Solving Mechanics IIT-JEE Homework: Find Min Mass & Time Between Collisions

[zorro]

Homework Statement

A ball of mass M moving with velocity v on a frictionless surface strikes the first of two identical balls, each of mass m = 2 kg, connected by a massless spring with spring constant k = 1 kg/s². Consider the collision to be central and elastic and essentially instantaneous. (see the attached fig.)

(a) Find the minimum value of the mass M for the incident ball to strike the system of two balls again.
(b) How much time will elapsed between the two collisions?

The Attempt at a Solution

This is my thought process. We use Newton' LOR and Conservation of momentum to find out the final absolute velocities of the masses involved in first collision. Next, we can find out the displacements of the connected masses and the mass M from the initial position.

These are my values up to the above analysis:

v1 = velocity of mass 'm' after collision
v2= " " " 'M' " "

v1= 2v/1+k
v2= v(1-k)/(1+k) where k=m/M

x1=2vt/(1+k)
x2= vt(1-k)/(1+k)

I don't have any idea after this :|
Any ideas?

 

[zorro]

Any genius in mechanics out there?

 

[hikaru1221]

Next time, you should post in the introductory physics forum. I think there are more instructors willing to help there
After the collision, change your reference frame to the frame of COM of the 2 masses m and see what happens

 

[zorro]

I think there are more instructors willing to help there

Yeah you are right, there are more instructors who are willing to help in IP, its a different thing that they are unable to help (I have no problem in posting in IP, but I had a bad experience with such type of questions there which require higher thinking skills)

After the collision, change your reference frame to the frame of COM of the 2 masses m and see what happens

Ok, then I will get the velocities of the two masses w.r.t the centre of mass. They move towards each other.

X_1,CM = vt/1+k , X_2,CM=-vt/1+k

Here ₂ stands for the second mass 'm'.

The two masses execute SHM. I have no idea how mass of M can be related here.

 

[hikaru1221]

Uhh, I think you're underestimating them
The 2 masses m's perform SHM, while M moves with constant velocity. Then write down the equations of coordinates of M and the mass m nearer to M. If they collide again then...
P.S.: I don't think x1 in post #1 is correct. It's supposed to be SHM, right?

 

[zorro]

If they collide again then...

the two co-ordinates are equal.

P.S.: I don't think x1 in post #1 is correct. It's supposed to be SHM, right?

Yeah, that's valid until the first collision of two connected masses.

I have a problem writing the equation for SHM.
X=Asin(ωt+ϕ)
here A = vt/1+k and ω=(2k/m)^1/2

Shall I take ϕ=0 at t=0? then equate the X above with the co-ordinate of bigger mass? I am getting stuck in this.

 

[hikaru1221]

Yeah, that's valid until the first collision of two connected masses.

Huh?
Before the first collision, the 2m don't even move, so how come x1 change with time before collision?

I have a problem writing the equation for SHM.
X=Asin(ωt+ϕ)
here A = vt/1+k and ω=(2k/m)^1/2

Shall I take ϕ=0 at t=0?

Why? That's correct by the way But you should reconsider A, or x1.

then equate the X above with the co-ordinate of bigger mass? I am getting stuck in this.

Yep.

 

[zorro]

Huh?
Before the first collision, the 2m don't even move, so how come x1 change with time before collision?

Sorry I misunderstood some thing.

Why? That's correct by the way But you should reconsider A, or x1.

Ah yes A is wrong. From energy conservation, I got A=vt/ω(1+k)

I am still confused if ϕ=0 or not.
At t=0, the mass m and M collide. This initial position of 'm' is not its mean position, right?
Its mean position occurs at the instant when the two smaller masses collide.

 

[hikaru1221]

Ah yes A is wrong. From energy conservation, I got A=vt/ω(1+k)

Why is there vt? Amplitude changes with time?

I am still confused if ϕ=0 or not.
At t=0, the mass m and M collide. This initial position of 'm' is not its mean position, right?
Its mean position occurs at the instant when the two smaller masses collide.

How do the 2 smaller masses collide? Aren't they connected by a spring?
I think you're misunderstanding the whole phenomenon.

 

[zorro]

Why is there vt? Amplitude changes with time?

Ignore 't'

How do the 2 smaller masses collide? Aren't they connected by a spring?
I think you're misunderstanding the whole phenomenon.

I used the wrong word there. I meant at the instant when the two smaller masses come to instantaneous rest from COM frame (maximum compression of the spring).

 

[hikaru1221]

Initial position = position at t=0
So phi doesn't have to be equal to 0. Just use the initial conditions to solve the problem:
_ At t=0, x_m = 0, v_m = -v1 (moving towards the other mass m)
_ We have: x_m = Asin(wt + phi), and so: v_m = Awcos(wt + phi).
_ Plug the values in and solve for A and phi. You have 2 unknowns with 2 equations, then you can solve it
And you will see that phi is not 0

 

[zorro]

Initial position = position at t=0
So phi doesn't have to be equal to 0. Just use the initial conditions to solve the problem:
_ At t=0, x_m = 0, v_m = -v1 (moving towards the other mass m)
_ We have: x_m = Asin(wt + phi), and so: v_m = Awcos(wt + phi).
_ Plug the values in and solve for A and phi. You have 2 unknowns with 2 equations, then you can solve it
And you will see that phi is not 0

Initial position = x_m=0 at t=0 so ϕ=0
I already got A earlier, = v/ω(1+k)

How do I put a constraint on M?

 

[hikaru1221]

Initial position = x_m=0 at t=0 so ϕ=0

What about phi = pi/2? Don't take the math in physics so lightly.

Now get the motion equation for M. Should the masses collide, the equation x_M = x_m must have solution. Find the condition.

 

[zorro]

What about phi = pi/2? Don't take the math in physics so lightly.

x_m = Asin(wt + phi)
x_m=0 at t=0

S0
0=Asin(phi)
If phi =pi/2 the equation is not valid!, so phi=0!

Do you get what I am talking about?
Xm = Asinwt + Xcm
XM=vt(1-k)/(1+k)

solving them, I got
sinwt = -wtk

How do I solve this equation?

 

[hikaru1221]

x_m = Asin(wt + phi)
x_m=0 at t=0

S0
0=Asin(phi)
If phi =pi/2 the equation is not valid!, so phi=0!

Do you get what I am talking about?

Oops, I'm sorry, typo, it should be phi = pi, not pi/2

XM=vt(1-k)/(1+k)

Is it? Have you switched to the reference frame of COM yet?

Let me go through this:

1/ Right after the collision, in the reference frame of ground:
v = v_2 + kv_1
v^2 = v_2^2 + kv_2^2
Solve this:
v_2 = (1-k)v/(1+k)
v_1 = 2v/(1+k)
v_{COM} = v_1/2 = v/(1+k)
However, notice that we implicitly assume the +ve is in the same direction as \vec{v} (or to the right, as in the picture).

2/ Switching to the reference frame of COM:
Now it's important to notice that the we can choose +ve to be either to the right or to the left. I'll choose it to the right, consistent with above.
v'_2 = v_2 - v_{COM} = -kv/(1+k)
v'_1 = v_1 - v_{COM} = v/(1+k)

Now choose the origin at the initial position of M and m right after the collision. We then have:
x_M = - kvt/(1+k)
x_m = Asin(wt+\phi)

The initial conditions for m are:
(1) x_m(0) = Asin\phi = 0
(2) v_m(0) = Awcos\phi = v/(1+k)
From (1), we have \phi is either 0 or \pi. For A>0, from (2), we can choose \phi = 0. So: A = v/w(1+k). Notice that I have to take into account v_m(0) and choose A>0, besides looking at x_m(0) in order to get phi = 0.
Therefore:
x_m = \frac{v}{w(1+k)}sin(wt)

Equate x_M and x_m:
- kvt/(1+k) = \frac{v}{w(1+k)}sin(wt)
kwt = - sin(wt) (***)

Now notice that in the reference frame of COM, v_M < 0 (M going away from m) and the smaller M is, the larger k be, the faster M "escapes" from m. Meanwhile, m is framed to perform SHM and cannot go anywhere farther than A. So in the extreme case, the next collision will occur at x = -A, or m already performs SHM for 3/4 of period, which means t = 3\pi /2w. Substitute this into (***), and we obtain M = ... That's the minimum M needed.

 

[zorro]

Is it? Have you switched to the reference frame of COM yet?

I switched back to the ground frame, that's why I added the Xcm term in Xm = Asinwt + Xcm

Now notice that in the reference frame of COM, v_M < 0 (M going away from m) and the smaller M is, the larger k be, the faster M "escapes" from m. Meanwhile, m is framed to perform SHM and cannot go anywhere farther than A. So in the extreme case, the next collision will occur at x = -A, or m already performs SHM for 3/4 of period, which means t = 3\pi /2w. Substitute this into (***), and we obtain M = ... That's the minimum M needed.

Yeah I had a problem there.
Its fine now.
I got the answer as 2m/3(pi) (approx.)

Thanks for your precious time and help