A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the origianl direction. The speed of each of the moving particle will be :

Question

A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the origianl direction. The speed of each of the moving particle will be : (A) v / (2 √2 ) (B) 2 √2 v (C)  √2 v (D)  v/ √2

Tardigrade Admin · Accepted Answer

Linear momentum conservation m 2v + 2m v = m × 0 + m (v'/√2) × 2 v' = 2 √2 v

$v / (22)$

$22 v$

$2 v$

$v / 2$

Linear momentum conservation
$m 2 v + 2 m v = m \times 0 + m \frac{v ^{'}}{2} \times 2$
$v^{'} = 22 v$

v/(22​)

22​v

2​v

v/2​

Linear momentum conservation m2v+2mv=m×0+m2​v′​×2 v′=22​v

$v / (22)$

$22 v$

$2 v$

$v / 2$

Linear momentum conservation
$m 2 v + 2 m v = m \times 0 + m \frac{v ^{'}}{2} \times 2$
$v^{'} = 22 v$