Centre of Mass 3 Question 25
28. Two balls, having linear momenta $\mathbf{p} _1=p \hat{\mathbf{i}}$ and $\mathbf{p} _2=-p \hat{\mathbf{i}}$, undergo a collision in free space. There is no external force acting on the balls. Let $\mathbf{p} _1^{\prime}$ and $\mathbf{p} _2^{\prime}$ be their final momenta. The following option figure (s) is (are) not allowed for any non-zero value of $p, a _1, a _2, b _1, b _2, c _1$ and $c _2$.
$(2008,4$ M)
(a) $\mathbf{p} _1^{\prime}=a _1 \hat{\mathbf{i}}+b _1 \hat{\mathbf{j}}+c _1 \hat{\mathbf{k}}, \mathbf{p} _2^{\prime}=a _2 \hat{\mathbf{i}}+b _2 \hat{\mathbf{j}}$
(b) $\mathbf{p} _1^{\prime}=c _1 \hat{\mathbf{k}}, \mathbf{p} _2^{\prime}=c _2 \hat{\mathbf{k}}$
(c) $\mathbf{p} _1^{\prime}=a _1 \hat{\mathbf{i}}+b _1 \hat{\mathbf{j}}+c _1 \hat{\mathbf{k}}, \mathbf{p} _2^{\prime}=a _2 \hat{\mathbf{i}}+b _2 \hat{\mathbf{j}}-c _1 \hat{\mathbf{k}}$
(d) $\mathbf{p} _1^{\prime}=a _1 \hat{\mathbf{i}}+b _1 \hat{\mathbf{j}}, \mathbf{p} _2^{\prime}=a _2 \hat{\mathbf{i}}+b _1 \hat{\mathbf{j}}$
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Answer:
Correct Answer: 28. $5 \times 10^{-3}$
Solution:
- Initial momentum of the system $\mathbf{p} _1+\mathbf{p} _2=0$
$\therefore$ Final momentum $\mathbf{p} _1^{\prime}+\mathbf{p} _2^{\prime}$ should also be zero.
Option (b) is allowed because if we put $c _1=-c _2 \neq 0$, $\mathbf{p} _1^{\prime}+\mathbf{p} _2^{\prime}$ will be zero. Similary, we can check other options.
