An object of mass `10 kg` is launched from the ground at `t = 0`, at an angle of `37^(@)` above the horizontal with a speed of `30 m//s`. At some time after its launch, an explosion splits the projectile into two pieces. One piece of mass `4 kg` is observed at (`105 m, 43 m`) at `t =2 s`. Find the location of second piece at `t = 2 s`?

Options

A. `(10,2)`
B. `(48,16)`
C. `(10,-2)`
D. information insufficient

Answer

D

Explanation

As only the gravity force is acting on the system, the centre of mass of the system follows a parabolic path.
At `t=2s`,
`x_(CM)=30cos37^(@)xx2=48m`
`y_(CM)=30sin37^(@)xx2-1/2xx10xx2^(2)=16m`
Let coordinates of the second piece, i.. `6kg` piece be `(x,y)`
Then
`x_(CM)=48=(6x+4xx105)10impliesx=10m`
`y_(CM)=16=(6y+4xx43)/10impliesy=-2m`
Negative value of `y` shows that the second piece collides the ground before `t=2s`.