Question

The balls, having linear momenta ${\overrightarrow{p}}_1=p\hat{i}$ and ${\overrightarrow{p}}_2=-p\hat{i}$, undergo a collision in free space. There is no extemal force acting on the balls. Let ${\overrightarrow{p}}_1$ and ${\overrightarrow{p}}_2$ be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of $p,a_1,a_2,b_1,b_2,c_1$ and $c_2$.

• A. ${\overrightarrow{p}}_1=a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$;
  ${\overrightarrow{p}}_2=a_{2}i+b_2\hat{j}$
• B. ${\overrightarrow{p}}_1=c_1\hat{k}$;
  ${\overrightarrow{p}}_2=c_2\hat{k}$
• C. ${\overrightarrow{p}}_1=a_{1}i+b_1\hat{j}+c_1\hat{k}$;
  ${\overrightarrow{p}}_2=a_2\hat{i}+b_2\hat{j}-c_1\hat{k}$
• D. ${\overrightarrow{p}}_1=a_1\hat{i}+b_1\hat{j}$;
  ${\overrightarrow{p}}_2=a_{2}i+b_1\hat{j}$

Answer 1

Answer: (A), (D)

The correct options are
A ${\overrightarrow{p}}_1=a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$;
${\overrightarrow{p}}_2=a_{2}i+b_2\hat{j}$
D ${\overrightarrow{p}}_1=a_1\hat{i}+b_1\hat{j}$;

${\overrightarrow{p}}_2=a_{2}i+b_1\hat{j}$
Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in $j$ and $k$ directions as both momenta are in $i$ direction. Hence options $A$ and $D$ are not possible.