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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Advanced 2008 · Paper 1 · Physics · Center of Mass, Momentum and Collision

Two balls, having linear momenta p _1=p i and p _2=-p i , undergo a...

Two balls, having linear momenta p→1=pi^\overrightarrow{\mathbf{p}}_1=p \hat{\mathbf{i}} and p→2=−pi^\overrightarrow{\mathbf{p}}_2=-p \hat{\mathbf{i}}, undergo a collision in free space. There is no external force acting on the balls. Let p→1′\overrightarrow{\mathbf{p}}_1^{\prime} and p→2′\overrightarrow{\mathbf{p}}_2^{\prime} 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

  1. Ap→1′=a1i^+b1j^+c1k^,p→2′=a2i^+b2j^\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}
  2. Bp→1′=c1k^,p→2′=c2k^\overrightarrow{\mathbf{p}}_1^{\prime}=c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=c_2 \hat{\mathbf{k}}
  3. Cp→1′=a1i^+b1j^+c1k^\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}
  4. Dp→1′=a1i^+b1j^\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}} p→2′=a2i^+b2j^−c1k^\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}-c_1 \hat{\mathbf{k}} p→2′=a2i^+b1j^\overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}

Correct answer

A. p→1′=a1i^+b1j^+c1k^,p→2′=a2i^+b2j^\overrightarrow{\mathbf{p}}_1^{\prime}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}}, \overrightarrow{\mathbf{p}}_2^{\prime}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}

Step-by-step solution

Initial momentum of the system p1→+p2→=0\overrightarrow{\mathbf{p}_1}+\overrightarrow{\mathbf{p}_2}=0 ∴\therefore Final momentum p1′→+p2′→\overrightarrow{\mathbf{p}_1^{\prime}}+\overrightarrow{\mathbf{p}_2^{\prime}} should also be zero. Option (b) is allowed because if we putc c1=−c2≠0,p1′→+p2′→c_1=-c_2 \neq 0, \overrightarrow{\mathbf{p}_1^{\prime}}+\overrightarrow{\mathbf{p}_2^{\prime}} will be zero. Similarly, we can check other options. ∴\therefore correct options are (a) and (d).
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