Question

Two uniform solid spheres of equal radii $R$ and mass $M$ and $4M$ have a centre to centre separation of $6R$. The two spheres are held fixed. A projectile of mass $m$ is projected from the surface of the sphere of mass $M$ directly towards the second sphere. Obtain an expression for the speed when it strikes the heavier sphere of mass $4M$.

Answer 1

Two uniform solid sphere

Radius R

Centre to centre separation $=6R$

Mass of sphere $1=M$

Mass of sphere $2=4M$

A projectile of mass m is projected from the sphere of mass M is directly towards centre of the second sphere.

By formula

Kinetic energy $=\frac{1}{2}m{v}^2$

Where

$v=$ velocity of the object and

$m=$ mass

Hence,

To find the gravitational potential energy potential $\left(GMM/R\right)$

Mass $+$ gravitational potential energy

Substituting the values

We get

$E_1=Kineticenergy-GMM/R-\frac{G4MM}{R}$

Consider $v=0$

Therefore

Kinetic energy$=0$

Substituiting

$n=--\frac{GMM}{2R}-\frac{G\times 4MM}{4R}$

By the law of conservation of energy

$R_1=n$

Hence,

$\left(\frac{1}{2}\right)mv-\frac{GMM}{R}-\frac{G4MM}{R}=-\frac{GMM}{2R}-\frac{G4MM}{4R}$

Calculate v

$v=\frac{3GM/5R}{I}\times \frac{1}{2}v=\frac{2GM}{5R}\times \frac{1}{2}$