Question

A bomb initially at rest explodes into three equally massive fragments. Right after the explosion, one of the fragments, travels in the positive y-direction with a velocity v1. The two fragments ravel downward, at angles +- 30degrees with respect to the negative y-axis.

What is the ratio, R, of the kinetic energy of fragment 1 kinetic energy of fragment 2 (ie. R=k1/k2)

Answer 1

you might want to use the momentum eq or the mechanical energy conservation.

Answer 2

Thats what ive been trying. I set the first kinetic energy equal to the sum of the other
(.5)m1(v1^2)=(.5)(m2)(v2cos30)^2+(.5)(m3)(v3cos30)^2
All the masses are equal so they can cancell out..but i then i get lost

Answer 3

Take the velocity of 2nd fragment to be v2 and for 3rd fragment as v3.
now apply the conservation of linear momentum in X and Y directions which are as follows -
In X direction :
\[m v _{2} \sin30 = m v _{3} \sin30 \]
which gives V2 = V3 ---------------------------------------------------(1)

Now in y direction :

\[m v _{1} = m v _{2} \cos30 + m v _{3} \cos 30 \]

cancelling out the common terms and putting v2 = v3 we get
\[v _{1} = \sqrt{3} \times v _{2} \]

Therefore
\[K= R _{1} \div R _{2}\]
\[= \left( 0.5 \times m v _{1}^{2} \right) \div \left( 0.5 \times m v _{2}^{2} \right)\]
and putting \[v _{1} = \sqrt{3} \times v _{2} \] wefinally get
K = 3 (Answer)