# Why does the ground always push up?

In the example of a bouncy ball being thrown at the ground, at some angle from the normal, $\theta$, we know that the ground will apply a force in the normal direction, causing the ball to rebound at $\theta$ degrees in the other direction. But why is it that the ground doesn't apply the force in the same direction as the incoming ball? Is this just the definition of an object hitting a surface?

To try to understand why this wouldn't be an appropriate model, I thought about something like an atom hitting a rigid lattice of atoms; but doing this I saw no reason why the atom, or ball in this case, should rebound the other way, nor why the force of the lattice on the ball should point in the normal direction.

Moreover, in the case of something like an elastic ball where there is a period of time where the ball has a forward velocity and is contacting the wall, why is it that friction does not slow it down (significantly?)? Is the time period too short?

• – user36790 Aug 24 '15 at 7:50

Let's simplify this problem a bit more. Suppose there is no air resistance and the ground is flat and smooth. If we throw this bouncy ball (which I'm assuming will undergo elastic collisions) at some angle to the ground, $\theta$, then it will have two components of velocity; that perpendicular to the ground, $v_y$, and that horizontal to the ground, $v_x$. When the ball strikes the ground, inertia would tend to make the ball remain at the same velocity. In the x-direction, this is perfectly fine and this tendency is unhindered. However, in the y-direction, the ground is right there preventing the ball from remaining at the same $v_y$. Therefore, the problem simplifies further to become the same as if the only component of the ball's velocity is $v_y$. It collides with the ground and, due to elastic collision, it is given a velocity equal to $-v_y$ after the collision. However, because there was absolutely nothing in the way that prevented the ball from maintaining its $v_x$, that simply doesn't change. The net result is that instead of $(v_x,v_y)$, we have $(v_x,-v_y)$, which make the ball fly off at the same angle to the ground.

The only way $v_x$ could have been affected during this collision is if there were some force acting upon it. In real life, friction from the ground would likely provide some small force affecting $v_x$. However, our smooth and flat ground can't change $v_x$ because the collision happens at a single point on the ground. To the ground, the ball comes in and hits it with some speed $v_y$ and then leaves with speed $-v_y$. There is literally no difference to the ground between our case and one where the ball is thrown straight down at the ground.

This is generalizable to all cases of a normal force. Friction can provide a limited force parallel to the surface; however, a normal force is always normal to the surface because (absent friction) there is simply nothing about the surface that can impede any motion tangent to it. The ground doesn't oppose horizontal forces because it isn't affected by or notice them. You can always break a force into its parallel and perpendicular forces and you'll see that the ground simply isn't in the way of parallel forces for it to oppose them (except with friction).

As for why friction doesn't provide a significant difference to the motion of a bouncy ball, sweber covered that fairly well. It is curious why they said an elastic ball collides inelastically, but no matter. The small point of contact, short timespan, and angular momentum of the ball all contribute to lessen any effects friction might otherwise impart. However, that does not mean they are insignificant. Friction can drastically affect the angle of the ball after collision. For instance, a backspinning ball can often return to you.

The point is that the ball gets a tangential hit by the ground. This changes the angular momentum of the ball.

Consider a ball thrown with a horizontal speed v. It should also not rotate.

Right before hitting the ground, the ball has an angular momentum of

$$L=mvr$$

This is a result of $\vec{L}=\vec{v}\times\vec{p}$, which is also valid for linear moving objects with respect to a resting observer. Also, the vertical movement of the ball is neglected here.

Now, the ball hits the ground (inelastic) in such a way, that the part touching the ground comes to rest. The ball will still move forward with a reduced speed v', but it will also rotate such that the point facing the ground does not move. This means the ball rotates with a radial speed of $v'$ or $\omega=\frac{v'}{r}$.

The overall momentum of the ball remains constant, because its mass is much lower than that of the earth. So:

$$L=mvr=mvr'+J\omega=mvr'+J\frac{v'}{r}$$

Now, $J=\beta mr^2$ ($\beta$ is that fraction defined by the shape of the body):

$$mvr=mvr'+\beta mv'r=(1+\beta)mv'r$$

$$\frac{v'}{v}=\frac{1}{1+\beta}$$

For a solid ball with $\beta=2/5$, you get $\frac{v'}{v}=\frac{5}{7}\approx0.71$
For a hollow ball with $\beta=2/3$, you get $\frac{v'}{v}=\frac{5}{7}\approx0.6$

Finally, the ball continues moving forward with 60-70% of its initial speed!

In reality, friction limits the maximum change of momentum, so the ball will slide while touching the ground and rotate slower after that. Then, more momentum is left for the forward movement, and the ball is faster. In an extreme case, there is no friction, the ball does not start to rotate, and does not change its speed.

From the side, the ball always flies a curve shaped like a V. A difference in angle is hard to see, and it also depends on the ratio of vertical and horizontal speed.

If you give the ball a rotation when throwing it, you will notice that it will move slower/faster after touching the ground. It's even possible that it comes back to you.