Practicing Success
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 : |
v \(\sqrt{2}\) v 2\(\sqrt{2}\) v 3\(\sqrt{2}\) v |
2\(\sqrt{2}\) v |
Momentum of the system would remain conserved. Initial momentum = 0 ; Final momentum should also be zero. Let masses be 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 |