## Impacts and Bouncing

When dealing with questions involving projectile motion so far, a typical scenario involved a particle being projected at a certain angle before eventually striking the ground.

###### \[\,\]Figure 7.1.1

Of course, that is not necessarily the end of the story. Any time one object strikes the ground, a wall or indeed any surface, an **impact** is said to have occurred. When that occurs, that object can then *bounce* off that surface, resulting in another round of projectile motion.

###### \[\,\]Figure 7.1.2

To start, let’s look at the simplified case of an object being dropped vertically from rest.

###### \[\,\]Figure 7.1.3

Let’s consider a system that is composed only of the bouncing object. For each bounce, i.e. at the moment of each impact between the object and the ground, do either of our conservation laws hold for this system?

Well, of course, momentum is certainly not conserved. The ground is applying an external force to the system and causing the system’s momentum to change. (The fact that the system reverses direction alone tells us that the velocity (and therefore momentum) of the system is not being conserved.)

What about energy? Does the object have the same energy after the impact as it did before?

Well, if it did, we would expect the ball to reach the same height that it was released from! Instead, with each bounce, the system typically transfers some energy to the ground in the form of heat, sound etc.

So, how do we proceed? How do we find out, for example, the maximum height of the object after the first bounce?

To do that, we need to know how *how much* energy has been lost, i.e. by how much the magnitude of the velocity has been reduced, due to the impact. Knowing the new, smaller velocity after the impact would then allow to find e.g. the new maximum height.

\[\,\]

You likely know from experience that different objects will bounce differently on the same surface.

For example, a football will bounce very high on football field whereas a bowling ball would bounce very little (if at all).

###### \[\,\]Figure 7.1.4

Likewise, that same football would bounce very little on the beach.

###### \[\,\]Figure 7.1.5

The amount of energy lost with each impact, i.e. the amount by which the magnitude of the velocity is reduced, is therefore dependent only on the materials that the object and surface are made from. This reduction in energy is quantified by the **coefficient of restitution** \(e\) in what is known as **Newton’s Law of Restitutio**n.

### Key Point

When an object with a velocity \(\vec{u}=u\vec{j}\) impacts the horizontal ground, it rebounds with a velocity \(\vec{v}=v\vec{j}\).

*Newton’s Law of Restitution* states that, as the velocity reverses direction on impact and, in general, is reduced in magnitude due to a loss of energy, these two velocities are related by

\begin{align}v=-eu\end{align}

where \(e\) is a constant known as the *coefficient of restitution *which depends on the materials that the surface and object are made from.

Note that therefore \(e\) always has a value of between (or equal to) \(0\) and \(1\) since \(u\) will always be greater than or equal to \(v\).

As an example, we may have that \(\vec{u} = -5 \vec{j}\mbox{ m/s}\) and \(\vec{v} = 3\vec{j}\mbox{ m/s}\).

###### \[\,\]Figure 7.1.6

The coefficients in that case are

\begin{align}u=-5 && v=3\end{align}

and therefore

\begin{align} e &= -\frac{v}{u} \\ &= -\frac{3}{-5} \\&=0.6 \end{align}

If \(e=0\), then Newton’s Law of Restitution gives \(v=0\), i.e. the ball does not bounce at all as all of the energy of the ball is transferred to the ground. Such an impact is said to be **inelastic**.

###### \[\,\]Figure 7.1.7

If \(e=1\), Newton’s Law of Restitution instead gives \(v=-u\), i.e. the ball perfectly bounces back to the height it was released from as it has kept all of its energy. Such an impact is instead said to be **elastic**. The energy of the ball is therefore conserved only in the case of an elastic impact, i.e. when \(e=1\).

###### \[\,\]Figure 7.1.8

In general, an impact can instead be neither elastic nor inelastic.

###### \[\,\]Figure 7.1.9

(Note also that we can use the same equation for \(e\) if the impact instead occurs between an object moving horizontally and a vertical wall.)

\[\,\]

In a typical exam question with 1D bounces, we have to break the question up into 3 stages:

\[\,\]

**Stage 1:**Ball is falling. Use the 1D SUVAT scalar equations to find the ball’s velocity when it reaches the ground.

**Stage 2:**Impact with the ground. Use Newton’s Law of Restitution to find the new velocity after impact.

**Stage 3:**Ball rises again. Use the 1D SUVAT equations again to find the SUVAT variable that we are being asked to find.

###### \[\,\]Figure 7.1.10

### Exam Tip

Be sure to keep track of the velocities correctly! For example, the final velocity in Stage 1 will be the initial velocity in Stage 2, and likewise for Stage 2 to Stage 3.

## Guided Example 7.1.1

An object is released from rest from a height of \(5\) metres.

The coefficient of restitution between the object and the ground is \(0.4\).

Find the maximum height of the object after the first bounce, correct to two decimal places.

First, we need to find the speed of the object when it reaches the ground using our usual method.

**Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!**

We are given the following

\begin{align} u = 0 && a = -9.8 && s=-5 \end{align}

and we are looking for \(v\).

\[\,\]

**Step 2: Choose the most suitable equation(s) either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.**

As we are not told \(t\) nor do we want to find \(t\), we should choose the SUVAT equation which does not have \(t\) in it, namely

\[v^2=u^2+2as\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain for

\begin{align}v^2 = 0^2+ 2(-9.8)(-5)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v^2 = 98\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v \approx \pm 9.90 \mbox{ m/s}\end{align}

As the object is moving downwards, we take the negative root, i.e. the velocity of the object just before impacting the ground is \(\vec{v} = -9.90 \vec{j}\mbox{ m/s}\).

This will then be the initial velocity during impact, i.e. \(\vec{u} = -9.90 \vec{j}\mbox{ m/s}\).

During impact, the object loses energy and reverses direction. Therefore, according to Newton’s Law of Restitution

\begin{align}v&=-eu\\&=-(0.4)(-9.90)\\&=3.96 \mbox{ m/s}\end{align}

i.e. the velocity of the object just after impacting the ground is \(\vec{v} = 3.96 \vec{j}\mbox{ m/s}\).

This will now be the object’s initial velocity when finding the maximum height reached after this first bounce, i.e. \(\vec{u} = 3.96 \vec{j}\mbox{ m/s}\)

Let us therefore now find the maximum height reached by the object, i.e. \(s\) when \(v=0\).

**Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!**

We are given the following

\begin{align} u = 3.96 && a = -9.8 && v=0\end{align}

and we are looking for \(s\).

\[\,\]

**Step 2: Choose the most suitable equation(s) either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.**

As we are not told \(t\) nor do we want to find \(t\), we should choose the SUVAT equation which does not have \(t\) in it, namely

\[v^2=u^2+2as\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain for

\begin{align}0^2 = 3.96^2 + 2(-9.8)s\end{align}

Rearranging for \(s\), we obtain

\begin{align}s&=-\frac{3.96^2}{2(-9.8)}\\&\approx 0.80 \mbox{ m}\end{align}

Therefore, the maximum height of the object after the first bounce is \(0.80\) metres.

What if the impact instead occurs in two dimensions?

###### \[\,\]Figure 7.1.11

In that case, we need to decompose the velocity vectors before and after impact.

###### \[\,\]Figure 7.1.12

Note that only one of the two components of the initial velocity vector, i.e. the \(y\) component, actually impacts the ground. The \(x\) component is parallel to the ground and therefore remains unchanged.

Hence, we apply the same logic as before, except that only the \(y\) component of the velocity vector will change.

### Key Point

When an object with a velocity \(\vec{u}=u_x\vec{i} + u_y\vec{j}\) impacts the horizontal ground, it rebounds with a velocity \(\vec{v}=v_x\vec{i} + v_y\vec{j}\).

According to *Newton’s Law of Restitution*, the velocity remains unchanged in the \(x\) direction, i.e. \(u_x=v_x\), and the amount by which the velocity changes in the \(y\) direction is determined by

\[v_y=-eu_y\]

Again, we can also apply the same logic if the ball instead strikes a vertical wall. In this case, it will instead be the \(y\) component of velocity that remains unchanged.

###### \[\,\]Figure 7.1.13

As was the case with 1D bounces, we again have to break the question up into 3 stages:

\[\,\]

**Stage 1:**Projectile is projected. Use the 2D SUVAT scalar equations to find the ball’s velocity when it reaches the ground.

**Stage 2:**Impact with the ground. Use Newton’s Law of Restitution to find the new velocity after impact.

**Stage 3:**Projectile is projected again. Use the 2D SUVAT scalar equations again to find the SUVAT variable that we are being asked to find.

###### \[\,\]Figure 7.1.14

## Guided Example 7.1.2

An object is projected with a speed of \(20\mbox{ m/s}\) at an angle of \(30^{\circ}\) relative to the horizontal ground.

The coefficient of restitution between the object and the ground is \(0.4\).

Find the maximum height of the object after the first bounce, correct to two decimal places.

First, we need to find the speed of the object when it reaches the ground using our usual method.

As the object is projected in the “northeast” direction, the initial velocity coefficients in component form are

\begin{align}u_x &= 20\cos 30^{\circ}\\&\approx 17.32\end{align}

\begin{align}u_y &= 20\sin 30^{\circ}\\&= 10\end{align}

As there is no acceleration in the \(x\) direction, the \(x\) component of the velocity remains unchanged when the object reaches the ground, i.e. \(v_x= 17.32\).

By symmetry, the \(y\) component has the same magnitude but the opposite direction, i.e. \(v_y=-10\).

We are free to simply state that on the exam. However, on this occasion, let’s prove it also just to be sure!

To find the \(y\) component, we need to find \(v_y\) when \(s_y=0\) using our usual method.

**Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!**

We are given the following

\begin{align} u_y = 10 && a_y=-9.8 && s_y=0 \end{align}

and we are looking for \(v_y\).

\[\,\]

**Step 2: Choose the most suitable equation(s) either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.**

As we are not told \(t\) nor do we want to find \(t\), we should choose the \(y\) component SUVAT equation which does not have \(t\) in it, namely

\[v_y^2=u_y^2+2a_ys_y\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}v_y^2 = 10^2 + 2(-9.8)(0)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_y^2=100\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_y=\pm 10\end{align}

As \(v_y\) is pointing in the negative \(y\) direction, we take the negative root, i.e. \(v_y=-10\) as expected.

Therefore, the velocity of the object when it impacts the ground is

\begin{align}\vec{v} &= v_x\vec{i}+v_y\vec{j}\\&=17.32\vec{i}-10\vec{j}\mbox{ m/s}\end{align}

This will then be the initial velocity vector during impact, i.e. \(\vec{u}=17.32\vec{i}-10\vec{j}\mbox{ m/s}\).

After impacting the ground, the object loses energy and therefore \(y\) component of the object’s velocity reduces in size and reverses direction.

Hence, according to Newton’s Law of Restitution

\begin{align}v_y&=-eu_y\\&=-(0.4)(-10)\\&=4 \mbox{ m/s}\end{align}

i.e. the velocity of the object just after impacting the ground is \(\vec{v} = 17.32 \vec{i} + 4 \vec{j}\mbox{ m/s}\).

This will now be the object’s initial velocity during its second round of projectile motion, i.e. \(\vec{u} = 17.32 \vec{i} + 4 \vec{j}\mbox{ m/s}\).

Let us therefore now find the maximum height reached by the object, i.e. \(s_y\) when \(v_y=0\).

We are given the following

\begin{align} u_y = 4 && a_y=-9.8 && v_y=0 \end{align}

and we are looking for \(s_y\).

\[\,\]

As we are not told \(t\) nor do we want to find \(t\), we should choose the \(y\) component SUVAT equation which does not have \(t\) in it, namely

\[v_y^2 = u_y^2+2a_ys_y\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}0^2 = 4^2+ 2(-9.8)s_y\end{align}

Rearranging for \(s_y\), this gives

\begin{align}s_y&=-\frac{4^2}{2(-9.8)}\\&\approx 0.82 \mbox{ m}\end{align}

Therefore, the maximum height of the object after the first bounce is \(0.82\) metres.