Question

A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2. Their velocities are in the ratio:
$\ A. \quad 8:1$
$\ B. \quad 6:1$
$\ C. \quad 4:1$
$\ D. \quad 2:1$

Answer 1

Hint: Since it is clearly mentioned in the question that the nucleus was initially at rest, hence net force acting on the nucleus must be zero. Now, from the law of conservation of linear momentum, if net external force acting on a body is zero then the momentum of the body will remain unchanged. Hence net momentum of the system before and after the event must be zero.

Formula used: Since $\sum { { F }_{ ext } }$=0
So, ${ P }_{ i }={ P }_{ f }$
But $P_{i}$ = 0 i.e. initial momentum of the body [ Initially body was at rest ]
$P_{f}$= ${ m }_{ 1 }\overrightarrow { { v }_{ 1 } }+ { m }_{ 2 }\overrightarrow { { v }_{ 2 } }$
Or ${ m }_{ 1 }\overrightarrow { { v }_{ 1 } }+ { m }_{ 2 }\overrightarrow { { v }_{ 2 } } =0$

Complete step by step solution:
Given, ratio of their radii = $\ 1:2$
i.e. $\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } = 1:2$
but$\rho= \dfrac mV$
so,$m= \rho \times V$=$\ \rho \times \dfrac {4} {3}\pi r^3$
Now, taking the ratio of both the masses [ so that density term gets cancel ]
$\dfrac { { m }_{ 1 } }{ { m }_{ 2 } }$=$\ \dfrac {\rho V_{1}} {\rho{V_{2}}}$=$\ \dfrac {V_{1}} {{V_{2}}}$=$\dfrac { \dfrac { 4\pi { { r }_{ 1 } }^{ 3 } }{ 3 } }{ \dfrac { 4\pi { { r }_{ 2 } }^{ 3 } }{ 3 } }$=$\dfrac { { r^3 }_{ 1 } }{ { r^3 }_{ 2 } }$=$\left( \dfrac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 3 }$=$\left( \dfrac { { 1 } }{ 2 } \right) ^{ 3 }$=$\ \dfrac 1 8$

Now as we require the ratio of velocities $\dfrac { { v }_{ 1 } }{ { v }_{ 2 } }$
Hence, from$\ { m }_{ 1 }\overrightarrow { { v }_{ 1 } }+ { m }_{ 2 }\overrightarrow { { v }_{ 2 } } =0$
${ m }_{ 1 }\overrightarrow { { v }_{ 1 } }= - { m }_{ 2 }\overrightarrow { { v }_{ 2 } }$
$\ \dfrac {|v_{1}|} {|{v_{2}}|}$=$\dfrac { { m }_{ 2 } }{ { m }_{ 1 } }$ [Neglecting negative sign as we are asked about magnitudes only]

As calculated $\dfrac { { m }_{ 1 } }{ { m }_{ 2 } }$=$\ \dfrac 18$
so,$\dfrac { { m }_{ 1 } }{ { m }_{ 2 } }$=$\ \dfrac 81$
hence, $\ \dfrac {v_1}{v_2}$=$\ \dfrac 81$

So, A. is the correct option.

Note: Students should always remember that momentum is only conserved if net external force acting on the centre of mass of the body is zero.
Students might make a mistake for the fact that being in motion need not always mean that net force is acting on the body. If the body is moving with uniform velocity, then also the net force is zero and momentum is conserved.
In some questions, it might be given that mass or nucleus is split into two equal parts. In that case, we will not take the ratio of radii, but we will take the ratio of masses of the fragments.