Question

A projectile launched upwards for maximum range R from point O on level ground, explodes into two fragments of equal masses when it was at maximum height. One of the fragments reaches back to point of projection O. The point where the other one reaches ground will be at a distance (Take point of projection as origin) 1) R from O 2) 2R from O 3) 4R from O 4) R /2from O

Answer 1

Even the projectile gets exploded into two or more fragments, the center of mass traces the trajectory path, so that it reaches at a horizontal distance R from point of projection, i.e., O.

Therefore, horizontal displacement of center of mass from O is R.

Here the first fragment reaches back to O, so it has zero displacement.

Let the displacement of the second fragment be $\sf{x}$ from O.

Let mass of each fragment be $\sf{m}$ since masses of both are same.

Thus the displacement of the center of mass (i.e., R) is given by,

$\longrightarrow\sf{R=\dfrac{m(0)+mx}{m+m}}$

$\longrightarrow\sf{R=\dfrac{mx}{2m}}$

$\longrightarrow\sf{R=\dfrac{x}{2}}$

$\longrightarrow\sf{\underline{\underline{x=2R}}}$

Therefore, the second fragment will reach at a distance 2R from O.

Hence (2) is the answer.