What is the angle between the velocity of and the normal to the racket?

[Amitayas Banerjee]

Homework Statement

A tennis ball falls at velocity v onto a heavy racket
and bounces back elastically. What does the racket’s velocity
u have to be to make the ball bounce back at a right angle to
its initial trajectory and not start spinning if it did not spin
before the bounce? What is the angle β between ⃗u and the
normal of the racket’s plane, if the corresponding angle for ⃗v is α?

Relevant Equations

For elastic collision, the relative velocities of approach and separation along the line of impact must be same.

My approach:

Let us take two orthogonal axes: x, parallel to the racket's plane and y, perpendicular to it. For the ball to not spin, the components of initial velocities of the racket and the ball along x-axis must be same. Also, as the line of collision is along the normal to the racket's plane, the initial and final velocity components of the ball along x-axis are same.

Therefore, initial velocity of the ball: $$|v|\sin \alpha \space i - |v|\cos \alpha \space j$$

and final velocity:
$$|v|\sin \alpha \space i+(|v|\cos \alpha + 2|u| \cos \beta)j$$

Also, as the components of initial velocities of the racket and the ball along x-axis must be same,
$$|u|\sin \alpha = |u| \sin \beta$$.
Now, as the initial and final velocities of the ball are perpendicular,
$$(|v|\sin \alpha \space i - |v|\cos \alpha \space j)⋅(|v|\sin \alpha \space i+(|v|\cos \alpha + 2|u| \cos \beta)j)=0$$

I am unable to proceed. What to do next?

 

[kuruman]

For elastic collision, the relative velocities of approach and separation along the line of impact must be same.

It also means that kinetic energy is conserved. What does this say about the relation between the incident speed and the reflected speed?