The balls, having linear momenta p 1 = p i and p ? 2 = - p i , undergo a collision in free space. There is no external force acting on the balls. Let p 1 ' and p 2 ' be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-z

Question

The balls, having linear momenta ${\vec{p}}_{1} = \vec{p} i$ and ${\vec{p}}_{2} = - \vec{p} i$ , undergo a collision in free space. There is no external force acting on the balls. Let ${\vec{p}}_{1}'$ and ${\vec{p}}_{2}'$ 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 $c_{2}$ . [2008]

Smart Achievers · Accepted Answer

(1, 4)

From law of conservation of linear momentum

The initial linear momentum of the system, $p \hat{i} - p \hat{i} = 0$

$∴$ Final linear momentum should also be zero i.e., $p_{1}' + p_{2}' = 0$

$Option 1:$

$p_{1}' + p_{2}' = (a_{1} + a_{2}) \hat{i} + (b_{1} + b_{2}) \hat{j} + c_{1} \hat{k} = Final momentum$

It is given that $a_{1}, b_{1}, c_{1}, a_{2}, b_{2}$ and $c_{2}$ have non-zero values.

If $a_{1} = x$ and $a_{2} = - x,$ also if $b_{1} = y$ and $b_{2} = - y$ , then the $\hat{i}$ and $\hat{j}$ components become zero. But the third term having $\hat{k}$ component is non-zero. This gives a definite final momentum to the system which violates conservation of linear momentum.

$∴$ Option 1 is wrong.

$Option 4:$

$p_{1}' + p_{2}' = (a_{1} + a_{2}) \hat{i} + 2 b_{1} \hat{j} \neq 0 because b_{1} \neq 0$

Following the same reasoning as above the option 4 is also wrong.

Solution

1 ${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{2} \hat{j}$

2 ${\vec{p}}_{1}' = c_{1} \hat{k}$
${\vec{p}}_{2}' = c_{2} \hat{k}$

3 ${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{2} \hat{j} - c_{1} \hat{k}$

4 ${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{1} \hat{j}$

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Solution

1 p→1'=a1i^+b1j^+c1k^ p→2'=a2i^+b2j^

2 p→1'=c1k^ p→2'=c2k^

3 p→1'=a1i^+b1j^+c1k^ p→2'=a2i^+b2j^-c1k^

4 p→1'=a1i^+b1j^ p→2'=a2i^+b1j^

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1 ${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{2} \hat{j}$

2 ${\vec{p}}_{1}' = c_{1} \hat{k}$
${\vec{p}}_{2}' = c_{2} \hat{k}$

3 ${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{2} \hat{j} - c_{1} \hat{k}$

4 ${\vec{p}}_{1}' = a_{1} \hat{i} + b_{1} \hat{j}$
${\vec{p}}_{2}' = a_{2} \hat{i} + b_{1} \hat{j}$