Question

A vessel at rest explodes into three pieces. Two pieces having equal masses fly off perpendicular to one another with the same velocity 30 metre per second. The third piece has three times mass of each of other piece. The magnitude and direction of the velocity of the third piece will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is A.

Initially, the vessel having mass m is at rest.

According to the question, x+x+3x=m ⟹x=0.2m

Thus the masses of the three fragments be 0.2m,0.2m and 0.6m

Let the third fragment fly off with velocity v making an angle ϕ.

Conservation of momentum in y-direction :

0.6m(vsinθ)=0.2m(30) .........(1)

Conservation of momentum in x-direction :

0.6m(vcosθ)=0.2m(30) .........(2)

Dividing (1) and (2) we get, tanθ=1

Therefore, $\theta ={45}^o$

$\varphi ={180}^o-{45}^o={135}^o$

Squaring and adding (1) and (2) we get,

$0.6m{v}^2(si{n}^2\theta +co{s}^2\theta )=\left[0.2m\right(30)+0.2m(30){]}^2$

∴ $0.6mv=0.2m\left(30\right)\times \sqrt{2}$

$\Rightarrow v=10\sqrt{2}$ m/s.