Solve Your Tennis Racket Problem with a Geometric Approach

[IamVector]

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

take one of the axes (say x) to be
perpendicular to the racket’s plane and the other one

(y) parallel to it. Absence of rotation means that the y-
components of the ball’s and racket’s velocities are equal

Is there any geometric approach??

 

[jbriggs444]

Is there any geometric approach??

Symmetry may be a fruitful approach. The arrival and departure trajectories should be similar.

 

[IamVector]

Symmetry may be a fruitful approach. The arrival and departure trajectories should be similar.

got u = v/2 cos α by algebraic approach and the angle. didn't got the geometric approach I think it will be easy by geometric approach.

 

[jbriggs444]

got u = v/2 cos α by algebraic approach and the angle? didn't got the geometric approach I think it will be easy by geometric approach.

Never mind. I misunderstood the problem. It's not an elastic collision per se. It's an elastic collision in the racket frame. Which means that in the ground frame, there is no symmetry to exploit.

 

[IamVector]

Never mind. I misunderstood the problem. It's not an elastic collision per se. It's an elastic collision in the racket frame. Which means that in the ground frame, there is no symmetry to exploit.

I saw the hint by geometric approach it says :
Geometric approach: draw a right trapezoid as follows:

we decompose ⃗v into parallel and perpendicular compon-
ents, ⃗v = ⃗vx + ⃗vy; let us mark points A, B and C so that

AB⃗ = ⃗vx and BC⃗ = ⃗vy (then, AC⃗ = ⃗v). Next we mark
points D, E and F so that CD⃗ = ⃗v ′

y = ⃗vy, DE⃗ = −⃗vx,

and EF⃗ = 2⃗ux; then, CF⃗ = ⃗v ′

y − ⃗vx + 2⃗ux ≡ ⃗v ′ and
AF⃗ = 2⃗vy + 2⃗ux ≡ 2⃗u. Due to the problem conditions,
∠ACF = 90◦

. Let us also mark point G as the centre of
AF; then, GC is both the median of the right trapezoid
ABDF (and hence, parallel to AB and the x-axis), and

the median of the triangle ACF. What is left to do, is ex-
pressing the hypotenuse of △ACF in terms of v = |AC|I found this very tricky to understand but it is considered to be more easy than algebraic one so please help.

 

[IamVector]

Never mind. I misunderstood the problem. It's not an elastic collision per se. It's an elastic collision in the racket frame. Which means that in the ground frame, there is no symmetry to exploit.

so what we can use??