Question

A particle of mass 4m, which is at rest, explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each, in mutually perpendicular directions. The momentum of the third fragment of mass 2m after explosion is

Answer 1

A particle of mass 4m , which is at rest explode into these fragments of mass, m, m and 2m .
all theses fragments are in mutually perpendicular directions .
let one fragment of mass m moves in x+ direction , 2nd fragment of mass m moves in y+ direction.

velocity of 1st fragment , $\vec{v_1}=v\hat{i}$
velocity of 2nd fragment, $\vec{v_2}=v\hat{j}$
Let the velocity of 3rd fragment is V

now from law of conservation of linear momentum, $P_i=P_f$

0 = $m(v\hat{i})+m(v\hat{j}) + 2m\vec{V}$
$=-\frac{1}{2}v(\hat{i}+\hat{j})$

so, momentum of 3rd fragment , P = (2m)V
= $-m(v\hat{i}+v\hat{j})$
magnitude of momentum, P = √2mv

Answer 2

Answer:

Explanation:

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