Tennis Kinematics Transient Analysis: A Ball Spin & Racket Collision Description by Miltiadis A. Boboulos - HTML preview

1_m 1− k_r()2

_{2 1+ k v} . (Eq. 6)

 

7.2.2 Straight collision of two bodies

Straight collision of two bodies is ava ilable when the straight line connecting their centres of gravity is a common normal at their point of contact and the direction of the speeds of their centres of gravity before the collision is along this normal.

To derive the necessary theoretical relations we shall consider two bo dies having masses m₁ and m₂, respectively. These bodies perform a translational movement and undergo straight collision at a certain moment of time. The speeds of the bodies before the strike are v₁ and v₂ and after the collision, u₁ and u₂, respectively [21]. The task we have here knowing the masses of the two bodies, their speeds before the collision and the coefficient of restitution is to determine the speeds after the collision and the impact impulse.

Using the theorem for the preservation of the amount of movement of a mechanical system we can write down the following expression:

 

m₁u₁+ m₂u₂= m₁v₁+ m₂v₂ (Eq. 7) This expression could be related to the expression we got after applying the first Newton law for the impact:

 

_{k =}u₁− u_{2 = −}u₁ −u_{2 . v1 − v2 v1 − v2}

 

From the set of expressions we obtained, we can determine the values of the speeds we are looking for:

 

u
₁ = v − 1+ k_m
^m − , (Eq. 8) _{1 +}m₂
u₂= v − 1+ k^m − (Eq. 9)

 

m₁ + m₂

A pair of equal impulses in opposite directions occurs between the two bodies. To determine their values it would be helpful if we used the basic equation for impact dynamics:

S₁= m₁u₁− m₁v₁. (Eq. 10) If we substitute u₁ with the relation we obtained previously for S, we get the following: