## Exercise Sets 7A & 7B

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These sets of questions will primarily test a student’s knowledge of the following lesson:

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**7.1 – Impacts and Bouncing**

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**Higher Level students:** attempt both sets.

**Ordinary Level students:** attempt the first set only.

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For each question, we recommend that each student uses the following approach until they feel confident that they fully understand the answer:

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**Answer \(\rightarrow\) Solution \(\rightarrow\) Walkthrough \(\rightarrow\) Video Walkthrough \(\rightarrow\) Teacher Chat**

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**Exercise Set 7A**

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These questions are considered **advanced Ordinary Level to beginner Higher Level** questions.

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We recommend that students move on to the next set of questions, or to the next lesson, only once they have answered \(\mathbf{3}\)** questions fully correct **within this set.

\[\,\]

If a student reaches the end of this set without getting \(3\) questions fully correct, we recommend that they first quickly **review the lesson stated above** before moving on to the next set of questions or to the next lesson.

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If a student still feels doubtful that they are fully prepared for these questions if they were to appear on their exam, we suggest that they **book a grind** with Mr. Kenny.

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## Question 1

An object is thrown vertically upwards with a speed of \(16\mbox{ m/s}\) from a height of \(2\) metres.

The coefficient of restitution between the object and the ground is \(0.5\).

Find the maximum height of the object after the first bounce.

\(3.76\mbox{ m}\)

**Before Impact**

\begin{align} u = 16 && a = -9.8 && s=-2 && v=\mathord{?} \end{align}

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

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

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

\begin{align}v^2 = 16^2+ 2(-9.8)(-2)\end{align}

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

\begin{align}v^2 = 295.2\end{align}

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

\begin{align}v \approx \pm 17.18 \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 \(\vec{v} = -17.18\mbox{ m/s}\vec{j}\).

\[\,\]

**Impact**

\begin{align}v&=-eu\\&=-(0.5)(-17.18)\\&=8.59 \mbox{ m/s}\end{align}

\[\,\]

**After Impact**

\begin{align} u = 8.59 && a = -9.8 && v=0 && s=\mathord{?}\end{align}

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

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

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

\begin{align}0^2 = 8.59^2 + 2(-9.8)s\end{align}

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

\begin{align}s&=-\frac{8.59^2}{2(-9.8)}\\&\approx 3.76 \mbox{ m}\end{align}

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 = 16 && a = -9.8 && s=-2 \end{align}

and we are looking for \(v\).

\[\,\]

**Step 2: Choose the most suitable equation(s) 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 = 16^2+ 2(-9.8)(-2)\end{align}

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

\begin{align}v^2 = 295.2\end{align}

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

\begin{align}v \approx \pm 17.18 \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 \(\vec{v} = -17.18 \vec{j}\mbox{ m/s}\).

This is then the initial velocity when impacting the ground, i.e. \(\vec{u} = -17.18 \vec{j}\mbox{ m/s}\). During the impact, the object loses energy and reverses direction, having a new speed of

\begin{align}v&=-eu\\&=-(0.5)(-17.18)\\&=8.59 \mbox{ m/s}\end{align}

i.e. the velocity of the object just after impacting the ground is \(\vec{v} = 8.59\vec{j}\mbox{ m/s}\).

This will now be the object’s initial velocity when finding the maximum height reached after this first bounce, i.e. \(\vec{u} = 8.59\vec{j}\mbox{ m/s}\).

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 = 8.59 && a = -9.8 && v=0\end{align}

and we are looking for \(s\).

\[\,\]

**Step 2: Choose the most suitable equation(s) 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 = 8.59^2 + 2(-9.8)s\end{align}

Rearranging for \(s\), we obtain

\begin{align}s&=-\frac{8.59^2}{2(-9.8)}\\&\approx 3.76 \mbox{ m}\end{align}

Therefore, the maximum height of the object after the first bounce is \(3.76\) metres.

## Question 2

A projectile of mass \(2\mbox{ kg}\) strikes the horizontal ground, lying along the \(x\) axis, with a speed \(3\vec{i}-8\vec{j}\mbox{ m/s}\) and bounces with a speed \(x\vec{i}+4\vec{j}\mbox{ m/s}\).

**(a)** What is the value of \(x\)?

**(b)** What is the coefficient of restitution between the object and the ground?

**(c)** What was the impulse imparted from the ground to the object?

**(d)** How much energy was lost during the impact?

**(a)** \(3\mbox{ m/s}\)

**(b) **\(0.5\)

**(c)** \(-1\vec{j}\mbox{ kg m/s}\)

**(d)** \(48\mbox{ J}\)

**(a)** Only the \(y\) component of the speed will change. Therefore, the value of \(x\) is \(3\).

**(b)**

\begin{align}e&=-\frac{v_y}{u_y}\\&=-\frac{4}{-8}\\&=\frac{1}{2}\end{align}

**(c)**

\begin{align}\vec{I} &= m\vec{v}-m\vec{u}\\&=(2)(3\vec{i}-8\vec{j})-(2)(3\vec{i}+4\vec{j})\\&=6\vec{i}-16\vec{j}-6\vec{i}-8\vec{j}\\&=-1\vec{j}\end{align}

**(d)**

\begin{align}E_i&=\frac{1}{2}mu_x^2 + \frac{1}{2}u_y^2\\&=\frac{1}{2}(2)(3^2)+\frac{1}{2}(2)(-8)^2\\&=9+64 = 73\mbox{ J}\end{align}

and

\begin{align}E_f&=\frac{1}{2}mv_x^2 + \frac{1}{2}v_y^2\\&=\frac{1}{2}(2)(3^2)+\frac{1}{2}(2)(4^2)\\&=9+16 = 25\mbox{ J}\end{align}

Therefore, the amount of energy lost during the impact is

\begin{align}E_f-E_i&=73-25\\&=48\mbox{ J}\end{align}

**(a)** Only the \(y\) component of the speed will change. Therefore, the value of \(x\) is \(3\).

**(b)** The coefficient of restitution \(e\) is given by

\begin{align}e&=-\frac{v_y}{u_y}\\&=-\frac{4}{-8}\\&=\frac{1}{2}\end{align}

**(c)** By definition, the impulse from the ground to the object is given by

\begin{align}\vec{I} &= m\vec{v}-m\vec{u}\\&=(2)(3\vec{i}-8\vec{j})-(2)(3\vec{i}+4\vec{j})\\&=6\vec{i}-16\vec{j}-6\vec{i}-8\vec{j}\\&=-1\vec{j}\end{align}

**(d)** If we set the reference location \(h=0\) as the horizontal ground, then the projectile’s potential energy just before and after impact is zero.

Hence, the energy of the projectile before impact is

\begin{align}E_i&=\frac{1}{2}mu_x^2 + \frac{1}{2}u_y^2\\&=\frac{1}{2}(2)(3^2)+\frac{1}{2}(2)(-8)^2\\&=9+64 = 73\mbox{ J}\end{align}

The energy of the projectile after impact is instead

\begin{align}E_f&=\frac{1}{2}mv_x^2 + \frac{1}{2}v_y^2\\&=\frac{1}{2}(2)(3^2)+\frac{1}{2}(2)(4^2)\\&=9+16 = 25\mbox{ J}\end{align}

Therefore, the amount of energy lost during the impact is

\begin{align}E_f-E_i&=73-25\\&=48\mbox{ J}\end{align}

## Question 3

An object is thrown vertically downwards with a speed \(u\) from a height of \(4\) metres.

The speed of the object immediately after impacting the ground is \(5\mbox{ m/s}\).

The coefficient of restitution between the object and the ground is \(0.5\).

What is the value of \(u\)?

\(1.41\mbox{ m/s}\)

**Impact**

\begin{align}u&=-\frac{v}{e}\\&=-\frac{5}{0.5}\\&=-10\end{align}

\[\,\]

**Before Impact**

\begin{align} v=-10 && a = -9.8 && s=-4 && v=\mathord{?} \end{align}

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

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

\begin{align}(-10)^2 = u^2+ 2(-9.8)(-5)\end{align}

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

\begin{align}u&=\pm \sqrt{(-10)^2-2(-9.8)(-5)}\\&\approx 1.41 \mbox{ m/s}\end{align}

The velocity of the object immediately after impact is \(\vec{v}=5\vec{j}\mbox{ m/s}\). Therefore, immediately before impact, it had speed of

\begin{align}u&=-\frac{v}{e}\\&=-\frac{5}{0.5}\\&=-10\end{align}

i.e. a velocity of \(-10\vec{j}\mbox{ m/s}\). Now that we know the speed that it strikes the ground with, we can find its initial speed using a SUVAT equation.

**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} v=-10 && a = -9.8 && s=-4 \end{align}

and we are looking for \(v\).

\[\,\]

**Step 3: Choose the most suitable equation(s) based on Step 2**

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 4: Insert the quantities into the equation(s)**

Subbing in our known values, we obtain for

\begin{align}(-10)^2 = u^2+ 2(-9.8)(-5)\end{align}

Rearranging for \(u\), this gives

\begin{align}u&=\pm \sqrt{(-10)^2-2(-9.8)(-5)}\\&\approx 1.41 \mbox{ m/s}\end{align}

## Question 4

A projectile strikes the ground with a speed \(2\vec{i}-6\vec{j}\mbox{ m/s}\) and bounces with a speed \(2\vec{i}+v\vec{j}\mbox{ m/s}\). The ground lies along the \(x\) axis.

The coefficient of friction between the object and the ground is \(\displaystyle\frac{1}{4}\).

**(a)** What is the value of \(v\)?

**(b)** What is the angle of the velocity vector after impact relative to the ground?

**(a)** \(1.5\mbox{ m/s}\)

**(b)** \(36.87^{\circ}\)

**(a)**

\begin{align}v_y&=-eu_y\\&=-\left(\frac{1}{4}\right)(-6)\\&=1.5\end{align}

and therefore \(v=1.5\).

**(b)**

\begin{align}\vec{v}=2\vec{i}+1.5\vec{j}\end{align}

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

\begin{align}\theta&= \tan^{-1}\left(\frac{v_y}{v_x}\right)\\&=\tan^{-1}\left(\frac{1.5}{2}\right)\\&\approx 36.87^{\circ}\end{align}

**(a)** This is determined by the coefficient of restitution as

\begin{align}v_y&=-eu_y\\&=-\left(\frac{1}{4}\right)(-6)\\&=1.5\end{align}

and therefore \(v=1.5\).

**(b)** We now know that the velocity after impact is

\begin{align}\vec{v}=2\vec{i}+1.5\vec{j}\end{align}

The angle that this vector is at relative to the horizontal is

\begin{align}\theta&= \tan^{-1}\left(\frac{v_y}{v_x}\right)\\&=\tan^{-1}\left(\frac{1.5}{2}\right)\\&\approx 36.87^{\circ}\end{align}

## Question 5

An object is projected with a speed of \(12\mbox{ m/s}\) at an angle of \(60^{\circ}\) relative to the horizontal ground.

The coefficient of restitution between the object and the ground is \(0.8\).

What is the maximum height of the object after the first bounce?

\(3.52\mbox{ m}\)

**Before Impact**

\begin{align}u_x &= u \cos A\\&=12 \cos 60^{\circ}\\&=6\end{align}

\begin{align}u_y &= u \sin A\\&=12 \sin 60^{\circ}\\&\approx 10.39\end{align}

\(v_x=6\) and \(v_y=-10.39\)

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

\(\vec{v} = 6\vec{i}-10.39\vec{j}\mbox{ m/s}\)

\[\,\]

**Impact**

\(\vec{u} = 6\vec{i}-10.39\vec{j}\mbox{ m/s}\)

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

\(v_x=6\)

and

\begin{align}v_y&=-eu_y\\&=-(0.8)(-10.39)\\&=\approx 8.31\end{align}

\(\vec{v}=6\vec{i}+8.31\vec{j}\mbox{ m/s}\)

\[\,\]

**After Impact**

\(\vec{u}=6\vec{i}+8.31\vec{j}\mbox{ m/s}\)

\begin{align} u_y = 8.31 && a_y=-9.8 && v_y=0 && s_y=H\end{align}

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

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

\begin{align}0^2 = 8.31^2+ 2(-9.8)H\end{align}

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

\begin{align}H &= \frac{8.31^2}{2(9.8)}\\&\approx 3.52 \mbox{ m}\end{align}

We first need to state the speed at which the object first strikes the ground. And to do that, we need to write the initial velocity vector in component form.

As the object is projected in the “northeast” direction, the coefficients of the initial velocity vector \(\vec{u}\) are

\begin{align}u_x &= u \cos A\\&=12 \cos 60^{\circ}\\&=6\end{align}

\begin{align}u_y &= u \sin A\\&=12 \sin 60^{\circ}\\&\approx 10.39\end{align}

As there is no acceleration in the \(x\) direction, the \(x\) component when the object lands remains unchanged, i.e. \(v_x=6\). Based on the symmetry, the \(y\) component will have the same magnitude but be in the opposite direction, i.e. \(v_y=-10.39\).

Hence, the velocity of the object just before impacting the ground is \(\vec{v} = 6\vec{i}-10.39\vec{j}\mbox{ m/s}\).

This is then the initial velocity during the bounce, i.e. \(\vec{u} = 6\vec{i}-10.39\vec{j}\mbox{ m/s}\).

The \(x\) component of the object’s velocity remains unchanged during the bounce, i.e. \(v_x=6\), and the \(y\) component is given by

\begin{align}v_y&=-eu_y\\&=-(0.8)(-10.39)\\&=\approx 8.31\end{align}

Hence, the initial velocity of the object for this second round of projection is \(\vec{u}=6\vec{i}+8.31\vec{j}\mbox{ m/s}\). We can now use this to find the maximum height of the projectile after the first bounce.

The maximum height corresponds to the value of \(s_y\) when \(v_y=0\). We should therefore use a \(y\) component SUVAT equation.

**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 = 8.31 && a_y=-9.8 && v_y=0 \end{align}

and we are looking for \(s_y=H\).

\[\,\]

**Step 2: Choose the most suitable equation(s) 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}0^2 = 8.31^2+ 2(-9.8)H\end{align}

Rearranging, this gives

\begin{align}H &= \frac{8.31^2}{2(9.8)}\\&\approx 3.52 \mbox{ m}\end{align}

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**Exercise Set 7B**

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These questions are considered **intermediate to advanced Higher Level** questions.

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We recommend that students move on to the next lesson only once they have answered \(\mathbf{3}\)** questions fully correct **within this set.

\[\,\]

If a student reaches the end of this set without getting \(3\) questions fully correct, we recommend that they first quickly **review the lesson stated above** before moving on to the next lesson.

\[\,\]

If a student still feels doubtful that they are fully prepared for these questions if they were to appear on their exam, we suggest that they **book a grind** with Mr. Kenny.

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## Question 1

An object impacts the ground with a velocity of \(-5\vec{j}\mbox{ m/s}\).

The object loses \(40\%\) of its energy each time it bounces.

What is the maximum height of the object

**(a)** after the first bounce?

**(b)** after the second bounce?

**(a)** \(0.76\mbox{ m}\)

**(b)** \(0.46\mbox{ m}\)

**(a)**

**First Impact**

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}m(-5)^2\\&=12.5m\end{align}

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

\begin{align}E_f&=0.6E_i\\&=(0.6)(12.5m)\\&=7.5m\end{align}

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

\begin{align}\frac{1}{2}mv^2=7.5m\end{align}

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

\begin{align}\frac{1}{2}v^2=7.5\end{align}

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

\begin{align}v&=\pm \sqrt{2(7.5)}\\&\approx \pm 3.87 \mbox{ m/s}\end{align}

As the object will be moving in the positive \(y\) direction after impact, we take the positive root, i.e. \(v=3.87\).

\[\,\]

**After First Impact**

\(\vec{u}=3.87\vec{j}\mbox{ m/s}\)

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

\begin{align} u = 3.87 && a = -9.8 && v=0 && s=\mathord{?}\end{align}

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

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

\begin{align}0^2 = 3.87^2 + 2(-9.8)s\end{align}

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

\begin{align}s&=-\frac{3.87^2}{2(-9.8)}\\&\approx 0.76 \mbox{ m}\end{align}

**(b)** By symmetry, the object will land just before impact with a velocity of \(\vec{v}=-3.87\vec{i}\mbox{ m/s}\).

\[\,\]

**Second Impact**

\(\vec{u}=-3.87\vec{i}\mbox{ m/s}\)

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}m(-3.87)^2\\&\approx 7.49m\end{align}

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

\begin{align}E_f&=0.6E_i\\&=(0.6)(7.49m)\\&\approx 4.49m \mbox{ J}\end{align}

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

\begin{align}\frac{1}{2}mv^2=4.49m\end{align}

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

\begin{align}\frac{1}{2}v^2=4.49\end{align}

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

\begin{align}v&=\pm \sqrt{2(4.49)}\\&\approx \pm 3.00 \mbox{ m/s}\end{align}

As the object will be moving in the positive \(y\) direction after impact, we take the positive root, i.e. \(v=3\).

\[\,\]

**After Second Impact**

\(\vec{u}=3\vec{i}\mbox{ m/s}\)

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

\begin{align} u = 3 && a = -9.8 && v=0 && s=\mathord{?}\end{align}

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

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

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

\begin{align}0^2 = 3^2 + 2(-9.8)s\end{align}

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

\begin{align}s&=-\frac{3^2}{2(-9.8)}\\&\approx 0.46 \mbox{ m}\end{align}

**(a)** Let the reference location \(h=0\) be the ground. Therefore, on impact, the object will only lose kinetic energy.

The energy of the object just before impact is therefore

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}m(-5)^2\\&=12.5m\end{align}

The energy of the object just after impact is instead

\begin{align}E_f&=0.6E_i\\&=(0.6)(12.5m)\\&=7.5m\end{align}

As this energy is only kinetic energy, the object’s speed after impact is related to this energy via

\begin{align}\frac{1}{2}mv^2=7.5m\end{align}

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

\begin{align}\frac{1}{2}v^2=7.5\end{align}

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

\begin{align}v&=\pm \sqrt{2(7.5)}\\&\approx \pm 3.87 \mbox{ m/s}\end{align}

As the object will be moving in the positive \(y\) direction after impact, we take the positive root, i.e. \(v=3.87\).

This will now be the object’s initial speed when finding the maximum height reached after this first bounce.

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

We are given the following

\begin{align} u = 3.87 && a = -9.8 && v=0\end{align}

and we are looking for \(s\).

\[\,\]

**Step 2: Choose the most suitable equation(s) based on Step 1.**

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

\[\,\]

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

Subbing in our known values, we obtain for

\begin{align}0^2 = 3.87^2 + 2(-9.8)s\end{align}

Rearranging for \(s\), we obtain

\begin{align}s&=-\frac{3.87^2}{2(-9.8)}\\&\approx 0.76 \mbox{ m}\end{align}

Therefore, the maximum height of the object after the first bounce is \(0.76\) metres.

**(b)** By symmetry, as the object was launched after the first bounce with a velocity of \(\vec{u}=3.87\vec{i}\mbox{ m/s}\), it will land just before impact with a velocity of \(\vec{v}=-3.87\vec{i}\mbox{ m/s}\).

This will then be the initial velocity just before impact, i.e. \(\vec{u}=-3.87\vec{i}\mbox{ m/s}\).

The energy of the object just before impact is therefore

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}m(-3.87)^2\\&\approx 7.49m\end{align}

The energy of the object just after impact is instead

\begin{align}E_f&=0.6E_i\\&=(0.6)(7.49m)\\&\approx 4.49m \mbox{ J}\end{align}

As this energy is only kinetic energy, the object’s speed after impact is related to this energy via

\begin{align}\frac{1}{2}mv^2=4.49m\end{align}

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

\begin{align}\frac{1}{2}v^2=4.49\end{align}

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

\begin{align}v&=\pm \sqrt{2(4.49)}\\&\approx \pm 3.00 \mbox{ m/s}\end{align}

As the object will be moving in the positive \(y\) direction after impact, we take the positive root, i.e. \(v=3\).

This will now be the object’s initial speed when finding the maximum height reached after this second bounce.

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

We are given the following

\begin{align} u = 3 && a = -9.8 && v=0\end{align}

and we are looking for \(s\).

\[\,\]

**Step 2: Choose the most suitable equation(s) based on Step 1.**

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

\[\,\]

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

Subbing in our known values, we obtain for

\begin{align}0^2 = 3^2 + 2(-9.8)s\end{align}

Rearranging for \(s\), we obtain

\begin{align}s&=-\frac{3^2}{2(-9.8)}\\&\approx 0.46 \mbox{ m}\end{align}

Therefore, the maximum height of the object after the second bounce is \(0.46\) metres.

## Question 2

An object is projected with a speed of \(10\mbox{ m/s}\) at an angle of \(30^{\circ}\) relative to the horizontal ground.

The coefficient of restitution between the object and the ground is \(0.6\).

How far is the object from its initial location when object impacts the ground for the second time?

\(14.11\mbox{ m}\)

\begin{align}u_x &= u \cos A\\&=10 \cos 30^{\circ}\\&\approx 8.66\end{align}

\begin{align}u_y &= u \sin A\\&=10 \sin 30^{\circ}\\&=5\end{align}

We will first find the range before the first bounce by finding the first time of flight.

\begin{align} u_y = 5 && a_y=-9.8 && s_y=0 && t=\mathord{?}\end{align}

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

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

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

\begin{align}0=5t +\frac{1}{2}(-9.8)t^2\end{align}

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

\begin{align}t\left(t-\frac{2(5)}{9.8}\right)=0\end{align}

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

\begin{align}t&=\frac{2(5)}{9.8}\\&\approx 1.02 \mbox{ s} \end{align}

We will now use this time of flight to find the range.

\begin{align} u_x = 8.66 && a_x=0 && t=1.02 && s_x=R\end{align}

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

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

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

\begin{align}R &= (8.66)(1.02)+\frac{1}{2}(0)(1.02^2)\\&\approx 8.83 \mbox{ m}\end{align}

\[\,\]

**Before Impact**

\(\vec{v} = 8.66\vec{i}-5\vec{j}\mbox{ m/s}\)

\[\,\]

**Impact**

\(\vec{u} = 8.66\vec{i}-5\vec{j}\mbox{ m/s}\)

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

\(v_x=8.66\)

and

\begin{align}v_y&=-eu_y\\&=-(0.6)(-5)\\&=3\end{align}

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

\(\vec{v}=8.66\vec{i}+3\vec{j}\mbox{ m/s}\)

\[\,\]

**After Impact**

\(\vec{u}=8.66\vec{i}+3\vec{j}\mbox{ m/s}\)

Let’s now find the time of flight between the first and second bounce.

\begin{align} u_y = 3 && a_y=-9.8 && s_y=0 && t=\mathord{?} \end{align}

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

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

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

\begin{align}0=3t +\frac{1}{2}(-9.8)t^2\end{align}

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

\begin{align}t\left(t-\frac{2(3)}{9.8}\right)=0\end{align}

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

\begin{align}t&=\frac{2(3)}{9.8}\\&\approx 0.61 \mbox{ s} \end{align}

Let’s now use this time of flight to find the range between the first and second bounce.

\begin{align} u_x = 8.66 && a_x=0 && t=0.61 && s_x=R \end{align}

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

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

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

\begin{align}R &= (8.66)(0.61)+\frac{1}{2}(0)(0.61^2)\\&\approx 5.28 \mbox{ m}\end{align}

The distance the object is from its initial location after it bounces for the second time is then the sum of these two ranges, i.e. \(8.83+5.28= 14.11\) metres.

First, we need to find how far the object is from its initial location when it bounces for the first time, i.e. we need to find the object’s range.

To do that, we need to write the initial velocity vector in component form. As the object is projected in the “northeast” direction, the coefficients of initial velocity vector \(\vec{u}\) are

\begin{align}u_x &= u \cos A\\&=10 \cos 30^{\circ}\\&\approx 8.66\end{align}

\begin{align}u_y &= u \sin A\\&=10 \sin 30^{\circ}\\&=5\end{align}

We wish to find the object’s range, i.e. the horizontal displacement coefficient \(s_x\) when \(s_y=0\).

As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.

We are given the following

\begin{align} u_y = 5 && a_y=-9.8 && s_y=0 \end{align}

and we are looking for \(t\).

\[\,\]

**Step 2: Choose the most suitable equation(s) based on Step 1.**

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

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

\[\,\]

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

Subbing in our known values, we obtain

\begin{align}0=5t +\frac{1}{2}(-9.8)t^2\end{align}

Factorising, this gives

\begin{align}t\left(t-\frac{2(5)}{9.8}\right)=0\end{align}

Therefore, either \(t=0\) or

\begin{align}t&=\frac{2(5)}{9.8}\\&\approx 1.02 \mbox{ s} \end{align}

The \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is \(t=1.02\mbox{ s}\).

We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.

We are given the following

\begin{align} u_x = 8.66 && a_x=0 && t=1.02 \end{align}

and we are looking for \(s_x=R\).

\[\,\]

**Step 2: Choose the most suitable equation(s) based on Step 1.**

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

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

\[\,\]

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

Subbing in our known values, we obtain

\begin{align}R &= (8.66)(1.02)+\frac{1}{2}(0)(1.02^2)\\&\approx 8.83 \mbox{ m}\end{align}

Next, we need to to determine the speed of the object after it bounces in order to find the second range after the first bounce.

To do that, we need to first state the speed at which the object first strikes the ground. As therefore is no acceleration in the \(x\) direction, the \(x\) component remains unchanged, i.e. \(v_x=8.66\). Based on the symmetry, the \(y\) component will have the same magnitude but be in the opposite direction, i.e. \(v_y=-5\).

Hence, the velocity of the object just before impacting the ground is \(\vec{v} = 8.66\vec{i}-5\vec{j}\mbox{ m/s}\).

This will then be the initial velocity during the bounce, i.e. \(\vec{u} = 8.66\vec{i}-5\vec{j}\mbox{ m/s}\)

During the bounce, the \(x\) component of the object’s velocity remains unchanged, i.e. \(u_x=8.66\), and the \(y\) component is given by

\begin{align}v_y&=-eu_y\\&=-(0.6)(-5)\\&=3\end{align}

Hence, the initial velocity of the object for this second round of projection is \(\vec{u}=8.66\vec{i}+3\vec{j}\mbox{ m/s}\). We now repeat our previous method to find the range between the first and second bounce.

As this statement defines an \(x\) component based on the condition of a \(y\) component, this suggests that we should first find \(t\) using a \(y\) component equation.

We are given the following

\begin{align} u_y = 3 && a_y=-9.8 && s_y=0 \end{align}

and we are looking for \(t\).

\[\,\]

**Step 2: Choose the most suitable equation(s) based on Step 1.**

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

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

\[\,\]

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

Subbing in our known values, we obtain

\begin{align}0=3t +\frac{1}{2}(-9.8)t^2\end{align}

Factorising, this gives

\begin{align}t\left(t-\frac{2(3)}{9.8}\right)=0\end{align}

Therefore, either \(t=0\) or

\begin{align}t&=\frac{2(3)}{9.8}\\&\approx 0.61 \mbox{ s} \end{align}

Again, the \(t=0\) is expected as \(s_y\) is also zero initially when the object is projected. However, we are instead interested in when it next hits the ground, i.e. the time of flight, which is \(t=1.02\mbox{ s}\).

We now wish to find what \(s_x =R\) is at this time. We should therefore now use an \(x\) component equation.

We are given the following

\begin{align} u_x = 8.66 && a_x=0 && t=0.61 \end{align}

and we are looking for \(s_x=R\).

\[\,\]

**Step 2: Choose the most suitable equation(s) based on Step 1.**

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

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

\[\,\]

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

Subbing in our known values, we obtain

\begin{align}R &= (8.66)(0.61)+\frac{1}{2}(0)(0.61^2)\\&\approx 5.28 \mbox{ m}\end{align}

The distance the object is from its initial location after it bounces for the second time is then the sum of these two ranges, i.e. \(8.83+5.28= 14.11\) metres.

## Question 3

An object is released from rest a height \(h\) above horizontal ground.

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

Show that the maximum height of the object after the first bounce is \(e^2h\).

The answer is already in the question!

**Before Impact**

\begin{align} u = 0 && a = -g && s=-h_1 && v=\mathord{?} \end{align}

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

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

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

\begin{align}v^2 = 0^2+ 2(-g)(-h_1)\end{align}

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

\begin{align}v^2 = 2gh_1\end{align}

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

\begin{align}v =\pm\sqrt{2gh_1}\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 \(\vec{v} = -\sqrt{2gh_1} \vec{j}\).

\[\,\]

**Impact**

\(\vec{u} = -\sqrt{2gh_1} \vec{j}\)

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

\begin{align}v&=-eu\\&=-e(-\sqrt{2gh_1})\\&=e\sqrt{2gh_1}\end{align}

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

\(\vec{v} = e\sqrt{2gh_1}\vec{j}\)

\[\,\]

**After Impact**

\(\vec{u} = e\sqrt{2gh_1}\vec{j}\)

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

\begin{align} u = e\sqrt{2gh_1} && a = -g && v=0 && s=h_2\end{align}

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

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

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

\begin{align}0^2 =(e\sqrt{2gh_1})^2 + 2(-g)(h_2)\end{align}

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

\begin{align}h_2&=-\frac{(e\sqrt{2gh_1})^2}{2(-g)}\\&=\frac{2e^2gh_1}{2g}\\&=e^2h_1\end{align}

as required.

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

We are given the following

\begin{align} u = 0 && a = -g && s=-h_1 \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.**

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

\[\,\]

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

Subbing in our known values, we obtain

\begin{align}v^2 = 0^2+ 2(-g)(-h_1)\end{align}

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

\begin{align}v^2 = 2gh_1\end{align}

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

\begin{align}v =\pm\sqrt{2gh_1}\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 \(\vec{v} = -\sqrt{2gh_1} \vec{j}\).

This will then be the initial velocity during impact, i.e. \(\vec{u} = -\sqrt{2gh_1} \vec{j}\).

The object loses energy and reverses direction during impact, having a new speed of

\begin{align}v&=-eu\\&=-e(-\sqrt{2gh_1})\\&=e\sqrt{2gh_1}\end{align}

i.e. the velocity of the object just after impacting the ground is \(\vec{v} = e\sqrt{2gh_1}\vec{j}\).

This will now be the object’s initial velocity when finding the maximum height reached after this first bounce, i.e. \(\vec{u} = e\sqrt{2gh_1}\vec{j}\)

Let us therefore now find the maximum height reached by the object, i.e. \(s=h_2\) when \(v=0\).

We are given the following

\begin{align} u = e\sqrt{2gh_1} && a = -g && v=0\end{align}

and we are looking for \(s=h_2\).

\[\,\]

**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.**

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

\[\,\]

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

Subbing in our known values, we obtain for

\begin{align}0^2 =(e\sqrt{2gh_1})^2 + 2(-g)(h_2)\end{align}

Rearranging for \(h_2\), we obtain

\begin{align}h_2&=-\frac{(e\sqrt{2gh_1})^2}{2(-g)}\\&=\frac{2e^2gh_1}{2g}\\&=e^2h_1\end{align}

as required.

## Question 4

An object of mass \(300\mbox{ g}\) is directed towards an angled wall as shown below.

The object strikes the wall with a speed of \(8\mbox{ m/s}\).

The coefficient of restitution between the object and the wall is \(0.25\).

**(a)** What is the speed of the object after impact?

**(b)** How much energy was lost during impact?

**(a) **\(7\mbox{ m/s}\)

**(b)** \(2.25\mbox{ J}\)

**(a)** We shall use rotated axes

with the initial velocity vector decomposed as below.

\begin{align}|\vec{u_x}| &= |\vec{u}| \cos A \\&= 8\cos 30^{\circ} \\&\approx 6.93\end{align}

\begin{align}|\vec{u_y}| &= |\vec{u}| \sin A \\&= 8\sin 30^{\circ} \\&=4\end{align}

##### \[\,\]

**Impact**

\(v_x = 6.93\)

and

\begin{align}v_y&=-eu_y\\&=-(0.25)(4)\\&=-1\end{align}

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

\begin{align}\vec{v} = 6.93\vec{i}-\vec{j}\end{align}

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

\begin{align}|\vec{v}|&=\sqrt{6.93^2+(-1)^2}\\&\approx 7 \mbox{ m/s}\end{align}

**(b)**

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}(0.3)(8^2)\\&=9.6 \mbox{ J}\end{align}

and

\begin{align}E_f&=\frac{1}{2}mv^2\\&=\frac{1}{2}(0.3)(7^2)\\&=7.35 \mbox{ J}\end{align}

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

\begin{align}E_f-E_i&=9.6-7.35\\&=2.25 \mbox{ J}\end{align}

**(a)** The easiest approach to this question is to use rotated axes such that the wall lies along the \(x\) axis.

In doing so, we now need to decompose the initial velocity vector just before impact.

**Step 1: Draw a labelled diagram of the triangle of decomposed vectors.**

**Step 2: Use the given angle \(\boldsymbol{\theta}\) to find the argument \(\boldsymbol{A}\) in this triangle.**

In this case, the angle \(\theta\) given in polar form is the angle of the vector relative to the positive \(x\) axis, i.e.

\begin{align}A = 30^{\circ}\end{align}

\[\,\]

**Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{u_x}| = |\vec{u}|\cos A} \) and \(\boldsymbol{|\vec{u_y}| = |\vec{u}|\sin A} \).**

\begin{align}|\vec{u_x}| &= |\vec{u}| \cos A \\&= 8\cos 30^{\circ} \\&\approx 6.93\end{align}

\begin{align}|\vec{u_y}| &= |\vec{u}| \sin A \\&= 8\sin 30^{\circ} \\&=4\end{align}

\[\,\]

**Step 4: Calculate \(\boldsymbol{u_x}\) and \(\boldsymbol{u_y}\) using \(\boldsymbol{|\vec{u_x}|}\) and \(\boldsymbol{|\vec{u_y}|}\), i.e. be careful with signs!**

According to the above diagram, \(\vec{u_x}\) is in the same direction as \(\vec{i}\) and \(u_y\) is in the *opposite* direction to \(\vec{j}\). Therefore

\begin{align}u_x =6.93 && u_y = -4\end{align}

\[\,\]

**Step 5: Write \(\boldsymbol{\vec{u} =u_x \vec{i} + u_y \vec{j}}\).**

\[\,\]

\[\vec{u}=6.93\vec{i} -4\vec{j}\]

During impact with the angled wall, the component along the wall, i.e. the \(x\) component, remains unchanged, and the \(y\) component reduces to

\begin{align}v_y&=-eu_y\\&=-(0.25)(4)\\&=-1\end{align}

Hence, the velocity of the object after impact is

\begin{align}\vec{v} = 6.93\vec{i}-\vec{j}\end{align}

We are instead asked for the speed of the object after impact, i.e. the magnitude of this vector. This is determined using Pythagoras’ theorem as

\begin{align}|\vec{v}|&=\sqrt{6.93^2+(-1)^2}\\&\approx 7 \mbox{ m/s}\end{align}

**(b)** If we set our reference location \(h=0\) as the position of the object at impact, then then potential energy just before and after impact is zero.

The initial kinetic energy of the object is therefore

\begin{align}E_i&=\frac{1}{2}mu^2\\&=\frac{1}{2}(0.3)(8^2)\\&=9.6 \mbox{ J}\end{align}

The final kinetic energy of the object is instead

\begin{align}E_f&=\frac{1}{2}mv^2\\&=\frac{1}{2}(0.3)(7^2)\\&=7.35 \mbox{ J}\end{align}

Therefore, the energy lost during impact is

\begin{align}E_f-E_i&=9.6-7.35\\&=2.25 \mbox{ J}\end{align}

## Question 5

An object is direct towards an angled wall.

The object strikes the wall horizontally with a speed \(u\).

The object moves vertically after impact with a speed \(v\).

The coefficient of restitution between the object and the wall is \(e\).

Show that:

**(a)** \(\tan A = \displaystyle\frac{v}{u}\)

**(b)** \(v^2=eu^2\)

**(c)** the fractional loss of energy during impact is \(e\)

**(a) **The answer is already in the question!

**(b) **The answer is already in the question!

**(c) **The answer is already in the question!

**(a)** We shall use rotated axes

with the initial velocity vector decomposed as below.

\begin{align}|\vec{u_x}| &=u \cos A\end{align}

\begin{align}|\vec{u_y}| &= u \sin A \end{align}

The final velocity vector is instead decomposed as

\begin{align}|\vec{v_x}| &=v \sin A\end{align}

\begin{align}|\vec{v_y}| &= v \cos A \end{align}

Hence, the velocity of the object just before and after impacting the angled wall is

\[\vec{u}=u\cos A\vec{i} -u\sin A\vec{j}\]

\[\vec{v}=v\sin A\vec{i} +v\cos A\vec{j}\]

After impacting the angled wall, the component along the wall, i.e. the \(x\) component, remains unchanged. Therefore

\begin{align}u \cos A = v\sin A\end{align}

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

\begin{align}\frac{\sin A}{\cos A}=\frac{u}{v}\end{align}

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

\begin{align}\tan A=\frac{u}{v}\end{align}

as required.

**(b)** The \(y\) component instead reduces to

\begin{align}v_y&=-eu_y\end{align}

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

\begin{align}v\cos A=-e(-u\sin A)\end{align}

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

\begin{align}\frac{\sin A}{\cos A}=\frac{v}{eu}\end{align}

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

\begin{align}\tan A=\frac{v}{eu}\end{align}

Setting our two equations for \(\tan A\) equal to each other gives

\begin{align}\frac{u}{v} = \frac{v}{eu}\end{align}

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

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

as required.

**(c)**

\begin{align}E_i=\frac{1}{2}mu^2\end{align}

and

\begin{align}E_f=\frac{1}{2}mv^2\end{align}

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

\begin{align}\frac{E_f}{E_i}&=\frac{\frac{1}{2}mv^2}{\frac{1}{2}mu^2}\\&=\frac{v^2}{u^2}\end{align}

Inserting the result from part (b), this reduces to

\begin{align}\frac{E_f}{E_i}=e\end{align}

as required.

**(a)** The easiest approach to this question is to use rotated axes such that the wall lies along the \(x\) axis.

In doing so, we now need to decompose the initial velocity vector just before impact.

**Step 1: Draw a labelled diagram of the triangle of decomposed vectors.**

**Step 2: Use the given angle \(\boldsymbol{\theta}\) to find the argument \(\boldsymbol{A}\) in this triangle.**

In this case, the angle \(\theta\) given in polar form is the angle of the vector relative to the positive \(x\) axis, i.e.

\begin{align}A = \theta\end{align}

\[\,\]

**Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{u_x}| = |\vec{u}|\cos A} \) and \(\boldsymbol{|\vec{u_y}| = |\vec{u}|\sin A} \).**

\begin{align}|\vec{u_x}| &=u \cos A\end{align}

\begin{align}|\vec{u_y}| &= u \sin A \end{align}

\[\,\]

**Step 4: Calculate \(\boldsymbol{u_x}\) and \(\boldsymbol{u_y}\) using \(\boldsymbol{|\vec{u_x}|}\) and \(\boldsymbol{|\vec{u_y}|}\), i.e. be careful with signs!**

According to the above diagram, \(\vec{u_x}\) is in the same direction as \(\vec{i}\) and \(u_y\) is in the *opposite* direction to \(\vec{j}\). Therefore

\begin{align}u_x =u\cos A && u_y = -u \sin A\end{align}

\[\,\]

**Step 5: Write \(\boldsymbol{\vec{u} =u_x \vec{i} + u_y \vec{j}}\).**

\[\vec{u}=u\cos A\vec{i} -u\sin A\vec{j}\]

We also need to decompose the final velocity vector just after impact.

**Step 1: Draw a labelled diagram of the triangle of decomposed vectors.**

**Step 2: Use the given angle \(\boldsymbol{\theta}\) to find the argument \(\boldsymbol{B}\) in this triangle.**

In this case, the angle \(\theta\) given in polar form is the angle of the vector relative to the positive \(x\) axis, i.e.

\begin{align}A = \theta\end{align}

\[\,\]

**Step 3: Calculate the magnitudes of the decomposed vectors using \(\boldsymbol{|\vec{v_x}| = |\vec{v}|\cos A} \) and \(\boldsymbol{|\vec{v_y}| = |\vec{v}|\sin A} \).**

\begin{align}|\vec{v_x}| &=v \sin A\end{align}

\begin{align}|\vec{v_y}| &= v \cos A \end{align}

\[\,\]

**Step 4: Calculate \(\boldsymbol{v_x}\) and \(\boldsymbol{v_y}\) using \(\boldsymbol{|\vec{v_x}|}\) and \(\boldsymbol{|\vec{v_y}|}\), i.e. be careful with signs!**

According to the above diagram, \(\vec{v_x}\) is in the same direction as \(\vec{i}\) and \(v_y\) is in the same direction as \(\vec{j}\). Therefore

\begin{align}v_x =v\sin A && v_y = v \cos A\end{align}

\[\,\]

**Step 5: Write \(\boldsymbol{\vec{v} =v_x \vec{i} + v_y \vec{j}}\).**

\[\vec{v}=v\sin A\vec{i} +v\cos A\vec{j}\]

Hence, the velocity of the object just before and after impacting the angled wall is

\[\vec{u}=u\cos A\vec{i} -u\sin A\vec{j}\]

\[\vec{v}=v\sin A\vec{i} +v\cos A\vec{j}\]

After impacting the angled wall, the component along the wall, i.e. the \(x\) component, remains unchanged. Therefore

\begin{align}u \cos A = v\sin A\end{align}

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

\begin{align}\frac{\sin A}{\cos A}=\frac{u}{v}\end{align}

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

\begin{align}\tan A=\frac{u}{v}\end{align}

as required.

**(b)** The \(y\) component instead reduces to

\begin{align}v_y&=-eu_y\end{align}

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

\begin{align}v\cos A=-e(-u\sin A)\end{align}

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

\begin{align}\frac{\sin A}{\cos A}=\frac{v}{eu}\end{align}

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

\begin{align}\tan A=\frac{v}{eu}\end{align}

Setting our two equations for \(\tan A\) equal to each other gives

\begin{align}\frac{u}{v} = \frac{v}{eu}\end{align}

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

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

as required.

**(c)** If we set \(h=0\) to be the location of the impact, then the potential energy of the object just before and after will be zero.

Hence, the object total energy of the object just before impact is

\begin{align}E_i=\frac{1}{2}mu^2\end{align}

and just after impact is

\begin{align}E_f=\frac{1}{2}mv^2\end{align}

Therefore, the *fractional* loss of energy during impact is

\begin{align}\frac{E_f}{E_i}&=\frac{\frac{1}{2}mv^2}{\frac{1}{2}mu^2}\\&=\frac{v^2}{u^2}\end{align}

Inserting the result from part (b), this reduces to

\begin{align}\frac{E_f}{E_i}=e\end{align}

as required.