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The 'e' Equation

In one dimens...

Question

In one dimensional elastic collision between a given incoming body A of mass $m_{1}$ and another body B of mass $m_{2}$ initially at rest.

B will recoil with greatest speed if

m_{2} ≪ m_{1}

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B will recoil with greatest Momentum if

m_{2} ≫ m_{1}

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B will recoil with greatest kinetic energy if

m_{1} = m_{2}

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B will recoil with greatest kinetic energy if

m_{2} = 2 m_{1}

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Solution

The correct option is B B will recoil with greatest speed if $m_{2} ≪ m_{1}$
Given, $m_{1}$ and $m_{2}$ be the masses of $A$ and $B$ ,
Let, $v_{1}$ and $v_{2}$ be the velocities of $A$ and $B$ ,
$p_{1}$ and $p_{2}$ be the momenta of A and B and $p = p_{1} = p_{2}$ ,
$E_{1}$ and $E_{2}$ be the kinetic energies of A and B,
Now, if in one dimensional elastic collision between a given incoming body A of mass $m_{1}$ and another body B of mass $m_{2}$ initially at rest, then
A will recoil with greatest speed if $m_{2} ≪ m_{1}$ since,
$p = m_{1} v_{1} = m_{2} v_{2}$

$\Rightarrow \frac{m_{1}}{m_{2}} = \frac{v_{2}}{v_{1}}$ ......(1)

$\Rightarrow m \propto \frac{1}{v}$

$B) B$ will recoil with greatest Momentum if $P_{1} = P_{2}$ after collision $A$ will be at rest

$C) B$ will not recoil with greatest kinetic energy if $m_{1} = m_{2}$ , since, $\frac{m_{2}}{m_{2}} = \frac{v_{2}}{v_{1}}$ ............from(1)

$\Rightarrow v_{1} = v_{2}$
$\Rightarrow E_{1} = E_{2}$

$D) B$ will not recoil with greatest kinetic energy if $m_{2} = 2 m_{1}$ , since,

$\frac{m_{1}}{2 m_{1}} = \frac{v_{2}}{v_{1}}$ ..........from(1)

$\Rightarrow 2 v_{2} = v_{1}$
$\Rightarrow v_{1} > v_{2}$
$\Rightarrow E_{1} > E_{2}$

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