Question

A shell projected from a level ground has a range r if it did not explode at the highest point the shell explodes into two fragments having masses in the ratio 1 is to 2 with each fragment moving horizontally immediately after the explosion if the writer fragment falls at the distance or two from the point of projection behind the point of projection the distance at which the other fragments falls from the point of projection is​

Answer 1

The distance at which the other fragment falls from the point of projection is 7R/4

Explanation:

Given:

Range of the projected shell = R

The shell explodes in two masses in the ratio 2:1 at the highest point

The lighter fragment falls at distance R/2 from the point of projection and behind it

To find out:

The distance from the point of projection of the second mass

Solution:

This question can be easily solved by the concept of centre of mass

Let the mass of the first (lighter) mass be m

Then the mass of the second shell = 2m

If there had been no explosion, the centre of mass would have followed the original path and it would have lied on the the range R

Therefore, after explosion the fragments will fall such that the centre of mass lie on the range R

Thus, if the distance of the second fragment be x from the point of projection then

$x_{cm}=R$

$x_{cm}=\frac{m\times (-R/2)+2mx}{m+2m}$

$\implies 3mR=-\frac{mR}{2}+2mx$

$\implies 6R=-R+4x$

$\implies x=\frac{7}{4}R$

Hope this answer is helpful.

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