A stationary particle explodes into two particles of masses m 1 and m 2 which move in opposite directions with velocities v 1 and v 2 . The ratio of their kinetic energies E 1 /E 2 is :

Center of Mass and Collision question

2003 · Q158

Center of Mass and Collision question

2003 · Q158

View written solutionFree

Correct answer: A

From conservation law of momentum, before collision and after collision linear momentum (p) will be same. That is,

initial momentum = final momentum.

$\Rightarrow$ 0 = m₁v₁ – m₂v₂ $\Rightarrow$ m₁v₁ = m₂v₂

p₁ = p₂

Now, $E = {{{p^2}} \over {2m}}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = {{p_1^2} \over {2{m_1}}} \times {{2{m_2}} \over {p_2^2}}$

$\Rightarrow {{{E_1}} \over {{E_2}}} = {{{m_2}} \over {{m_1}}}\left[ {{p_1} = {p_2}} \right]$

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Center of Mass and Collision question

2003 · Q158

Center of Mass and Collision question

2003 · Q158

View written solutionFree

Correct answer: A

From conservation law of momentum, before collision and after collision linear momentum (p) will be same. That is,

initial momentum = final momentum.

$\Rightarrow$ 0 = m₁v₁ – m₂v₂ $\Rightarrow$ m₁v₁ = m₂v₂

p₁ = p₂

Now, $E = {{{p^2}} \over {2m}}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = {{p_1^2} \over {2{m_1}}} \times {{2{m_2}} \over {p_2^2}}$

$\Rightarrow {{{E_1}} \over {{E_2}}} = {{{m_2}} \over {{m_1}}}\left[ {{p_1} = {p_2}} \right]$