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Physics 5 Marks Questions

Model Paper 8 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 6 questions.

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Question 15 Marks
A marble rolls along a table at a constant speed of $1.00 m / s$ and then falls off the edge of the table to the floor 1.00 m below,
i. How long does the marble take to reach the floor?
ii. At what horizontal distance from the edge of the table does the marble land?
iii. What is its velocity as it strikes the floor?
Answer
As the marble was rolling on the table, therefore it has horizontal velocity and it will act as a projectile as soon as it leaves the edge of the table and fall freely under the effect of gravity.
Since, the marble is initially moving horizontally, $v _{ y 0}=0$ and $v _{ x 0}=1.00 m / s$. We must consider the origin to be at the edge of the table, so that $x _0= y _0=0$
Image

i. $t=?$ and $y=-1.00 m$
$
\begin{aligned}
& \therefore y=\frac{-1}{2} gt^2 \\
& \Rightarrow t=\sqrt{\frac{-2 y}{g}}=\sqrt{\frac{(-2)(-1.00)}{9.8}}=0.452 s
\end{aligned}
$
ii. $x =$ ?, when $t =0.452 s$
$
\therefore x=v_{x 0} t=1.00 \times 0.452 s=0.452 m
$
iii. $v =$ ?, $\theta=$ ? at $t =0.452 s$
The $x$-component of velocity is constant throughout the motion,
$
v_{x}=v_{x 0}=1.00 m / s
$
The $y$-component of velocity is given by
$
\begin{aligned}
& v_{y}=v_{y 0}-gt=0-9.8 \times 0.452=-4.43 m / s \\
& \therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{(1.00)^2+(-4.43)^2}=4.54 m / s \text {, the magnitude of the resultant velocity of the motion. } \\
& \theta=\tan ^{-1}\left|\frac{v_y}{v_x}\right|=\frac{4.43}{1.00}=77.3^{\circ}
\end{aligned}
$
As the marble hits the floor, its velocity is $4.54 m / s$ directed $77.3^{\circ}$ downward with respect to the horizontal.
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Question 25 Marks
Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation is equal to the average potential energy over the same period. At what distance from the mean position is the kinetic energy in simple harmonic oscillator equal to potential energy?
Answer
Let $x = A \sin \omega t$
Then, $v =\frac{d x}{d t}= A \omega \cos (\omega t )$
K. E. $=\frac{1}{2} mv ^2$
Average kinetic energy is
$
\frac{1}{T} \int_0^{T} \frac{1}{2} mv^2 dt=\frac{1}{4} mA^2 \omega^2
$
Average potential energy is
$
\frac{1}{T} \int_0^{T} \frac{1}{2} kx^2 dt=\frac{1}{4} mA^2 \omega^2
$
Let x be the distance where KE is equal to PE . Kinetic energy of particles executing SHM is $KE =\frac{1}{2} m \omega^2\left( a ^2- x ^2\right)$
Potential energy of particles is given by $PE =\frac{1}{2} m \omega^2 x ^2$ Let x be the distance where $KE = PE$
$
\begin{aligned}
& \frac{1}{2} m \omega^2\left(a^2-x^2\right)=\frac{1}{2} m \omega^2 x^2 \\
& \left(a^2-x^2\right)=x^2 \\
& x=\frac{a}{\sqrt{2}}
\end{aligned}
$
Kinetic energy equal to its potential energy at $x =\frac{a}{\sqrt{2}}$
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Question 35 Marks
State the law of conservation of angular momentum and illustrate it with the example of planetary motion.
Answer
Law of conservation of angular momentum. Suppose the external torque acting on a rigid body due to external forces is zero.
Then
$
\tau=\frac{d L}{d t}=0
$
Hence, L = constant
So when the total external torque acting on a rigid body is zero, the total angular momentum of the body is conserved. This is the law of conservation of angular momentum.
Clearly, when $\tau=0, L=I \omega=$ constant or $I_1 \omega_1=I_2 \omega_2$
This means that when no external torque is acting, the angular velocity co of the body can be increased or decreased by decreasing or increasing the moment of inertia of the body
Illustrations of the law of conservation of angular momentum:
i. Planetary motion: The angular velocity of a planet revolving in an elliptical orbit around the sun increases, when it comes closer to the sun because its moment of inertia about the axis through the sun decreases. When it goes far away from the sun, its moment of inertia increases and hence angular velocity decreases so as to conserve angular momentum.
ii. A man carrying heavy weights in his hands and standing on a rotating turn-table can change the angular speed of the turntable. As shown in Fig., if a person stands on a turn-table with some heavyweights in his hands stretched out and the table is rotated slowly, his angular speed at once increases, as he draws his hands to his chest. The moment of inertia of man and weights taken together decreases, as he draws his arms inward. As moment of inertia decreases, the angular speed increases so as to conserve total angular momentum.
Image

iii. A diver jumping from a spring board exhibits somersaults in air before touching the water surface. After leaving the spring board, a diver curls his body by pulling his arms and legs towards the centre of his body. This decreases his moment of inertia and he spins fast in midair. Just before hitting the water surface, he stretches out his arms. This decreases his moment of inertia and the diver enters water at a gentle speed.
Image

iv. An ice-skater or a ballet dancer can increase her angular velocity by folding her arms and bringing the stretched leg close to the other leg. When she stretches her hands and a leg outward [Fig.], her moment of inertia increases and hence angular speed decreases to conserve angular momentum. When she folds her arms and brings the stretched leg close to the other leg [Fig.], her moment of inertia decreases and hence angular speed increases.
v . The speed of the inner layers of the whirlwind in a tarnado is alarmingly high. The angular velocity of air in a tarnado increases as it goes towards the centre. This is because as the air moves towards the centre, its moment of inertia (I) decreases and to conserve angular momentum $( L = I \omega)$, the angular velocity co increases.
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Question 45 Marks
A particle of mass $m$ is released from point P at $x = x _0$ on the X -axis from origin O and falls vertically along the Y-axis, as shown in Fig.
Image

i. Find the torque $t$ acting on the particle at a time $t$ when it is at point Q with respect to O .
ii. Find the angular momentum L of the particle about O at this time t .
iii. Show that $\tau=\frac{d L}{d t}$ in this example.
Answer
i. The force of gravity, $F = mg$ produces the torque $\tau$. Let $\vec{r}$ be the position vector of Q . Then the magnitude of the torque is given by
$
\begin{aligned}
& \tau=r F \sin \theta \\
& =r \times m g \times \frac{x_0}{r}=m g x_0 \quad\left[\because \sin \theta=\frac{x_0}{r}\right]
\end{aligned}
$
The direction of the torque is directed into the plane of paper and perpendicular to it, as shown by $\otimes$.
ii. The magnitude of the angular momentum is $L=r p \sin \theta=r m v \sin \theta$
But the velocity v at point Q is given by $v = u + at =0+ gt = gt$
$
\therefore \quad L=r m g t . \frac{x_0}{r}=m g x_0 t
$
The direction of angular momentum is the same as that of torque.
iii. Now $L = mgx _0 t$
Differentiating both sides with respect to $f$, we get
$
\frac{d L}{d t}=\frac{d}{d t}\left(m g x_0 t\right)=m g x_0=\tau
$
Hence the relation $\tau=\frac{d L}{d}$ holds in this example.
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Question 55 Marks
On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ after every 500 m . Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer

Image
The path followed by the motorist is a regular hexagon with side 500 m , as shown in the given figure.
Let the motorist start from point P . The motorist takes the third turn at S .
Magnitude of displacement $= PS = PV + VS =500+500=1000 m(\because P V=Q R, V S=S R)$
Total path length, $d _1=P Q+Q R+R S=500+500+500=1500 m$
The motorist take the sixth turn at point P , which is the starting point
$\therefore$ Magnitude of displacement $=0$
Total path length, $d _2= PQ + QR + RS + ST + TU + UP$
$
d_2=500+500+500+500+500=3000 m
$
The motorist takes the eight turn at point R
$\therefore$ Magnitude of displacement $= PR$
$
\begin{aligned}
& P R=\sqrt{P Q^2+Q R^2+2(P Q) \cdot(Q R) \cos 60^{\circ}} \\
& P R=\sqrt{500^2+500^2+\left(2 \times 500 \times 500 \times \cos 60^{\circ}\right)} \\
& P R=\sqrt{250000+250000+\left(500000 \times \frac{1}{2}\right)} \\
& P R=866.03 m \\
& \beta=\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)=30^{\circ}
\end{aligned}
$
Therefore, the magnitude of displacement is 866.03 m at an angle of $30^{\circ}$ with PR.
Total path length $=$ Circumference of the hexagon $+ PQ + QR$
Total path length $=6 \times 500+500+500=4000 m$
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table
TurnMagnitude of displacement (m)Total path length (m)
Third10001500
Sixth03000
Eighth$866.03 ; 30^{\circ}$4000
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Question 65 Marks
Find the total energy of the particle executing S.H.M. and show graphically the variation of P.E. and K.E. with time in S.H.M. What is the frequency of these energies with respect to the frequency of the particle executing S.H.M?
Answer
The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude.
$
\begin{aligned}
& \frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2 \\
& E=\frac{1}{2} m \omega^2 A^2
\end{aligned}
$
Hence, the total energy of the particle in SHM is constant and it is independent of the instantaneous displacement. Relationship between potential energy, kinetic energy, and time in Simple Harmonic Motion at $t =0$, when $x = \pm A$.
Image
The frequency of kinetic energy is twice the frequency of SHM.
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5 Marks Questions - Physics STD 11 Science Questions - Vidyadip