Hi,
http://www.tennis.com/gear/2013/01/c.../#.UQwqtx1bNes
Law of Conservation of Momentum:
What that means is that the two momentum (of the racket and the ball, after the impact) have to be equal and with a heavier racket, the ball's speed will be increased, while the racket will not slow down as much (being more stable in your hands, as opposed to lighter frames).
Now that got me thinking? Which one of the two would be affected more?
The crux of the problem being:
1. Would we be hitting much harder with a heavier racket?
or
2. Would the heavier racket be drastically more stable?
What's your guess?
So, I've dusted off my rusty physics (ahem, googled it:
http://en.wikipedia.org/wiki/Elastic_collision) and the answer is:
I’ve done some calculations (based on the equations bellow) and it seems that by adding 100g to a 500g racket:
2. We’d be hitting about 18% harder and the heavier racket(by 100g) would be about 13% more stable.
A bit more realistic calculations, with the ball at around 60g and considering that the pros hit in average around 75-85 mph and the hardest at around 100 mph, so they probably are able to swing the racket faster, say at 10 m/s:
Example A:
Ball: mass = 0.06 kg , velocity = 30 m/s (around 60 mph)
Racket: mass = 0.5 kg, velocity = -10 m/s
After collision:
Ball: velocity = -41.4 m/s (about 93 mph)
Racket: velocity = - 1.42 m/s (The racket continues to swing forward at around 14% of the original speed)
--------------------------------------------------------------
Example B:
Ball: mass = 0.06 kg , velocity = 30 m/s (around 60 mph)
Racket: mass = 0.6 kg, velocity = -10 m/s
After collision:
Ball: velocity = -50 m/s (around 100 mph an increase of 18% when hitting with the heavier racket)
Racket: velocity = -2.7 m/s (The heavier racket continues to swing forward at around 27% of the original speed. Hence more stable, less impact on the arm and inducing longer swings/easier to go more through the ball as opposed to coming over too soon)
A comparison with (60g balls in both cases), seems to give (for a 1 m/s swing aka drop shot):
Example A:
Ball: mass = 0.06 kg , velocity = 30 m/s
Racket: mass = 0.5 kg, velocity = -1 m/s
After collision:
Ball: velocity = -25.35 m/s (you are hitting slower then the incoming ball)
Racket: velocity =5.6 m/s (your arm is pushed back)
--------------------------------------------------------------
Example B:
Ball: mass = 0.06 kg , velocity = 30 m/s
Racket: mass = 0.6 kg, velocity = -1 m/s
After collision:
Ball: velocity = -26.36 m/s (you are hitting just 4% faster)
Racket: velocity = 4.6 m/s (your raquet is 22% more stable)
That means that at slow swings (aka drop shots), one just gets stability out of heavier raquets…No surprise there
Equations
One-dimensional Newtonian
Let m1 and m2 be the masses, u1 and u2 the velocities before collision, and v1and v2 the velocities after collision.
The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision, and is expressed by the equation
m1u1 +m2u2 = m1v1+m2v2
Likewise, the conservation of the total kinetic energy is expressed by the equation
m1u1u1/2 +m2u2u2/2= m1v1v1/2+m2v2v2/2
NOTE: The collision is probably not totally elastic, but rather around 85% (according to USTA), so this science is simplified....
These equations may be solved directly to find vi when ui are known or vice versa. However, the algebra involved can be cumbersome[dubious – discuss]. An alternative solution is to first change the frame of reference such that one of the known velocities is zero. The unknown velocities in the new frame of reference can then be determined and followed by a conversion back to the original frame of reference to reach the same result. Once one of the unknown velocities is determined, the other can be found by symmetry.
Solving these simultaneous equations for vi we get:
v1= (u1(m1-m2)+2m2u2)/(m1+m2)
v2=(u2(m2-m1)+2m1u1)/(m1+m2)
HTH,
Fintft@yahoo.com