5 Marks Questions

Question 15 Marks

The figure shows three vectors $\overrightarrow{O A}, \overrightarrow{O B}$ and $\overrightarrow{O C}$ which are equal in magnitude (say, F ). Determine the direction of $\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}$

Answer

Given, $|\vec{O} A|=|\vec{O} B|=|\vec{O} C|=\vec{F}$
Angles are $30^{\circ}, 45^{\circ}$ and $60^{\circ}$.
Resolve all the vector components individually

$
R_{x}= R _{x_1}+ R _{x_2}+ R _{x_3}
$
Sum of vectors in x -direction (i.e. $R _{ x }$ ) and sum of vactors in y -direction (i.e. $R _{ y }$ )
$
\begin{aligned}
& R_{X}=A \cos 30^{\circ}+B \cos 60^{\circ}-C \cos 45^{\circ} \\
& =\frac{F \sqrt{3}}{2}+\frac{F}{2}-\frac{F}{\sqrt{2}}[\because A=B=C=F] \\
& =\frac{F}{2 \sqrt{2}}(\sqrt{6}+\sqrt{2}-2) \\
& R_{y}=A \sin 30^{\circ}+C \cos 45^{\circ}-B \sin 60^{\circ} \\
& =\frac{F}{2}+\frac{F}{\sqrt{2}}-\frac{F \sqrt{3}}{2}=\frac{F}{2 \sqrt{2}}(\sqrt{2}+2-\sqrt{6})
\end{aligned}
$
Determination of magnitude,
$
\begin{aligned}
& R=\sqrt{R_x^2+R_y^2} \\
& =\sqrt{\left[\frac{F}{2 \sqrt{2}}(\sqrt{6}+\sqrt{2}-2)\right]^2+\left[\frac{F}{2 \sqrt{2}}(\sqrt{2}+2-\sqrt{6})\right]^2} \\
& =\sqrt{F^2(0.435)+F^2(0.116)} \\
& =\sqrt{F^2(0.550)}+F \sqrt{0.550} \\
& \Rightarrow R=0.74 F
\end{aligned}
$
Determination of direction
$
\begin{aligned}
& \tan \theta=\frac{R_y}{R_x}=\frac{\frac{F}{2 \sqrt{2}}(\sqrt{2}+2-\sqrt{6})}{\frac{F}{2 \sqrt{2}}(\sqrt{6}+\sqrt{2}-2)} \\
& =\frac{0.97}{1.85} \approx 0.524 \\
& \theta \approx 27.65
\end{aligned}
$
This is the angle which R makes with x -axis.

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Question 25 Marks

A spring having with a spring constant $1200 N m ^{-1}$ is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Take the position of mass when the spring is unstreched as $x=0$, and the direction from left to right as the positive direction of $x$-axis. Give $x$ as a function of time $t$ for the oscillating mass if at the moment we start the stopwatch $(t=0)$, the mass is
a. at the mean position,
b. at the maximum stretched position, and
c. at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer

The functions have the same frequency and amplitude, but different initial phases.
Given:
Distance travelled by the mass sideways is given by, $A =2.0 cm$
Force constant of the spring is given by, $k=1200 N m ^{-1}$
Mass, m is given by $=3 kg$
Angular frequency of oscillation is given by:
$
\begin{aligned}
& \omega=\sqrt{\frac{\text { spring cons } \tan t}{\text { mass }}} \\
& \Rightarrow \omega=\sqrt{\frac{k}{m}} \\
& =\sqrt{\frac{1200}{3}}=\sqrt{400}=20 rads^{-1}
\end{aligned}
$
a. When the mass is at the mean position, the initial phase is 0.
Displacement is given by,
$
\Rightarrow x=A \sin \omega t=2 \sin 20 t
$
b. At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is $\frac{\pi}{2}$. hence, Displacement is given by,
$
\begin{aligned}
& \Rightarrow x=A \sin \left(\omega t+\frac{\pi}{2}\right) \\
& =2 \sin \left(20 t+\frac{\pi}{2}\right)=2 \cos 20 t
\end{aligned}
$
c. At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is $\frac{3 \pi}{2}$.
Displacement is given by,
$
\begin{aligned}
& \Rightarrow x=A \sin \left(\omega t+\frac{3 \pi}{2}\right) \\
& =2 \sin \left(20 t+\frac{3 \pi}{2}\right)=-2 \cos 20 t
\end{aligned}
$
d. The functions have the same frequency $\left(\frac{20}{2 \pi} H z\right)$ and amplitude $(2 cm)$, but initial phases are different $\left(0, \frac{\pi}{2}, \frac{3 \pi}{2}\right)$.

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Question 35 Marks

Define moment of inertia of a body. Give its units and dimensions. What is the physical significance of moment of inertia?

Answer

The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration, which is the sum of the product of the mass of every particle with its square of the distance from the axis of rotation.
The units of M. l are $kgm ^2$ and its dimensional formula is $\left[ M ^1 L^2 T^0\right]$.
The moment of inertia has the same physical significance as a mass in translational motion. The mass of a body is used to calculate inertia in translational motion.

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Question 45 Marks

Four particles of masses $4 kg, 7 kg, 3 kg$ and 5 kg are respectively located at the four corners $A , B , C$ and D of a square of side 1 m , as shown in Fig. Calculate the moment of inertia of the system about

i. an axis passing through the point of intersection of the diagonals and perpendicular to the plane of the square,
ii. the side $A B$, and
iii. the diagonal BD.

Answer

Here $AB = BC = CD = DA =1 m$
$
O A=O B=O C=O D=\frac{1}{2} \sqrt{1^2+1^2}=\frac{1}{\sqrt{2}} m
$
i. M.I. of the system about an axis through O and perpendicular to the plane of the square,
$
\begin{aligned}
& I=4(OA)^2+2(O B)+3(OC)^2+5(OD)^2 \\
& =(4+2+3+5) \times\left(\frac{1}{\sqrt{2}}\right)^2=14 \times \frac{1}{2}=7 kg m^2
\end{aligned}
$
ii. M.I. of the system about the side AB ,
$
I=3(BC)^2+5(AD)^2=3 \times 1+5 \times 1=8 kg m^2
$
iii. M.I. of the system about the diagonal BD,
$
\begin{aligned}
& I=4(OA)^2+3(OC)^2 \\
& =4 \times \frac{1}{2}+3 \times \frac{1}{2}=3.5 kg m^2
\end{aligned}
$

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Question 55 Marks

At what angle should a body be projected with a velocity $24 ms^{-1}$ just to pass over the obstacle 16 m high at a horizontal distance of 32 m ? Take $g =10 ms^{-2}$.

Answer

As shown in figure, if point of projection is taken as the origin of the coordinate system, the projected body must pass through a point having coordinates ( $32 m, 16 m$ ). If $u$ be the initial velocity of the projectile and $\theta$ the angle of projection, then Horizontal component of initial velocity, $u _{ x }= u \cos \theta$
Vertical component of initial velocity, $u _{ y }= u \sin \theta$

If the body passes through point P after time t , then horizontal distance covered,
$x =( u \cos \theta) t$
or $32=(24 \cos \theta) t$ $\ldots(i)$
Similarly, vertical distance covered,
$y=(u \sin \theta) t-\frac{1}{2} g t^2$
or $16=(24 \sin \theta) t-\frac{1}{2} \times 10 \times t^2$
From equation (i), $t=\frac{32}{24 \cos \theta}$
Putting this value of $t$ in equation (ii), we get
$16=(24 \sin \theta) \frac{32}{24 \cos \theta}-\frac{1}{2} \times 10 \times\left(\frac{32}{24 \cos \theta}\right)^2$
or $16=32 \tan \theta-5 \times \frac{16}{9 \cos ^2 \theta}$
or $1=2 \tan \theta-\frac{5}{9} \sec ^2 \theta$
or $9=18 \tan \theta-5\left(1+\tan ^2 \theta\right)$
or $5 \tan ^2 \theta-18 \tan \theta+14=0$
$
\begin{aligned}
& \therefore \tan \theta=\frac{18 \pm \sqrt{(18)^2-4 \times 5 \times 14}}{10} \\
& =2.462 \text { or } 1.37
\end{aligned}
$
Hence $\theta=67^{\circ} 54^{\prime}$ or $48^{\circ} 40^{\prime}$

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Question 65 Marks

The motion of a particle executing simple harmonic motion is described by the displacement function, $x ( t )= A \cos (\omega t$ $+\phi)$ If the initial $(t=0)$ position of the particle is 1 cm and its initial velocity is $\omega cm / s$, then what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi s ^{-1}$. If instead of the cosine function, we choose the sine function to describe the $SHM , x = B \sin (\omega t +\phi)$, then what are the amplitude and initial phase of the particle with the above initial conditions?

Answer

Given, displacement equation $x ( t )= A \cos (\omega t+\phi) \ldots$ (i)
At $t =0 ; x (0)=1 cm$, velocity of the particle $v=\omega cm / s$
Angular frequency $\omega=\pi s ^{-1}$
$\Rightarrow 1=A \cos (\omega t+\phi)$
For $t =0,1= A \cos \phi$$\ldots(i)$
$
\begin{aligned}
& \text { Now, } v(t)=\frac{d x(t)}{d t}=\frac{d}{d t} A \cos (\omega t+\phi) \\
& =-A \omega \sin (\omega t+\phi)
\end{aligned}
$
Again at $t =0, v=\omega cm / s$
$
\begin{aligned}
& \Rightarrow \omega=-A \omega \sin \phi \\
& \Rightarrow-1=A \sin \phi \ldots(ii)
\end{aligned}
$
Squaring and adding eqs.(i) and (ii),
$
\begin{aligned}
& A^2 \cos ^2 \phi+A^2 \sin ^2 \phi=(1)^2+(-1)^2 \\
& A^2=2 \Rightarrow A= \pm \sqrt{2} cm
\end{aligned}
$
Hence, the amplitude of the $SHM =\sqrt{2} cm$
Dividing Eq. (ii) by (i), we get
$
\begin{aligned}
& \frac{A \sin \phi}{A \cos \phi}=\frac{-1}{1} \text { or } \tan \phi=-1 \\
& \Rightarrow \phi=-\frac{\pi}{4} \text { or } \frac{7 \pi}{4}
\end{aligned}
$
Now, if instead of cosine, we choose the sine function in the displacement equation, then $x ( t )= B \sin (\omega t+\alpha)$
At $t =0, x =1 cm, \Rightarrow 1=B \sin (0+\alpha)$
or $B \sin \alpha=1$ $\ldots(iii)$
Velocity $v ( t )=\frac{d x(t)}{d t}=\frac{d}{d t}[B \sin (\omega t+\alpha)]$
$
=+B \omega \cos (\omega t+\alpha)
$
Again at $t =0, v ( t )=\omega cm / s$
$B \cos \alpha=+1$$\ldots(iv)$
Squaring and adding Eqs.(iii) and (iv),
$
\begin{aligned}
& B^2 \sin ^2 \alpha+B^2 \cos ^2 \alpha=(1)^2+(+1)^2 \\
& \Rightarrow B^2 \sin ^2 \alpha+B^2 \cos ^2 \alpha=2 \\
& B^2\left(\sin ^2 \alpha+\cos ^2 \alpha\right)=2 \\
& B^2 1=2 \Rightarrow B= \pm \sqrt{2} cm
\end{aligned}
$
Hence, amplitude of the simple harmonic motion in both types of trigonometric wave equation expression $=\sqrt{2} cm$ Dividing Eq. (iii) by (iv), we get
$\frac{B \sin \alpha}{B \cos \alpha}=\frac{1}{1}$ or $\tan \alpha=1$
$\therefore \alpha=\frac{\pi}{4}$, only the phase angle differs for sine and cosine wave equation.

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Physics 5 Marks Questions

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