Question

a stationary particle explores into two particles of masses X and Y which move in opposite directions with velocities V1 and V2 the ratio of their kinetic energy E1:E2 is

Answer 1

ratio of thr kinetic energy will be 1 becoz energy will be conserved

Answer 2

According to the given data, conservation of momentum is to be applied on the given situation to form the equation and then be compared to rule out the relation.

Thereby, $Initial \quad momentum = Final \quad momentum$

$\Rightarrow Initial \quad momentum = Zero\quad as \quad mass \quad at \quad rest.$

$\Rightarrow Final \quad momentum = m _2 v _2 - m _1 v _1$

$\Rightarrow 0 = m _2 v _2 - m _1 v _1$

$\Rightarrow m _1 v _1 = m _2 v _2$

$\Rightarrow \frac { v_{ 1 } }{ v_{ 2 } } =\frac { m_{ 2 } }{ m_{ 1 } }$

$\Rightarrow \frac { E_{ 1 } }{ E_{ 2 } } =\frac { \frac { 1 }{ 2 } m_{ 1 }{ v_{ 1 } }^{ 2 } }{ \frac { 1 }{ 2 } m_{ 2 }{ v_{ 2 } }^{ 2 } }$        

$\Rightarrow \frac { E_{ 1 } }{ E_{ 2 } } =\frac { \frac { { { P }_{ 1 } }^{ 2 } }{ 2m_{ 1 } } }{ \frac { { { P }_{ 2 } }^{ 2 } }{ 2m_{ 2 } } }$

$As\quad m_{ 1 }v_{ 1 }=m_{ 2 }v_{ 2 }$

$\left[ \therefore { P }_{ 1 }={ P }_{ 2 }\quad \right]$

$\Rightarrow \frac { E_{ 1 } }{ E_{ 2 } } =\frac { \frac { 1 }{ 2m_{ 1 } } }{ \frac { 1 }{ 2m_{ 2 } } }$

$\Rightarrow \frac { E_{ 1 } }{ E_{ 2 } } =\frac { 1 }{ 2m_{ 1 } } \times \frac { 2m_{ 2 } }{ 1 }$

$\Rightarrow \frac { E_{ 1 } }{ E_{ 2 } } =\frac { m_{ 2 } }{ m_{ 1 } }$