# A mass M breaks into two pieces in the ratio 1 : 3 while at rest. If the heavier has a speed of v, the speed of the lighter is

Given,

A mass M breaks into two pieces in the ratio 1 : 3

Let the common mass fraction be x, then

one part will have mass x while the other will have mass 3x, such that

x+3x = M

So, x = M/4

Initially, the mass is at rest, i.e., initial momentum of the mass is 0.

After breaking, the heavier mass 3x = 3M/4 moves with a velocity v.

Let u be the velocity with which the smaller mass x = M/4 moves be u.

Then, from conservation of linear momentum, we get

$xu+3xv=0\phantom{\rule{0ex}{0ex}}\frac{Mu}{4}=-\frac{3Mv}{4}\phantom{\rule{0ex}{0ex}}u=-3v$

So, the speed of the lighter mass is 3 times the speed of the heavier mass and the lighter mass moves in a direction opposite to that of the heavier mass.

Regards

**
**