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 1:2, with each fragment moving horizontally immediately after the explosion. If the lighter fragment falls at
a distance from the point of projection, behind the point of projection, the distance at which the other fragments falls from the
point of projection is
(A)
R
2
(B)
2R
3
(C)
4R
3
(D)
7R
4​

Answer 1

Answer:

4R/3. is the answer I think

Answer 2

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

Therefore, option (D) is correct.

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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