Question

A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of heaviest fragment in m/s will be(a) 7√2(b) 5√2(c) 3√2(d) √2

Answer 1

Body of Mass = 5kg ( Given)

Directional Speed = 21m/s ( Given)

Since 5 kg body explodes into three fragments of ratio 1:1:3, thus the masses of the fragments will be 1, 1,3 kg.  The magnitude of resultant momentum of two fragments each of mass 1 kg, moving with velocity 21 m/s, in perpendicular directions is =

The magnitude of resultant momentum of two fragments each of mass 1 kg, moving with velocity 21 m/s, in perpendicular directions is

√(m1v1)²+(m2v2)²

m′v′= √(21)²+(21)²

= 21√2

According to law of conservation of linear momentum

m3v3 =m′v ′=21√2or3 v3 = 21√2

orv3= 7√2m/s