Top spin

Relevant wiki: Friction

Without friction, the normal force $N$ exerted by the court would give an impulse to the ball in the vertical direction so that the vertical component of the velocity is the same as before, but the direction is opposite (elastic collision). The friction force is horizontal, therefore it will not influence the vertical motion. Therefore we can conclude that in this collision the magnitude of the vertical impulse is $2mv \sin \alpha$ , where $m$ is the mass, $v$ is the velocity and $\alpha$ is the angle of incidence.

During the collision the friction force exerts a horizontal impulse. Since the ball has a top spin, this force will point forward, causing the horizontal component of the velocity to increase from its original value of $v \cos \alpha$ . To see this, assume the collision takes a time of $\Delta t$ . If the vertical force is $N$ and the horizontal force is $F=\mu N$ , than the vertical impulse is $N \Delta t$ and the horizontal impulse is $\mu N \Delta t$ . The difference between the final and initial momentum is equal to the impulse. For the horizontal component we get

$m v_x' -mv_x = \mu N \Delta t$

For the vertical components

$2 mv_y = N \Delta t$

because in the vertical direction the collision is elastic. If we eliminate $N \Delta t$ and divide by $m$ we get

$v_x' -v_x = 2\mu v_y$

The angle after the collision is $\theta$ and the magnitude of the velocity $v'$ . If we express the components in terms of angles and magnitudes we get

$v \sin \alpha= v' \sin \theta$ , because in this direction the collision is elastic

$v \cos \alpha +2\mu v \sin \theta = v' \cos \theta$ , because the friction force increases the velocity of the ball, as seen above.

Therefore

$\tan \theta =\frac{v_y'}{v_x'}= \frac{\sin \alpha}{\cos \alpha +2 \mu \sin \alpha}$ .

With the numbers in the problem we get $\theta \approx 26^o$

Notes:

• For a weaker top spin it is possible that the ball stops spinning and simply rolls on the surface at some time during the collision. In that case the angle of emergence will be larger. If the ball has no spin at all, it will start rotating during the time it is in contact with the court. This will reduce the horizontal component of the velocity and the ball will emerge at an angle larger than $30^o$

• The coefficient of friction greatly depends on the surface of the court. The value used here is relatively small compared to the measured values on real tennis courts. For example according to measurements by the International Tennis Federation $\mu$ is between 0.4 and 0.9.

Hello All: Just subtract 15 (friction pct) from 100 to get 85 - 15 is the percent (%) loss of ball power. So the ball after the bounce has just 85% of original power. Then multiply 30 x .85 = 26.5.

Relevant wiki: Conservation of Momentum

An approximate answer is given by taking the horizontal component of velocity as-is, and multiplying the vertical component of the velocity by (1 - u); this gives a new vector cos(30) i + 0.85 * sin(30) j. tan(x) = 0.85 sin(30) / cos(30).