A projectile of mass 3m is projected from ground with velocity `20sqrt2 m//s` at `45^@`. At highest point it explodes into two pieces. One of mass 2m and the other of mass m. Both the pieces fly off horizontally in opposite directions. Mass 2m falls at a distance of `100m` from point of projection. Find the distance of second mass from point of projection where it strikes the ground.
`(g=10m//s^2)`

Answer

C, D

Explanation

Range of the projectile in the absence of explosion
`R=(u^2sin2 theta)/(g)=((20sqrt2)^2sin90^@)/(10)=80m`
The path of centre of mass of projectile will not change, i.e. `x_(COM)` is still 80 m. Now, from the definition of centre of mass
`x_(COM)=(m_1x_1+m_2x_2)/(m_1+m_2)`
or `80=((m)(x_1)+(2m)(100))/(m+2m)`
Solving this equation, we get `x_1=40m`
Therefore, the mass m will fall at a distance `x_1=40cm` from point of projection.