Using a tennis ball as an example object, if one ball weighs 1 ounce and the other is 2 ounces, and both are struck at 100 mph on the same trajectory, would there be any difference in the deceleration rate between the 2 different mass balls? (All other things about the two balls being equal). For instance, would the elapsed time for the ball to travel let's say 100 feet be different or the same?

FOLLOW UP QUESTION (based on answer #1): If it's true that the lighter ball would decelerate faster (and consequently take longer to travel a given distance), then what would be the difference in the initial velocity of the two balls (one 1 ounce, the other 2 ounces) if both are struck with the same implement with the same force (let's say a tennis racket travelling 100 mph).

I'm assuming the lighter ball would have a higher initial speed. If so, would the higher initial speed of the lighter ball offset the increased deceleration rate for the lighter ball. In practical terms: with the initial speed being different, which ball would arrive first in the above example of 100 feet, the lighter ball or the heavier ball?

If not too difficult, could you explain how this relationship (first the initial speed difference and then the total travel time difference for 100 feet) would be calculated?


They have the same drag force, the lighter one will decelerate faster

  • 2
    $\begingroup$ So the elapsed time to travel 100 feet would be longer for the lighter ball. - just to finish answering the question.. $\endgroup$ – Floris Jun 15 '14 at 3:12

When an object (ball) of mass $m$ at rest is struck elastically by another object of mass $M$ traveling with initial velocity $v$, then the velocity after impact is given by


Two limiting cases: $m=M$, maximum transfer of energy (the racket stops and the ball travels with the velocity of the racket); when $m<<M$, the final velocity is twice the initial velocity (but the racket maintains most of its energy).

With the values given you are in an intermediate regime - the lighter ball will travel faster immediately after impact but it will slow down faster. The math for this is complicated in 2D - but we can make some progress in 1D.

Drag force on a sphere is given roughly by

$$F= \frac12 \rho v^2 A C_D$$

Where $\rho= 1.2 kg/m^3, C_D=0.47, A= 0.0035 m^2$, so $F=0.002 v^2$.

The acceleration $a=F/m$ so as you can see the lighter ball will decelerate more quickly. The equations of motion become

$$\frac{dv}{dt}=-kv^2\\ \frac{dv}{v^2}=-k\cdot dt\\ \frac{1}{v}=kt + \frac{1}{v_0}\\ v(t)=\frac{1}{kt+\frac{1}{v_0}}\\ x(t)=\frac{1}{k}\log(v_0 k t +1)$$

Where $k=0.002/m$ - it depends on the mass of the ball

The typical mass of a tennis ball is about 58 gram and typical velocity about 30 m/s. A racket has a weight around 250 - 300 gram, so the extra speed you get for the lighter ball is small - but the deceleration is real.

Putting in round numbers:

$$v_{racket} = 20 m/s$$

$$v_1= 20\frac{600}{360} = 33 m/s\\ v_2 = 20\frac{600}{330} = 36 m/s$$

Plotting a graph of velocity and position as a function of time for a 1 oz and 2 oz (nominal) ball, I get for the position:

enter image description here

and for the ball speed:

enter image description here

This confirms that the lighter ball will initially go a little bit faster - but that drag will quickly eliminate its advantage over any but the shortest distances. I will add graphs of the above function of x(t) when I get near a computer (hard to do on a phone...)


If the force applied to each ball is the same then it will provide the same impuls to both, then using J = M(v-u) it can be seen that the lighter ball will reach a higher maximum speed.

Following this, the ball's are now experiencing a drag force, depending on the assumptions made we can either assume that the drag force is constant, or that it is a function of the ball's velocity F = f(v) depending on the complexity of the model we are using.

Either way using Newtons Second Law, F = MA we can determine the acceleration on each of the balls. From here it is a case of using calculus to determine the time taken to travel a specific distance. In this case integrating twice, from here it is simply a case of subbing in numbers. I hope that this helped.

  • $\begingroup$ Same force would have to be applied for the same time for your first sentence to be true... $\endgroup$ – Floris Mar 6 '15 at 15:03

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