Question

a point mass m=2 kg breaks apart into m1 and m2. the final masses are placed at a distance of 2 m. the force between the masses will be maximum if :
a.m1=1.5kg, m2=0.5kg
b.m1=1 kg, m2=1 kg
c.m1=0.25kg, m2=1.75kg
d.m1=0.75kg, m2=1.25kg​

Answer 1

Answer : m1 = 1kg and m2= 1kg

Option b is correct.

Explanation:

both have mass ratio 1:1.

Answer 2

Given:

point mass m=2 kg breaks apart into m1 and m2. the final masses are placed at a distance of 2 m.

To find:

Value of m1 and m2 such that force is max ?

Calculation:

This can be understood by the concept of maxima :

Let a mass M be broken into 2 masses as m and (M-m).

So, gravitational force :

$\sf \therefore \: F = \dfrac{Gm(M - m)}{ {r}^{2} }$

$\sf \implies \: F = \dfrac{G \bigg( {m}^{2} - Mm \bigg)}{ {r}^{2} }$

$\sf \implies \: \dfrac{dF}{dm}= \dfrac{G}{ {r}^{2} } \bigg \{ \dfrac{d ({m}^{2} -Mm ) }{dm} \bigg \}$

$\sf \implies \: \dfrac{dF}{dm}= \dfrac{G}{ {r}^{2} } \bigg \{ 2m-M \bigg \}$

For Maxima , dF/dm = 0

$\sf \implies \: \dfrac{dF}{dm}= \dfrac{G}{ {r}^{2} } \bigg \{ 2m-M \bigg \} = 0$

$\sf \implies \: 2m-M= 0$

$\sf \implies \: m = \dfrac{M}{2}$

So, max force is when the initial mass is equally divided.

In our question , initial mass is 2 kg , so maximum force will be obtained when the m1 = m2 = 1 kg.

m1 = 1 kg and m2 = 1 kg.