Question

heavy nucleus at rest breaks into two fragments which fly off with velocity 8 is to 1 ratio of the radii of the fragment is:

Answer 1

Answer:

1:2

Explanation:

as the nucleus will break into mass m1

m2 and suppose the velocities are v1

and V2.

m1v1= m2v2

m1/m2 = v1/v2

therefore the velocities are given in

the question . so

1/8 = m1/m2

v1/v2= m1 /m2

r1^3/r2^3= 1/8

r1/r2= 1/2

therefore I got ur answer

Answer 2

Answer:

The ratio of the the radii of the fragments is 1 : 2.

Explanation:

(Note; The nucleus is like a sphere)

Initially the nucleus is at rest which implies that the net initial momentum is zero. So, the net final momentum also must be zero. Now, we can write it as;

                    Pi = Pf = 0

Let the mass of first fragment be m₁, second fragment be m₂, let the velocities be v₁ and v₂ respectively. Now; we can write momentum as;

                    Pi = Pf = 0

            P₁ + P₂ = Pf = 0

                  P₁  + P₂ = 0

                           P₁ = - P₂

                       m₁v₁ = - m₂v₂

             $\frac{m1}{m2} = \frac{v1}{v2} = \frac{1}{8} \, (given)$

             $\frac{m1}{m2} = \frac{1}{8}$            

       $\frac{p \frac{4}{3}\pi R1^{3} }{p\frac{4}{3} \pi R2^{3} } = \frac{1}{8}$

(since, mass = density × volume)

            $\frac{R1^{3} }{R2^{3} } = \frac{1}{8}$

              $\frac{R1}{R2} = \sqrt[3]{\frac{1}{8} }$

              $\frac{R1}{R2} = \frac{1}{2}$

 Hope it helps! ;-))