Question

A body of mass 5 m initially at rest explodes into 3 fragments with mass ratio 3:1:1. two of fragments each of mass ‘m' are found to move with a speed of 60 m/s is mutually perpendicular directions. the velocity of third fragment is

Answer 1

The correct answer is
20 square root 2

Hope this helps you
By jerman

Answer 2

The third fragment’s velocity is $\bold{20 \sqrt{2}}$

Given:

The initial momentum of body = 5m × 0 = 0

Final momentum of the fragments = 60mi +60 mj +3m V\

{Assuming the velocity of the mass “3m” to be V}

Solution:

Now applying the “Law of conservation of momentum”,

Momentum before and after the collision are same.

So, Initial momentum = final momentum

m(60i) + m(60j) + 3m(v) = 0

$V=-\frac{60 i+60 j}{3}$

= - (20i +20j)

Therefore, $|v|=\sqrt{\left(20^{2}+20^{2}\right)}$

$v=20 \sqrt{2}$

And the velocity of third fragment is$20 \sqrt{2}.$