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

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

Tardigrade Admin · Accepted Answer

Since 5 kg body explodes into three fragments with masses in the ratio 1: 1: 3 thus, masses of fragments will be 1 kg , 1 kg and 3 kg respectively. The magnitude of resultant momentum of two fragments each of mass 1 kg, moving with velocity 21 m / s, in perpendicular directions is √(m1 v1)2+(m2 v2)2 m prime v prime=√(21)2+(21)2=21 √2 kg m / s According to law of conservation of linear momentum m3 v3=m prime v prime=21 √2 or 3 v3=21 √2 or v3=7 √2 m / s

Q. A body of mass $5 k g$ 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

$72$

$52$

$32$

$2$

Q. A body of mass 5kg 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 21m/s. The velocity of heaviest fragment in m/s will be

72​

52​

32​

2​

Q. A body of mass $5 k g$ 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

$72$

$52$

$32$

$2$