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Question: Two particles A & B of masses m and 2m respectively are connected through a spring in its natural le...

Two particles A & B of masses m and 2m respectively are connected through a spring in its natural length. They are projected directly away from each other along spring with the same speed. Mark the correct statement(s):

A

At the moment of maximum distance between them, the particles are moving with same speed in same direction

B

At the moment of maximum distance between them, the particles are moving with same speed in opposite direction

C

Minimum individual speed of the particles A & B are both zero

D

Minimum individual speed of the particles A & B are both non-zero

Answer

A, C

Explanation

Solution

The problem involves two particles of different masses connected by a spring, undergoing motion under internal forces. We need to analyze their motion using conservation laws.

1. Conservation of Linear Momentum and Center of Mass Velocity:

Let the mass of particle A be mA=mm_A = m and particle B be mB=2mm_B = 2m.
Let the initial speed of projection be vv. Since they are projected directly away from each other along the spring, if A moves in the positive direction, B moves in the negative direction.
Initial velocity of A: v⃗A,i=vi^\vec{v}_{A,i} = v \hat{i}
Initial velocity of B: v⃗B,i=−vi^\vec{v}_{B,i} = -v \hat{i}

The total initial linear momentum of the system is:
Pi=mAv⃗A,i+mBv⃗B,i=m(vi^)+2m(−vi^)=mvi^−2mvi^=−mvi^P_i = m_A \vec{v}_{A,i} + m_B \vec{v}_{B,i} = m(v \hat{i}) + 2m(-v \hat{i}) = mv \hat{i} - 2mv \hat{i} = -mv \hat{i}

Since there are no external forces acting on the system, the total linear momentum is conserved.
The velocity of the center of mass (vCMv_{CM}) remains constant:
v⃗CM=mAv⃗A,i+mBv⃗B,imA+mB=mvi^−2mvi^m+2m=−mvi^3m=−v3i^\vec{v}_{CM} = \frac{m_A \vec{v}_{A,i} + m_B \vec{v}_{B,i}}{m_A + m_B} = \frac{mv \hat{i} - 2mv \hat{i}}{m + 2m} = \frac{-mv \hat{i}}{3m} = -\frac{v}{3} \hat{i}

2. Analysis of Statement (A) and (B) - At the moment of maximum distance:

When the distance between the particles is maximum (i.e., the spring is maximally extended), their relative velocity must be momentarily zero. This means both particles are moving with the same velocity: v⃗A=v⃗B=v⃗′\vec{v}_A = \vec{v}_B = \vec{v}'.

Applying conservation of momentum at this instant:
Pf=(mA+mB)v⃗′=(m+2m)v⃗′=3mv⃗′P_f = (m_A + m_B) \vec{v}' = (m + 2m) \vec{v}' = 3m \vec{v}'
Since Pf=Pi=−mvi^P_f = P_i = -mv \hat{i}:
3mv⃗′=−mvi^3m \vec{v}' = -mv \hat{i}
v⃗′=−v3i^\vec{v}' = -\frac{v}{3} \hat{i}

So, at the moment of maximum distance, both particles are moving with the same speed (v/3v/3) in the same direction (negative x-direction, which is the initial direction of B).
Therefore, statement (A) is correct and statement (B) is incorrect.

3. Analysis of Statement (C) and (D) - Minimum individual speed of the particles:

To find the minimum individual speed, we can express the velocities of the particles in the center of mass (CM) frame.
Velocity of a particle in the lab frame = Velocity of CM + Velocity of particle in CM frame.
v⃗A=v⃗CM+v⃗A,CM\vec{v}_A = \vec{v}_{CM} + \vec{v}_{A,CM}
v⃗B=v⃗CM+v⃗B,CM\vec{v}_B = \vec{v}_{CM} + \vec{v}_{B,CM}

Initial velocities in CM frame:
v⃗A,CM,i=v⃗A,i−v⃗CM=vi^−(−v3i^)=4v3i^\vec{v}_{A,CM,i} = \vec{v}_{A,i} - \vec{v}_{CM} = v \hat{i} - (-\frac{v}{3} \hat{i}) = \frac{4v}{3} \hat{i}
v⃗B,CM,i=v⃗B,i−v⃗CM=−vi^−(−v3i^)=−2v3i^\vec{v}_{B,CM,i} = \vec{v}_{B,i} - \vec{v}_{CM} = -v \hat{i} - (-\frac{v}{3} \hat{i}) = -\frac{2v}{3} \hat{i}

In the CM frame, the particles oscillate. The velocities in the CM frame will vary sinusoidally. Let the angular frequency of oscillation be ω\omega.
v⃗A,CM(t)=4v3cos⁡(ωt)\vec{v}_{A,CM}(t) = \frac{4v}{3} \cos(\omega t) (assuming ϕ=0\phi=0 for initial phase)
v⃗B,CM(t)=−2v3cos⁡(ωt)\vec{v}_{B,CM}(t) = -\frac{2v}{3} \cos(\omega t) (since mAv⃗A,CM+mBv⃗B,CM=0m_A \vec{v}_{A,CM} + m_B \vec{v}_{B,CM} = 0)

Now, let's find the absolute velocities in the lab frame:
v⃗A(t)=v⃗CM+v⃗A,CM(t)=−v3i^+4v3cos⁡(ωt)i^=v3(−1+4cos⁡(ωt))i^\vec{v}_A(t) = \vec{v}_{CM} + \vec{v}_{A,CM}(t) = -\frac{v}{3} \hat{i} + \frac{4v}{3} \cos(\omega t) \hat{i} = \frac{v}{3} (-1 + 4 \cos(\omega t)) \hat{i}
v⃗B(t)=v⃗CM+v⃗B,CM(t)=−v3i^−2v3cos⁡(ωt)i^=v3(−1−2cos⁡(ωt))i^\vec{v}_B(t) = \vec{v}_{CM} + \vec{v}_{B,CM}(t) = -\frac{v}{3} \hat{i} - \frac{2v}{3} \cos(\omega t) \hat{i} = \frac{v}{3} (-1 - 2 \cos(\omega t)) \hat{i}

For the minimum individual speed of particle A to be zero, we need vA(t)=0v_A(t) = 0:
v3(−1+4cos⁡(ωt))=0\frac{v}{3} (-1 + 4 \cos(\omega t)) = 0
−1+4cos⁡(ωt)=0  ⟹  cos⁡(ωt)=14-1 + 4 \cos(\omega t) = 0 \implies \cos(\omega t) = \frac{1}{4}
Since cos⁡(ωt)=1/4\cos(\omega t) = 1/4 is a physically possible value (between -1 and 1), particle A can momentarily come to rest.
Thus, the minimum individual speed of particle A is zero.

For the minimum individual speed of particle B to be zero, we need vB(t)=0v_B(t) = 0:
v3(−1−2cos⁡(ωt))=0\frac{v}{3} (-1 - 2 \cos(\omega t)) = 0
−1−2cos⁡(ωt)=0  ⟹  cos⁡(ωt)=−12-1 - 2 \cos(\omega t) = 0 \implies \cos(\omega t) = -\frac{1}{2}
Since cos⁡(ωt)=−1/2\cos(\omega t) = -1/2 is a physically possible value (between -1 and 1), particle B can momentarily come to rest.
Thus, the minimum individual speed of particle B is zero.

Therefore, statement (C) is correct and statement (D) is incorrect.

Conclusion:
Both statements (A) and (C) are correct.