestermiller]

Adding that latter condition, I finally obtained:
tg(x)=x
where x=wt
That gives t~4.49/w

This result for x is approximately correct. So, m/M=-cos(4.49) is the solution you have been looking for. That completes the solution.