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:

CONCEPT:

• The momentum is equal to the product of the mass and velocity and it is written as;

           p = mv

           Here we have p as the momentum, m is the mass, and v is the  velocity

• According to the conservation of momentum when the external force is equal to zero, momentum remains constant and the mathematical representation of the conservation of momentum is written as;

          m₁u₁ + m₂u₂+... = m₁v₁ + m₂v₂+....

Here m₁, and m₂ be the masses of the two bodies, u₁, and u₂ be the initial velocity, and v₁, and v₂ be the final velocity.

CALCULATION:

A shell of mass m is at rest initially and it explodes into three fragments having mass in the ratio  2 ∶ 2 ∶ 1 as shown in the figure below;

Here we have the ratio of the masses as; 2: 2: 1

According to the conservation of momentum, there is only one mass that explodes into three different fragments, therefore the initial and final momentum is written as;

mu = m₁v₁ + m₁v₂ + m₁v₃    -----(1)

Here initial velocity, u =0 because the mass is at rest we have,

m(0) = m1v1 + m1v2 + m1v3 

When it explodes the masses are going to different - different directions as we see in the figure, therefore its velocity is 

0 = $m_1{v}_1(-\hat{i})+m_1{v}_2(-\hat{j})+m_1{\overrightarrow{v}}_3$

Now, on putting the masses we have;

 $\frac{2m}{5}{v}_1(-\hat{i})+\frac{2m}{5}{v}_2(-\hat{j})+\frac{2m}{5}{\overrightarrow{v}}_3$ = 0

⇒ $-2m{v}_1\hat{i}-2m{v}_2\hat{j}+2m{\overrightarrow{v}}_3$ =0

⇒ $v_3=\sqrt{(2v{)}^2+(2v{)}^2}$

⇒ $v_3=\sqrt{8(v{)}^2}$

⇒ v₃ = $2\sqrt{2}v$

Hence option 4) is the correct answer.