Question

A small particle travelling with a velocity v collides elastically with a smooth spherical body of equal mass and of radius r initially kept at rest. The centre of this spherical body is located a distance (<r) away from the direction of motion of the particle. Find the final velocities of the two bodies.

Answer 1

The velocity of the particle tangent to the sphere remains unchanged as no force acts along the tangent.

Thus,

$v^{\prime }sin\varphi =vsin\theta ...\left(1\right)$

The momentum along the normal at point of impact.

$-mvcos\theta =m{v}^{\prime }cos\varphi -mV$

$-vcos\theta =v^{\prime }cos\varphi -V..\left(2\right)$

Also as the collision is elastic,

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v^{\prime }}^2+\frac{1}{2}m{V}^2$

$v^2={v^{\prime }}^2+V^2...\left(3\right)$

Thus,

Squaring (1) and (2) and adding,

${v^{\prime }}^2=v^2{sin}^2\theta +V^2+v^2{cos}^2\theta -2vVcos\theta$

Replacing ${v^{\prime }}^2$ from (3).

$v^2-V^2=v^2{sin}^2\theta +V^2+v^2{cos}^2\theta -2vVcos\theta v^2-V^2=v^2+V^2+-2vVcos\theta 2V^2=2Vvcos\theta V=vcos\theta$

From the figure,

$cos\theta =\frac{\sqrt{r^2-{\rho }^2}}{r}$

Thus,

$V=v\frac{\sqrt{r^2-{\rho }^2}}{r}$

and

$v^{\prime 2}=v^2-V^2$

therefore,

$v^{\prime 2}=v^2-v^2{cos}^2\theta =v^2{sin}^2\theta v^{\prime }=vsin\theta v^{\prime }=v\frac{\rho }{r}$