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PhyMedAdv 2008 • Paper 1

Allowed Final Momenta in Isolated Two Body Collision

Two balls, having linear momenta p⃗1=pi^\vec{p}_1 = p\hat{i} and p⃗2=−pi^\vec{p}_2 = -p\hat{i}, undergo a collision in free space. There is no external force acting on the balls. Let p⃗1′\vec{p}_1' and p⃗2′\vec{p}_2' be their final momenta. The following option(s) is(are) NOT ALLOWED for any non-zero value of p,a1,a2,b1,b2,c1p, a_1, a_2, b_1, b_2, c_1 and c2c_2.

Options

A

p⃗1′=a1i^+b1j^+c1k^\vec{p}_1' = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}
p⃗2′=a2i^+b2j^\vec{p}_2' = a_2\hat{i} + b_2\hat{j}

Correct
B

p⃗1′=c1k^\vec{p}_1' = c_1\hat{k}
p⃗2′=c2k^\vec{p}_2' = c_2\hat{k}

C

p⃗1′=a1i^+b1j^+c1k^\vec{p}_1' = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}
p⃗2′=a2i^+b2j^−c1k^\vec{p}_2' = a_2\hat{i} + b_2\hat{j} - c_1\hat{k}

D

p⃗1′=a1i^+b1j^\vec{p}_1' = a_1\hat{i} + b_1\hat{j}
p⃗2′=a2i^+b1j^\vec{p}_2' = a_2\hat{i} + b_1\hat{j}

Correct

Topics & Concepts

Work, Energy and PowerCollision

Step-by-Step Solution

To determine which of the given options are not allowed, we apply the principle of conservation of linear momentum.

The two balls are in free space with no external forces acting on them (F⃗ext=0⃗\vec{F}_{\text{ext}} = \vec{0}). Therefore, the total linear momentum of the system must be conserved during the collision:
P⃗i=P⃗f\vec{P}_i = \vec{P}_f

The initial total momentum of the system is:
P⃗i=p⃗1+p⃗2=pi^+(−pi^)=0⃗\vec{P}_i = \vec{p}_1 + \vec{p}_2 = p\hat{i} + (-p\hat{i}) = \vec{0}

Thus, for any physically allowed collision, the total final momentum must also be zero:
P⃗f=p⃗1′+p⃗2′=0⃗\vec{P}_f = \vec{p}_1' + \vec{p}_2' = \vec{0}

We are given that all the parameters p,a1,a2,b1,b2,c1,p, a_1, a_2, b_1, b_2, c_1, and c2c_2 are non-zero. Let us examine each option:

  1. Option (A):
    p⃗1′=a1i^+b1j^+c1k^\vec{p}_1' = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}
    p⃗2′=a2i^+b2j^\vec{p}_2' = a_2\hat{i} + b_2\hat{j}
    The total final momentum is:
    P⃗f=(a1+a2)i^+(b1+b2)j^+c1k^\vec{P}_f = (a_1 + a_2)\hat{i} + (b_1 + b_2)\hat{j} + c_1\hat{k}
    For P⃗f=0⃗\vec{P}_f = \vec{0}, the k^\hat{k}-component requires:
    c1=0c_1 = 0
    However, c1c_1 is strictly non-zero (c1≠0c_1 \neq 0), which violates the conservation of momentum. Therefore, Option (A) is NOT ALLOWED.

  2. Option (B):
    p⃗1′=c1k^\vec{p}_1' = c_1\hat{k}
    p⃗2′=c2k^\vec{p}_2' = c_2\hat{k}
    The total final momentum is:
    P⃗f=(c1+c2)k^\vec{P}_f = (c_1 + c_2)\hat{k}
    For P⃗f=0⃗\vec{P}_f = \vec{0}, we need:
    c1+c2=0  ⟹  c2=−c1c_1 + c_2 = 0 \implies c_2 = -c_1
    Since non-zero values can satisfy c2=−c1≠0c_2 = -c_1 \neq 0, this collision is allowed.

  3. Option (C):
    p⃗1′=a1i^+b1j^+c1k^\vec{p}_1' = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}
    p⃗2′=a2i^+b2j^−c1k^\vec{p}_2' = a_2\hat{i} + b_2\hat{j} - c_1\hat{k}
    The total final momentum is:
    P⃗f=(a1+a2)i^+(b1+b2)j^+(c1−c1)k^=(a1+a2)i^+(b1+b2)j^\vec{P}_f = (a_1 + a_2)\hat{i} + (b_1 + b_2)\hat{j} + (c_1 - c_1)\hat{k} = (a_1 + a_2)\hat{i} + (b_1 + b_2)\hat{j}
    For P⃗f=0⃗\vec{P}_f = \vec{0}, we require:
    a2=−a1andb2=−b1a_2 = -a_1 \quad \text{and} \quad b_2 = -b_1
    These conditions can be satisfied with non-zero values of a1,a2,b1,b2,a_1, a_2, b_1, b_2, and c1c_1. Thus, this collision is allowed.

  4. Option (D):
    p⃗1′=a1i^+b1j^\vec{p}_1' = a_1\hat{i} + b_1\hat{j}
    p⃗2′=a2i^+b1j^\vec{p}_2' = a_2\hat{i} + b_1\hat{j}
    The total final momentum is:
    P⃗f=(a1+a2)i^+2b1j^\vec{P}_f = (a_1 + a_2)\hat{i} + 2b_1\hat{j}
    For P⃗f=0⃗\vec{P}_f = \vec{0}, the j^\hat{j}-component requires:
    2b1=0  ⟹  b1=02b_1 = 0 \implies b_1 = 0
    Since b1b_1 is strictly non-zero (b1≠0b_1 \neq 0), the total final momentum cannot be zero. Therefore, Option (D) is NOT ALLOWED.

Thus, the options that are NOT ALLOWED are (A) and (D).

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