\text{Chapter 4}
\text{Chapter 4}
\text{Chapter 4}
\text{Chapter 4}
\text{A particle is accelerated by air over a horizontal $xy$-plane with a constant acceleration}
\text{ $\vec{a}=$ $(4.00 \hat{\imath}+8.00 \hat{\jmath}) \mathrm{m} / \mathrm{s}^2$. At time $t=0$, the velocity is $(4.00 \mathrm{~m} / \mathrm{s}) \hat{\imath}$.}
\text{ What is the speed of the particle at $t=1.65 \mathrm{~s}$ ?}
\text{A) 16.9 m/s}\\
\text{B) 16.2 m/s}\\
\text{C) 32.1 m/s}\\
\text{D) 20.4 m/s}\\
\text{E) 14.6 m/s}
\text{A particle is accelerated by air over a horizontal $xy$-plane with a constant acceleration}
\text{ $\vec{a}=$ $(4.00 \hat{\imath}+8.00 \hat{\jmath}) \mathrm{m} / \mathrm{s}^2$. At time $t=0$, the velocity is $(4.00 \mathrm{~m} / \mathrm{s}) \hat{\imath}$.}
\text{ What is the speed of the particle at $t=1.65 \mathrm{~s}$ ?}
\text{\textcolor{red}{A) 16.9 m/s}}\\
\text{B) 16.2 m/s}\\
\text{C) 32.1 m/s}\\
\text{D) 20.4 m/s}\\
\text{E) 14.6 m/s}
v_x=v_x^0+a_x t
v_y=v_y^0+a_y t
v_x=4+4 . 1.65=10.6 \text{ m/s}
v_y=0+8 1.65=13.2 \text{ m/s}
\displaystyle v=\sqrt{v_x^2+v_y^2}=\sqrt{10.6^2+13.2^2}=16.9 \text{ m/s}
\text{Motion with constant acceleration along the two axis:}
\Rightarrow
\text{A projectile is fired from level ground with a speed of $12.0 \mathrm{~m} / \mathrm{s}$ at a target located at a height }
\text{\textcolor{black}{A) }}55.1^0\\
\text{B) }33.2^0\\
\text{C) }41.5^0\\
\text{D) }64.3^0\\
\text{E) }44.7^0
\text{$h=5.00 \mathrm{~m}$ above the ground as shown in Figure 2. The projectile's velocity to be horizontal}
\text{ at the instant it reaches the target. }
\text{At what angle above the horizontal must the projectile be fired?}
\text{A projectile is fired from level ground with a speed of $12.0 \mathrm{~m} / \mathrm{s}$ at a target located at a height }
\text{\textcolor{red}{A) }}55.1^0\\
\text{B) }33.2^0\\
\text{C) }41.5^0\\
\text{D) }64.3^0\\
\text{E) }44.7^0
\text{$h=5.00 \mathrm{~m}$ above the ground as shown in Figure 2. The projectile's velocity to be horizontal}
\text{ at the instant it reaches the target. }
\text{At what angle above the horizontal must the projectile be fired?}
\displaystyle \theta=\sin^{-1}0.82=55.1^0
\text{The velocity at the target is horizontal. This means that } v_y=0
\text{We conclude that the target is the \textcolor{red}{maximum height} we can reach}
\displaystyle h=\frac{v_0^2 \sin^2\theta}{2 g}
\displaystyle \sin\theta=\sqrt{\frac{2 g h}{v_0^2 }}
\Rightarrow
\Rightarrow
\displaystyle \sin\theta=\sqrt{\frac{2\times9.8\times5}{12^2 }}=0.82
\Rightarrow
\text{A ball thrown horizontally at $2.5 \mathrm{~m} / \mathrm{s}$ travels a horizontal distance of $1.6 \mathrm{~m}$ before hitting }
\text{\textcolor{black}{A) }}3.7\text {m}\\
\text{B) }4.1\text {m}\\
\text{C) }3.2\text {m}\\
\text{D) }2.0\text {m}\\
\text{E) }1.7\text {m}
\text{the ground. From what height was the ball thrown?}
\text{(old exam)}
\text{A ball thrown horizontally at $2.5 \mathrm{~m} / \mathrm{s}$ travels a horizontal distance of $1.6 \mathrm{~m}$ before hitting }
\text{\textcolor{black}{A) }}3.7\text {m}\\
\text{B) }4.1\text {m}\\
\text{C) }3.2\text {m}\\
\text{\textcolor{red}{D) }}2.0\text {m}\\
\text{E) }1.7\text {m}
\text{the ground. From what height was the ball thrown?}
\displaystyle y=x\tan\theta+\frac{1}{2}g \frac{x^2}{(v_0 \cos\theta)^2}
x_0=0,y_0=0, \theta=0
\text{We choose a reference at the point the object was fired.}
\Rightarrow
\displaystyle y=-\frac{1}{2}9.8 \frac{1.6^2}{(2.5)^2}=-2.0 \text{ m}
\text{The object hits the ground with }y=-2.0 \text{ m} ,\text{ so the height is } 2.0\text{ m}
\text{(old exam)}
\text{An object falls from a bridge that is $45 \mathrm{~m}$ above the water. It falls directly into a small boat,}
\textcolor{black}{A) \hspace{1mm} 4.0\text { m/s}}\\
\text{B) }8.0\text { m/s}\\
\text{C) }6.0\text { m/s}\\
\text{\textcolor{black}{D) }}5.0\text { m/s}\\
\text{E) }2.5\text { m/s}
\text{(old exam)}
\text{ moving with constant speed $v$ as shown in Figure 2. The boat was $12 \mathrm{~m}$ away from the point }
\text{of impact (the point at which the object falls on the boat) when the object was released.}
\text{ What is the speed $v$ of the boat? [Ignore air resistance.]}
\text{An object falls from a bridge that is $45 \mathrm{~m}$ above the water. It falls directly into a small boat,}
\textcolor{red}{A) \hspace{1mm} 4.0\text { m/s}}\\
\text{B) }8.0\text { m/s}\\
\text{C) }6.0\text { m/s}\\
\text{\textcolor{black}{D) }}5.0\text { m/s}\\
\text{E) }2.5\text { m/s}
\text{(old exam)}
\text{ moving with constant speed $v$ as shown in Figure 2. The boat was $12 \mathrm{~m}$ away from the point }
\text{of impact (the point at which the object falls on the boat) when the object was released.}
\text{ What is the speed $v$ of the boat? [Ignore air resistance.]}
y-y_0=v_0t+\frac{1}{2}at^2
x-x_0=v t
\text{For the ball:}
\text{Let us choose the origin of the frame at the position of the ballon the boat}
\Rightarrow
0-h=-\frac{1}{2}gt^2
\Rightarrow
\displaystyle t=\sqrt{\frac{2h}{g}}=\frac{2 45}{9.8}=3.0s
\text{For the boat:}
\Rightarrow
0-x_0=v t
\displaystyle v=-x_0/t=12/3.0=4.0 \text{ m/s}
\text{After flying for $15 \mathrm{~min}$ in a wind blowing $44 \mathrm{~km} / \mathrm{h}$ at an angle of $30^{\circ}$ south of east, an airplane
}
\textcolor{black}{\text{A) $245 \mathrm{~km} / \mathrm{h}$}}\\
\text{B) $38.1 \mathrm{~km} / \mathrm{h}$}\\
\text{C) $202 \mathrm{~km} / \mathrm{h}$}\\
\text{D) $220 \mathrm{~km} / \mathrm{h}$}\\
\text{E) $44.0 \mathrm{~km} / \mathrm{h}$}
\text{(old exam)}
\text{pilot is over a town that is $55 \mathrm{~km}$ due north of the starting point.}
\text{ What is the speed of the airplane relative to the wind? }
\text{After flying for $15 \mathrm{~min}$ in a wind blowing $44 \mathrm{~km} / \mathrm{h}$ at an angle of $30^{\circ}$ south of east, an airplane
}
\textcolor{red}{\text{A) $245 \mathrm{~km} / \mathrm{h}$}}\\
\text{B) $38.1 \mathrm{~km} / \mathrm{h}$}\\
\text{C) $202 \mathrm{~km} / \mathrm{h}$}\\
\text{D) $220 \mathrm{~km} / \mathrm{h}$}\\
\text{E) $44.0 \mathrm{~km} / \mathrm{h}$}
\text{(old exam)}
\text{pilot is over a town that is $55 \mathrm{~km}$ due north of the starting point.}
\text{ What is the speed of the airplane relative to the wind? }
\vec{V}_\text{PG}=\vec{V}_\text{PW}+\vec{V}_\text{WG}
\text{We project over the x-axis}
{V}_\text{PW}^x={V}_\text{PG}^x-{V}_\text{WG}^x
\vec{V}_\text{PW}=\vec{V}_\text{PG}-\vec{V}_\text{WG}
{V}_\text{PW}^x=0-44 \cos(-30^0)=-22\sqrt{3} \text{ km/h}
\text{We project over the y-axis}
{V}_\text{PW}^y={V}_\text{PG}^y-{V}_\text{WG}^y
{V}_\text{PW}^y=220-44 \sin(-30^0)=242\text{ km/h}
V_\text{PW}=\sqrt{(22\sqrt{3})^2+242^2}=245 \text{ km/h}
{V}_\text{PG}=55\text{ km}/0.25\text{h}=220 \text{ km/h}
\text{A particle $P$ travels with constant speed on a circle of radius $r=3.00 \mathrm{~m}$ as shown in Figure 2 }
\text{and completes one revolution in $20.0 \mathrm{~s}$. The particle passes through $O$ at time $t=0$.}
\text{With respect to $O$, find the particle's position vector and acceleration at the time $t=15 \mathrm{~s}$. }
\text{(old exam)}
\text{A particle $P$ travels with constant speed on a circle of radius $r=3.00 \mathrm{~m}$ as shown in Figure 2 }
\text{and completes one revolution in $20.0 \mathrm{~s}$. The particle passes through $O$ at time $t=0$.}
\text{With respect to $O$, find the particle's position vector and acceleration at the time $t=15 \mathrm{~s}$. }
\text{The distance travelled on the circle is}
d=v t =\frac{2\pi r}{T} t
d=r\theta
\text{The length of an arc of circle is:}
\displaystyle \theta=2 \pi\frac{t}{T}=\frac{3 \pi}{2}= 270^0
\text{We deduce that:}
\text{The position is therefore:}
x=r \cos\theta=3.00\cos(270^0)=-3.00\text{ m}
y=r \cos\theta=3.00\sin(270^0)=3.00 \text{ m}
\displaystyle a=\frac{v^2}{r}=\frac{(2\pi r/T)^2}{r}=\frac{(2\pi)^2 r}{T^2}=(2\times 3.14)^2/20^2 \times3=0.296 \text{ $m/s^2$}
\text{The particle is at $270^0$. The acceleration points out towards the center. So,}
\displaystyle \vec{a}=-0.296 \hspace{1mm}\hat{ i}\text{ $m/s^2$}
\vec{r}=-3.00 \hat{i}+3.00 \hat{j} (m)
\text{(old exam)}
\text{Challenge:}
\text{An object is thrown with a speed $v_0$ with an angle $\theta$ from the horizontal. It reaches }
\text{a maximum height $H=5 $ m and a range $R=10 $ m}
\text{Find $v_0$ and $\theta$.}
\displaystyle H=\frac{v_0^2 \sin^2\theta}{2 g}
\displaystyle R=\frac{v_0^2 \sin2\theta}{2 g}
\displaystyle \frac{H}{R}=\frac{\sin^2 \theta}{\sin2\theta}=\frac{\sin^2 \theta}{2\sin\theta\cos\theta}=\frac{\tan\theta}{2}
\displaystyle \theta=\tan^{-1}\Big(\frac{2H}{R}\Big)
\displaystyle \theta=\tan^{-1}\Big(\frac{2\times 5}{10}\Big)=45^0
\Rightarrow
\Rightarrow
\Rightarrow
\text{An object is thrown with a speed $v_0$ with an angle $\theta$ from the horizontal. It reaches }
\text{a maximum height $H=5 $ m and a range $R=10 $ m}
\text{Find $v_0$ and $\theta$.}
\displaystyle H=\frac{v_0^2 \sin^2\theta}{2 g}
\displaystyle R=\frac{v_0^2 \sin2\theta}{2 g}
\displaystyle \frac{H}{R}=\frac{\sin^2 \theta}{\sin2\theta}=\frac{\sin^2 \theta}{2\sin\theta\cos\theta}=\frac{\tan\theta}{2}
\displaystyle \theta=\tan^{-1}\Big(\frac{2H}{R}\Big)
\displaystyle \theta=\tan^{-1}\Big(\frac{2\times 5}{10}\Big)=45^0
\Rightarrow
\Rightarrow
\Rightarrow
\text{To be updated}