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QJEEAdv08Ph889f: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ph889f Solution: Two balls, having linear momenta and , undergo a collision in…

Two balls, having linear momenta p→1=pi^{\overrightarrow p _1} = p\widehat i and p→2=−pi^{\overrightarrow p _2} = - p\widehat i , undergo a collision in free space. There is no external force acting on the balls. Let p′→1{\overrightarrow {p'} _1} and p′→2{\overrightarrow {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 c2{c_2} :

Multiple Choice Options

  • A p′→1=a1i^+b1j^+c1k^;p′→2=a2i^+b2j^{\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k;{\overrightarrow {p'} _2} = {a_2}\widehat i + {b_2}\widehat j
  • B p′→1=c1k^;p′→2=c2k^{\overrightarrow {p'} _1} = {c_1}\widehat k;{\overrightarrow {p'} _2} = {c_2}\widehat k
  • C p′→1=a1i^+b1j^+c1k^;p′→2=a2i^+b2j^−c1k^{\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k;{\overrightarrow {p'} _2} = {a_2}\widehat i + {b_2}\widehat j - {c_1}\widehat k
  • D p′→1=a1i^+b1j^;p′→2=a2i^+b1j^{\overrightarrow {p'} _1} = {a_1}\widehat i + {b_1}\widehat j;{\overrightarrow {p'} _2} = {a_2}\widehat i + {b_1}\widehat j

Step-by-Step Verified Solution

In free space, there is no external force. Hence linear momentum of the system is conserved. Initial and final linear momentum of the system are

amp;p⃗i=p⃗1+p⃗2=p^−ı^ı^=0→,amp;p⃗f=p⃗1′+p⃗2′.\begin{aligned}& \vec{p}_i=\vec{p}_1+\vec{p}_2=\hat{p}-\hat{\imath} \hat{\imath}=\overrightarrow{0}, \\\\& \vec{p}_f=\vec{p}_1^{\prime}+\vec{p}_2^{\prime} .\end{aligned}

The conservation of linear momentum, p⃗i=p⃗f\vec{p}_i=\vec{p}_f gives

p⃗1′+p⃗2′=0→.\vec{p}_1^{\prime}+\vec{p}_2^{\prime}=\overrightarrow{0} .

In case (A)(A) ,

p⃗1′+p⃗2′amp;=(a1+a2)ı^+(b1+b2)ȷ^+c1k^amp;≠0→,(∵c1≠0).\begin{aligned}\vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(a_1+a_2\right) \hat{\imath}+\left(b_1+b_2\right) \hat{\jmath}+c_1 \hat{k} \\\\& \neq \overrightarrow{0}, \quad\left(\because c_1 \neq 0\right) .\end{aligned}

In case (B)(B) ,

p⃗1′+p⃗2′amp;=(c1+c2)k^amp;=0→,( if c2=−c1).\begin{aligned}\vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(c_1+c_2\right) \hat{k} \\\\& =\overrightarrow{0}, \quad\left(\text { if } c_2=-c_1\right) .\end{aligned}

In case (C)(\mathrm{C}) ,

p⃗1′+p⃗2′amp;=(a1+a2)ı^+(b1+b2)ȷ^amp;=0→,( if a2=−a1 and b2=−b1).\begin{aligned}\vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(a_1+a_2\right) \hat{\imath}+\left(b_1+b_2\right) \hat{\jmath} \\\\& =\overrightarrow{0}, \quad\left(\text { if } a_2=-a_1 \text { and } b_2=-b_1\right) .\end{aligned}

In case (D),

p⃗1′+p⃗2′amp;=(a1+a2)ı^+2b1ȷ^amp;≠0→,(∵b1≠0).\begin{aligned}\vec{p}_1^{\prime}+\vec{p}_2^{\prime} & =\left(a_1+a_2\right) \hat{\imath}+2 b_1 \hat{\jmath} \\\\& \neq \overrightarrow{0}, \quad\left(\because b_1 \neq 0\right) .\end{aligned}

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