A shell is fired from a cannon with a speed of `100 m//s` at an angle `30^(@)`with the vertical (`y`-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio `1:2`. The lighter fragment moves vertically upwards with an initial speed of `200 m//s`. What is the speed of the heavier fragment at the time of explosion?

Explanation

The velocity of shell at the highest point is `v=u sin theta= 100 xx sin 30^(@) = 50 m//s`. Let in be the mass of the shell. Then the mass of the lighter fragment is `m//3` and that of heavier fragment is `2m//3`. Initial momentum of the shell before explosion is
`mv=50m`
As no external force is acting on the shell, we can conserve momentum of the shell before and after its explosion.
In `x`-direction
`mv=(2m)/3v_(2)costhetaimpliesv_(2)costheta=3/2v`..............i
in `y`- direction
`0=m/3v_(1)-(2m)/3v_(2)sinthetaimpliesv_(2)sintheta=(v_(1))/2`........ii ltbr `Squaring and adding eqn i and ii we get
`v_(2)^(2)=9/4v^(2)+1/4v_(1)^(2)`
`implies v_(2)=1/2sqrt(9v^(2)+v_(1)^(2))=1/2sqrt(9xx2500+40000)=125m//s`