A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is :
Answer & explanation
Correct answer: option 3
By Conservation of Momentum :
If initial momentum = 0 $\Rightarrow$ Final momentum should also be zero.
If the masses are 2m, 2m, and m,
Momentum along x-direction = 2mv i ; Momentum along y-direction = 2mv j
Net momentum : m v'= \(\sqrt{(2mv)^2 + (2mv)^2}\)
v' = 2\(\sqrt{2}\) v