3 Marks Question - Physics STD 11 Science Questions - Vidyadip

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Question 13 Marks

Which of the following potential energy curves in cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Answer

The potential energy of a system of two masses is inversely proportional to the separation between them. In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero (i.e., V(r) = 0) when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

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Question 23 Marks

A body of mass 0.5kg travels in a straight line with velocity $\text{v}=\text{ax}^{3/2}$ where $\text{a}=5\text{m}^{-1/2}\text{s}^{-1}.$ What is the work done by the net force during its displacement from x = 0 to x = 2m?

Answer

Given,
Mass of the body, m = 0.5kg
Velocity $\text{v} =\text{ax}^{(3/2)}\ ...(\text{i})$
Acceleration $\text{a}=5\text{m}^{–(1/2)}\text{s}^{–1}$
Initial velocity, u(at x = 0) = 0
Final velocity v(at x = 2m)
$=\text{a}×2^{(3/2)} = 5\times2^{(3/2)}=10\sqrt{2}\text{m/s}$
Work done, W = Increase in kinetic energy
$=\frac{1}{2}\text{m}(\text{v}^2–\text{u}^2)$
$=\frac{1}{2} \times0.5\big [(10\sqrt{2})^2 – (0)^2\big]$
$=\Big(\frac{1}{2}\times 0.5\times10\times10\times2\Big)$
$=50\text{J}$

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Question 33 Marks

A trolley of mass 300kg carrying a sandbag of 25kg is moving uniformly with a speed of 27km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kgs^-1. What is the speed of the trolley after the entire sand bag is empty?

Answer

As the trolley carrying the sand bag is moving uniformly, therefore, external force on the system = 0. When the sand leaks out, it does not lead to the application of any external force on the trolley. Hence, the speed of the trolley shall not change.

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Question 43 Marks

Consider the decay of a free neutron at rest: n → p + e^- Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus.

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e^-, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e^- + v]

Answer

The decay process of free neutron at rest is given as,
n → p + e^-
From Einstein’s mass-energy relation, we have the energy of electron as $\Delta\text{mc}^2$
Where,
$\Delta\text{m}$ = Mass defect = Mass of neutron - (Mass of proton + Mass of electron)
c = Speed of light
$\Delta\text{m}$ and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the $\beta$-decay of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly explains the continuous energy distribution.

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Question 53 Marks

The potential energy function for a particle executing linear simple harmonic motion is given by $\text{V(x)} = \frac{\text{kx}^2}{2},$ where k is the force constant of the oscillator. For k = 0.5Nm^-1, the graph of V(x) versus x is shown in Show that a particle of total energy 1J moving under this potential must ‘turn back’ when it reaches $\text{x} =\pm 2\text{m}.$

Answer

Given,

Potential energy for a particle executing linear simple harmonic motion is,

$\text{V(x)}=\frac{1}{2}\text{kx}^2$

Where, $\text{k}=\frac{1}{2}\text{N/m}$

Total energy of particle E = 1J

Since at extreme position total energy is potential energy,

$\therefore\text{V}=\frac{1}{2}\text{kx}^2=1\text{J}$

$\Rightarrow\text{x}=\pm\sqrt{\frac{2}{\text{k}}}=\pm\sqrt{4}$

$\Rightarrow\text{x}=\pm2\text{m}$

Hence the results.

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Question 63 Marks

The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Answer

When two bodies of same mass undergo elastic collisions their velocities get interchanged. In the given situation, bob A is moving with certain speed and bob B is at rest. Therefore, after collision, bob A comes to rest and the bob B starts moving with the speed of bob A. The whole momentum of bob A will get transfer to bob B and so bob A, will not rise at all after the second collision. Therefore bob A will come to rest and bob B will keep on moving.

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Question 73 Marks

A molecule in a gas container hits a horizontal wall with speed 200ms^-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Answer

The momentum of the gas molecule remains conserved whether the collision is elastic or inelastic. The gas molecule moves with a velocity of 200m/s and strikes the stationary wall of the container, rebounding with the same speed. It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

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Question 83 Marks

To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass $1000 kg$ moving with a speed $18.0 km / h$ on a smooth road and colliding with a horizontally mounted spring of spring constant $5.25 \times 10^3 N m ^{-1}$. What is the maximum compression of the spring?

Answer

At maximum compression the kinetic energy of the car is converted entirely into the potential energy of the spring.
The kinetic energy of the moving car is
$
\begin{array}{l}
K=\frac{1}{2} m v^2 \\
=\frac{1}{2} \times 10^3 \times 5 \times 5 \\
K=1.25 \times 10^4 J
\end{array}
$
where we have converted $18 km h ^{-1}$ to $5 m s ^{-1}$ [It is useful to remember that $36 km h ^{-1}=10 m s ^{-1}$. At maximum compression $x_m$, the potential energy $V$ of the spring is equal to the kinetic energy $K$ of the moving car from the principle of conservation of mechanical energy.
$
V=\frac{1}{2} k x_m^2
$
$
=1.25 \times 10^4 J
$
We obtain
$
x_m=2.00 m
$
We note that we have idealised the situation. The spring is considered to be massless. The surface has been considered to possess negligible friction.

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Question 93 Marks

A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate the work done by the two forces over 20 m.

Answer

The plot of the applied force is shown in Fig. 5.4. At $x=20 m , F=50 N (\neq 0)$. We are given that the frictional force $f$ is $| f |=50 N$. It opposes motion and acts in a direction opposite to $F$. It is therefore, shown on the negative side of the force axis.
The work done by the woman is
$W_F \rightarrow$ area of the rectangle $ABCD +$ area of the trapezium CEID
$
\begin{array}{l}
W_F=100 \times 10+\frac{1}{2}(100+50) \times 10 \\
=1000+750 \\
=1750 J
\end{array}
$
The work done by the frictional force is
$
\begin{array}{l}
W_f \rightarrow \text { area of the rectangle AGHI } \\
W_f=(-50) \times 20 \\
=-1000 J
\end{array}
$
The area on the negative side of the force axis has a negative sign.

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Question 103 Marks

It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass $1.00 g$ falling from a height $1.00 km$. It hits the ground with a speed of $50.0 m s ^{-1}$. (a) What is the work done by the gravitational force? What is the work done by the unknown resistive force?

Answer

(a) The change in kinetic energy of the drop is
$
\begin{array}{l}
\Delta K=\frac{1}{2} m v^2-0 \\
=\frac{1}{2} \times 10^{-3} \times 50 \times 50 \\
=1.25 J
\end{array}
$
where we have assumed that the drop is initially at rest.
Assuming that $g$ is a constant with a value $10 m / s ^2$, the work done by the gravitational force is,
$
\begin{aligned}
W_g & =m g h \\
& =10^{-3} \times 10 \times 10^3 \\
& =10.0 J
\end{aligned}
$
(b) From the work-energy theorem
$
\Delta K=W_g+W_r
$
where $W_r$ is the work done by the resistive force on the raindrop. Thus
$
\begin{aligned}
W_r & =\Delta K-W_g \\
& =1.25-10 \\
& =-8.75 J
\end{aligned}
$
is negative.

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Question 113 Marks

Consider the collision depicted in Fig. 5.10 to be between two billiard balls with equal masses $m_1=m_2$. The first ball is called the cue while the second ball is called the target. The billiard player wants to 'sink' the target ball in a corner pocket, which is at an angle $\theta_2=37$. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain $\theta_1$.

Answer

From momentum conservation, since the masses are equal
$
\begin{array}{l}
v _{1 i }= v _{1 f }+ v _{2 f } \\
\text { or } \quad v_{1 i}{ }^2=\left( v _{1 f}+ v _{2 f}\right) \cdot\left( v _{1 f}+ v _{2 f}\right) \\
=v_{1 f}{ }^2+v_{2 f}{ }^2+2 v _{1 f} \cdot v _{2 f}
\end{array}
$
$
=\left\{v_{1 f}{ }^2+v_{2 f}{ }^2+2 v_{1 f} v_{2 f} \cos \left(\theta_1+37^{\circ}\right)\right\} \quad \quad \text{(5.31)}
$
Since the collision is elastic and $m_1=m_2$ it follows from conservation of kinetic energy that
$
v_{1 i}{ }^2=v_{1 f}{ }^2+v_{2 f}{ }^2 \quad \quad \text{(5.32)}
$
Comparing Eqs. (5.31) and (5.32), we get
$
\cos \left(\theta_1+37^{\circ}\right)=0
$
or $\theta_1+37^{\circ}=90^{\circ}$
Thus, $\theta_1=53^{\circ}$
This proves the following result : when two equal masses undergo a glancing elastic collision with one of them at rest, after the collision, they will move at right angles to each other.

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Question 123 Marks

Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically $10^7 m s ^{-1}$ ) must be slowed to $10^3 m s ^{-1}$ so that it can have a high probability of interacting with isotope ${ }_{92}^{235} U$ and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water $\left( D _2 O \right)$ or graphite, is called a moderator.

Answer

The initial kinetic energy of the neutron is
$
K_{1 i}=\frac{1}{2} m_1 v_{1 i}^2
$
while its final kinetic energy from Eq. (5.26)
$
K_{1 f}=\frac{1}{2} m_1 v_{1 f}^2=\frac{1}{2} m_1\left(\frac{m_1-m_2}{m_1+m_2}\right)^2 v_{1 i}^2
$
The fractional kinetic energy lost is
$
f_1=\frac{K_{1 f}}{K_{1 i }}=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2
$
while the fractional kinetic energy gained by the moderating nuclei $K_{2 f} / K_{l i}$ is
$
\begin{array}{c}
f_2=1-f_1 \text { (elastic collision) } \\
=\frac{4 m_1 m_2}{\left(m_1+m_2\right)^2}
\end{array}
$
One can also verify this result by substituting from Eq. (5.27).
For deuterium $m_2=2 m_1$ and we obtain $f_1=1 / 9$ while $f_2=8 / 9$. Almost $90 \%$ of the neutron's energy is transferred to deuterium. For carbon $f_1=71.6 \%$ and $f_2=28.4 \%$. In practice, however, this number is smaller since head-on collisions are rare.

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Question 133 Marks

In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact).

Kinetic energy.
Total linear momentum?

Give reason for your answer in each case.

Answer

When two billiard balls collide each other then their linear momentum and kinetic energy remains conserved. Because here it is considered that there is not any non conservative force (like air resistance/ friction on surface etc.) and speed of ball is not so high so that they deformed on collision.

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Question 143 Marks

Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?

Answer

When elevator is descending then it is not its free fall under gravity it descends with uniform speed. Power is required to decrease the velocity due to free fall.
Power of motor or system of an elevator is constant and a limited or specified power can stop the speed of freely falling of passenger along with elevator.

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Question 153 Marks

A stone is dropped from a height ‘h'. Prove that the energy at any point in its path is mgh.

Answer

Let a stone of mass m be dropped from a point A at a height h.
P.E. at A = mgh
K.E. = 0
Total energy at A = mgh
As it reaches B, it would have lost some P.E. and gained
K.E. Velocity on reaching

P.E at B = mg (h - x)
$\text{K.E}=\frac{1}{2}\text{mv}_\text{B}^2=\frac{1}{2}\text{m}.2\text{gx}=\text{mgx}$
Total energy at
B = mg (h – x) + mgx = mgh
On reaching the ground C the mass must have gained a velocity $\sqrt{2\text{gx}}$ and the P.E must be zero.
P.E. at C = 0
K.E. at $\text{C}=\frac{1}{2}\text{mv}_\text{c}^2=\frac{1}{2}\text{m}(2\text{gh)}=\text{mgh}$
Total energy at C = mgh
Thus' it is proved that the total energy at any point in its path is mgh.

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Question 163 Marks

Light in certain cases may be considered as a stream of particles called photons. Each photon has a linear momentum $\frac{\text{h}}{\lambda}$ where h is the Planck's constant and $\lambda$ is the wavelength of the light. A beam of light of wavelength $\lambda$ is incident on a plane mirror at an angle of incidence $\theta.$ Calculate the change in the linear momentum of a photon as the beam is reflected by the mirror.

Answer

$\overrightarrow{\text{P}}_{\text{incidence}}=\Big(\frac{\text{h}}{\lambda}\Big)\cos\theta\hat{\text{i}}-\Big(\frac{\text{h}}{\lambda}\Big)\sin\theta\hat{\text{j}}$
$\overrightarrow{\text{P}}_{\text{Reflected}}=-\Big(\frac{\text{h}}{\lambda}\Big)\cos\theta\hat{\text{i}}-\Big(\frac{\text{h}}{\lambda}\Big)\sin\theta\hat{\text{j}}$
The change in momentum will be only in the x-axis direction. i.e.

$|\triangle\text{P}|=\Big(\frac{\text{h}}{\lambda}\Big)\cos\theta-\Big(\frac{\text{h}}{\lambda}\Big)\cos\theta=\Big(\frac{\text{2h}}{\lambda}\Big)\cos\theta$

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Question 173 Marks

A block of mass 2kg is pulled up on a smooth incline of angle 30^° with horizontal. If the block moves with an acceleration of 1m/ s², find the power delivered by the pulling force at a time 4 seconds after motion starts. What is the/ frac delivered during these four seconds after the motion starts?

Answer

The forces acting on the block are shown in the figure.
Resolving forces parallel to incline, $\text{F}-\text{mg}\sin\theta=\text{ma}$
$\text{F}-\text{mg}\sin\theta+\text{ma}$
= 2 × 9.8 × sin 30^° + 2 × 1
= 11.8N
The velocity after 4 seconds = u + at
= 0 + 1 4 = 4m/ s
Power delivered by force at t = 4 seconds
= Force × velocity = 11.8N × 4s = 47.2W
The displacement during 4 seconds is given by
$\text{v}^2=\text{u}^2+2\text{ as}$
$\Rightarrow\text{v}^2=0+2\times1\times\text{s}$
s = 8m
Work done in 4 seconds = Force x distance = 11.8 × 8 = 94.4J
$\therefore$ Average power delivered = $\frac{\text{work done}}{\text{time}}=\frac{94.4}{4}$
= 23.6W.

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Question 183 Marks

Define power. Obtain an expression for it in terms of force and velocity.

Answer

Power The rate of doing work is called power.
$\text{P}=^{\ \ \text{Lt}}_{\Delta\text{t}\rightarrow0}\frac{\Delta\text{W}}{\Delta\text{t}}=\frac{\text{d}\text{W}}{\text{d}\text{t}}$
$=\frac{\text{d}}{\text{d}\text{t}}(\vec{\text{F}}.\vec{\text{S})}=\vec{\text{F}.}\frac{\text{d}\vec{\text{s}}}{\text{dt}}$
$\text{P}=\vec{\text{F}.}\vec{\text{V}}$
$\Rightarrow\text{P}=\text{Fv}$

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Question 193 Marks

Draw a graph showing variation of potential energy, kinetic energy and the total energy of a body freely falling on Earth from a height h.

Answer

Graphs depicting variation of:

Gravitational potential energy (P.E.).
Inetic energy (K.E.), and.
The total sum of potential and kinetic energies for a freely falling body are as shown in adjoining Fig. From the graphs, it is clear that.

Gravitational potential energy decreases as the body falls downwards and is zero at the Earth.
Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground.
The sum of kinetic and potential energies remains constant at all points during its free fall.

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Question 203 Marks

A small block of mass is pressed against a horizontal spring fixed at one end to compress the spring through 5.0cm. When released the block moves horizontally till it leaves the spring. Where will it hit the ground at a distance 2m below the slab?
[k = 100N/ m, m = 100g)

Answer

$\frac{1}{2}\text{kx}^2=\frac{1}{2}\text{mv}^2$
$\text{v}=\sqrt{\frac{\text{k}}{\text{m}}}\text{x}^2=\sqrt{\frac{100}{100}\times\frac{25\times10^{-4}}{10^{-3}}}$
$=\sqrt{2.5}\text{ ms}^{-1}$
Height = 2m, $\text{t}=\sqrt{\frac{2\text{h}}{\text{g}}}=\sqrt{0.4}$
Horizontal length conered $=\sqrt{0.4}\times\sqrt{2.5}=1\text{m }$

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Question 213 Marks

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1m. What is the speed with which the bob arrives at the lower most point given that it dissipates 10% of it initial energy against air resistance. (g = 10m/ s^-2).

Answer

Length of pendulum = 1m

Potential energy (P.E.) at A = mgh

Kinetic energy (K.E.) at $\text{B}=\frac{1}{2}\text{mv}^2$

As 10% of P.E dissipated againt air resistance so,

$\frac{1}{2}\text{m}\text{v}^2=90\%\text{ of P.E}$

$\Rightarrow\frac{1}{2}\text{m}\text{v}^2=\frac{90}{100}\times\text{mgh}$

$\Rightarrow\text{v}^2= 1.8\times1\times10$

$\Rightarrow\text{v}=\sqrt{18}\text{ms}^{-1}$

$\therefore\text{The speed is }\sqrt{18}\text{ ms}^{-1}$

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Question 223 Marks

What is Einstein's energy-mass equivalence? Explain the energy output of a mass on the basis of this relationship. Give two situations where this relationship can be employed. Also calculate energy released from complete annihilation of 1g matter.

Answer

Einstein's energy-mass equivalence relationship states that mass and energy are equivalent. The equivalence relation is given by E = mc² where 'm' is the mass that disappears.

In the case of sun, four hydrogen (lighter) nuclei fuse to form a helium nucleus whose mass is less than the sum of masses of four hydrogen nuclei. This mass difference, called the mass defect $\Delta\text{m}$, it is the source of energy which is released by the sun.

Two other examples are:

Atom bomb based on uncontrolled nuclear fission process.
To provide electrical energy as in nuclear power plant by controlled nuclear fission process
In chemical reactions.

Annihilation of 1g of matter is

E = mc²

=(10^-3) × (3 × 10⁸)² = 9 × 10¹³J

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Question 233 Marks

Which of the following potential energy curves in cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Answer

The potential energy of a system of two masses is inversely proportional to the separation between them. In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero (i.e., V(r) = 0) when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

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Question 243 Marks

In a children's park, there is a slide which has a total length of 10m and a height of 8.0m (figure). Vertical ladder are provided to reach the top. A boy weighing 200N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. Find

The work done by the ladder on the boy as he goes up.
The work done by the slide on the boy as he comes down. Neglect any work done by forces inside the body of the boy.

Answer

$\ell=10\text{m},\text{h}=8\text{m},\text{mg}=200\text{N}$

$\text{f}=200\times\frac{3}{10}=60\text{N}$

Work done by the ladder on the boy is zero when the boy is going up because the work is done by the boy himself.
Work done against frictional force, $\text{W}=\mu\text{RS}=\text{f}\ell$

$=(-60)\times10=-600\text{J}$

Work done by the forces inside the boy is,

$\text{W}_\text{b}=(\text{mg}\sin\theta)\times10$

$=200\times\frac{8}{10}\times10=1600\text{J}$

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Question 253 Marks

State and explain Work-Energy theorem.

Answer

Work done on a body is reflected as change in kinetic energy, according to Work-Energy theorem.
$\text{W}=\int\text{Fdx}=\int\text{m}\frac{\text{dv}}{\text{df}} \text{dx}$
$=\int\text{mv}\text{ dv}=\Big|\frac{1}{2}\text{mv}^2\Big|^\text{vf}_\text{vi}$
Work done $=\frac{1}{2}\text{m}(\text{v}^2_\text{f}-\text{v}_\text{i}^2)$
$=\frac{1}{2}\text{mv}_\text{f}^2-\frac{1}{2}\text{mv}_\text{i}^2$
Work done = Change in kinetic energy.

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Question 263 Marks

A ball falls on the ground from a height of 2.0m and rebounds up to a height of 1.5m. Find the coefficient of restitution.

Answer

Let the velocity of the ball falling from height h₁ be u (when it approaches the ground).
Velocity on the ground $\text{u}=\sqrt{2\text{gh}}_1$
Let the velocity of ball when it separates from the ground be v. (Assuming it goes up to height h₂)
$\Rightarrow\text{v}=\sqrt{2\text{gh}}_2$
$=\sqrt{2\times9.8\times1.5}$
Let the coefficient of restituti be e.
We khow, v = eu
$\Rightarrow\text{e}=\frac{\sqrt{2\times9.8\times1.5}}{\sqrt{2\times9.8\times2}}=\frac{\sqrt{3}}{2}$
Hence, the coefficient of restitution is $\frac{\sqrt{3}}{2}$

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Question 273 Marks

A simple pendulum consists of a 50cm long string connected to a 100g ball. The ball is pulled aside so that the string makes an angle of 37° with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

Answer

From the figure, $\cos\theta=\frac{\text{AC}}{\text{AB}}$

$\Rightarrow\text{AC}=\text{AB}\cos\theta$

$\Rightarrow(0.5)\times(0.8)=0.4$

So, CD = (0.5) - (0.4) = (0.1)m

Energy at D = energy at B

$\frac{1}{2}\text{mv}^2=\text{mg}(\text{CD})$

$\text{v}^2=2\times10\times(0.1)=2$

So, the tension is given by,

$\text{T}=\frac{\text{mv}^2}{\text{r}}+\text{mg}$

$\Rightarrow(0.1)\Big(\frac{2}{0.5}+10\Big)=1.4\text{N}$

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Question 283 Marks

Find the work done in pulling and pushing a roller through 100m horizontally when a force of 1500N is acting along a chain making an angle of 60^° with ground. Assume the floor to be smooth.

Answer

Here, forec F = 1500N
and displacement, s = 100m
$\theta=60^\circ$
$\therefore$ Work done,
$\text{W}=\text{Fs}\cos\theta$
$=1500\times100\times\cos60^\circ$
$=1500\times100\times\frac{1}{2}$
$=75000\text{J}$
$=75\text{KJ}$

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Question 293 Marks

A ball of mass m, moving with a speed 2v₀, collides inelastically (e > 0) with an identical ball at rest. Show that:
For head-on collision, both the balls move forward.

Answer

Let the v₁, v₂ are the velocities of the two balls after the collision. Now by the principle of law of conservation of momentum.
$\text{mv}_0=\text{mv}_1+\text{mv}_2$
$2\text{v}_0=\text{v}_1+\text{v}_2\ ...(\text{i})$
$\text{e}=\frac{\text{v}_2-\text{v}_1}{\text{v}_0+\text{v}_0}$
$\text{v}_2-\text{v}_1=2\text{ev}_0$
From (i) $\text{v}_1=-\text{v}_2+2\text{v}_0$
$\text{v}_1=-\text{v}_1-2\text{ev}_0+2\text{v}_0$
$2\text{v}_1=2\text{v}_0-2\text{ev}_0$
$\text{v}_1=\text{v}_0(1-\text{e})$
$\because\ \text{e}<1$
v₀ is positive so the direction of v₁ is same as v₀ or v₁ is in forward direction. Hence proved.

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Question 303 Marks

An inclined plane $(\theta)$ has its topmost point at a height 'h'. Prove that the work done to bring a mass to the ground level either vertically or along the inclined surface is equal and is mgh. Also prove that the velocity of the mass at the lowermost point is $\sqrt{2\text{gh}}$.

Answer

In path I.
$\text{a}=\sin\theta,\text{u}=0$

$\text{v}^2=0+2\text{g}\sin\theta\text{ l}$
$=2\text{ g}\sin\theta\frac{\text{h}}{\sin\theta}=2\text{gh}$
Change in Kinetic energy
$= \frac{1}{2}\text{mv}^2-0=\frac{1}{2}\text{m}2\text{gh}=\text{mgh}$
Work done = Change in K.E. = mgh
In path II. Gravitational force does work.
Work done = mgh
This will be converted into kinetic energy.
$\frac{1}{2}\text{mv}^2=\text{mgh},\text{v}=\sqrt{2\text{gh}}$

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Question 313 Marks

A ball bounces to 80% of its original height. What fraction of its mechanical energy is lost in each bounce?

Answer

Let the ball fall from a height, h then
K.E. of ball at the time of just striking the ground = P.E. of ball at height h
323 Marks

A small block of mass 'm' is pressed against a horizontal spring fixed at one end to compress the spring through 5.0cm. When released the block moves horizontally till it leaves the spring. Where will it hit the ground at a distance 2m below the slab?

$[\text{k}=100\frac{\text{N}}{\text{m}}=100\text{g}]$

Answer

Here $\frac{1}{2}\text{kx}^2=\frac{1}{2}\text{m}\upsilon^2$
$\upsilon=\sqrt{\frac{\text{k}}{\text{m}}\text{x}^2}$
$=\sqrt{\frac{100}{100}\times\frac{25\times10^{-4}}{10^{-3}}}\text{ms}^{-1}$
$=\sqrt{2.5}\text{ms}^{-1}$
Height = 2m; $\text{t}=\sqrt{\frac{2\text{h}}{\text{g}}}$
$=\sqrt{0.4}\text{ms}^{-1}$

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Question 333 Marks

When you lift a box from the floor and put it on an almirah the potential energy of the box increases, but there is no change in its kinetic energy. Is it a violation of conservation of energy?

Answer

No Work done in lifting the box increases the potential energy of the box. During lifting at every point, the force applied by us on the box in the upward direction is equal to the gravitational force acting on the box in the downward direction. Therefore, there is no change in the velocity of the box. As a result, the kinetic energy of the box will not change.

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Question 343 Marks

A body of mass 2kg is at rest at a height of 10m above the ground. Calculate its potential energy and kinetic energy after it has fallen through half the height. Also find the velocity at this instant.

Answer

Total energy at

B = kinetic energy + potential energy

= 0 + mgh = 2 × 9.8 × 10 = 196J

As it descends half the height, it loses potential energy which is given by

$=\text{mg}\frac{\text{h}}{2}=\frac{1}{2}\text{mgh}=98\text{J}$

$\therefore$ Its potential energy at C = (196 - 98) = 98J

The loss of potential energy = gain in kinetic energy

= 196 - 98 = 98 J

But $\text{K.E.}=\frac{1}{2}\text{m}\text{v}^2$

$\therefore\frac{1}{2}\times2\times\text{v}^2$

$=98\Rightarrow\text{v}^2=98$

$\text{v}=7\sqrt{2}\text{m/ s}$

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Question 353 Marks

A body of mass 0.5kg travels in a straight line with velocity $\text{v}=\text{ax}^{3/2}$ where $\text{a}=5\text{m}^{-1/2}\text{s}^{-1}.$ What is the work done by the net force during its displacement from x = 0 to x = 2m?

Answer

Given,
Mass of the body, m = 0.5kg
Velocity $\text{v} =\text{ax}^{(3/2)}\ ...(\text{i})$
Acceleration $\text{a}=5\text{m}^{–(1/2)}\text{s}^{–1}$
Initial velocity, u(at x = 0) = 0
Final velocity v(at x = 2m)
$=\text{a}×2^{(3/2)} = 5\times2^{(3/2)}=10\sqrt{2}\text{m/s}$
Work done, W = Increase in kinetic energy
$=\frac{1}{2}\text{m}(\text{v}^2–\text{u}^2)$
$=\frac{1}{2} \times0.5\big [(10\sqrt{2})^2 – (0)^2\big]$
$=\Big(\frac{1}{2}\times 0.5\times10\times10\times2\Big)$
$=50\text{J}$

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Question 363 Marks

Define an electron volt. Express it in terms of joule.

Answer

Electron Volt. Electron volt is the amount of energy possessed by an electron in falling through a potential difference of 1 volt.
m = 4Kg, u = 0,
F = 16N, t = 10 Sec
F = ma, 16 = 4a, a = 4 m/ s²
V = u + at = 0 + 4 (10)
V = 40m/ s
K.E. of the body at end of 10 sec.
$\text{E}=\frac{1}{2}\text{mv}^2$
$=\frac{1}{2}\times4\times(40)^2$
$=3200\text{J}$

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Question 373 Marks

How will you find work done by a variable force? What is the significance of F-x graph?

Answer

Work done is the product of force and displacement.
When a variable force acts on the body, the displacement x is split into small parts dx and the work is estimated for each part.
Since force is continuous in variation, the net work done will be $\int\text{Fdx}$ in the limits x = x_i to x = x_f.

Since $\int\text{Fdx}$ geometrically is the area below the graph of F with x, the area below signifies work done in the displacement by the variable force acting.

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Question 383 Marks

What is an elastic collision? What will happen, when

A heavy body collides with a light mass at rest.
A light body collides with a heavy mass at rest.

Answer

Elastic collision is one in which both momentum and energy are conserved. We know, when two bodies m₁ and m₂ moving with velocities u₁ and u₂ collide, their velocities become,

$\text{v}_1=\frac{(\text{m}_1-\text{m}_2)\text{u}_1+2\text{m}_2\text{u}_2}{\text{(m}_1+\text{m}_2}$

$\text{v}_2=\frac{\text{m}_2-\text{m}_1)\text{u}_2+2\text{m}_1\text{u}_1}{(\text{m}_1+\text{m}_2)}$

if m₁ >> m₂ and u₂ = 0, we get

$\text{v}_1=\frac{\text{m}_1\text{u}_1}{\text{m}_1}=\text{u}_1,\text{v}_2=2\text{u}_1$

If m₁ >> m₂ and u₁ = 0, we get

$\text{v}_1=\frac{2\text{m}_2\text{u}_2}{\text{m}_1}, $ as m₂ << m₁

_{$\therefore\text{v}_1\approx0$}

_{$\text{v}_2=-\frac{\text{m}_1\text{u}_2}{\text{m}_1}=-\text{u}_2$}

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Question 393 Marks

What are the conditions so that transfer of kinetic energy is maximum during a collision?

Answer

For maximum transfer of kinetic energy during a collision, following conditions should be fulfilled.

The collision should be a head on collision.
The collision should be perfectly elastic.
The target body should be at rest.
The mass of the striking body and the target body should be exactly same.

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Question 403 Marks

Two masses, one 'n' times as heavy as the other, have equal kinetic energy. What is the ratio of their momenta?

Answer

Since $\text{p}=\sqrt{2\text{mE}_\text{k}}$
$\text{E}_\text{k}=\frac{\text{p}^2}{2\text{m}}$
As E_k is constant
$\therefore\text{p}\propto\sqrt{\text{m}}$
Cleariy, $\frac{\text{p}_1}{\text{p}_2}=\frac{\sqrt{\text{nm}}}{\sqrt{\text{m}}}=\frac{\sqrt{\text{n}}}{1}$
$\Rightarrow\text{p}_1:\text{p}_2=\sqrt{\text{n}}:1$

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Question 413 Marks

A trolley of mass 300kg carrying a sandbag of 25kg is moving uniformly with a speed of 27km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kgs^-1. What is the speed of the trolley after the entire sand bag is empty?

Answer

As the trolley carrying the sand bag is moving uniformly, therefore, external force on the system = 0. When the sand leaks out, it does not lead to the application of any external force on the trolley. Hence, the speed of the trolley shall not change.

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Question 423 Marks

A sphere of mass 'm' moving with a velocity u hits another stationary sphere of same mass at rest. If e is the coefficient of restitution. Find the ratio of the velocities of two spheres after the collision.

Answer

According to law of conservation of momentaum,
$\text{mu}=\text{m}\upsilon_1+\text{m}\upsilon_2$
$\upsilon_1+\upsilon_2=\text{u}$
Also $\upsilon_1-\upsilon_2=-\text{eu}$
Solving eqn. (i) and (ii), we get
$\upsilon_1=\frac{\text{u}(1-\text{e})}{2}$ and $\upsilon_2=\frac{\text{u}(1+\text{e})}{2}$
The ratio of velocities, $\frac{\upsilon_1}{\upsilon_2}=\Big(\frac{1-\text{e}}{1+\text{e}}\Big)$.

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Question 433 Marks

A block of mass 30.0kg is being brought down by a chain. If the block acquires a speed of 40.0cm/s in dropping down 2.00m, find the work done by the chain during the process.

Answer

Given m = 30kg, v = 40cm/sec = 0.4m/sec, s = 2m

From the free body diagram, the force given by the chain is,

F = (ma - mg) = m(a - g) [where a = acceleration of the block]

$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}$

$=\frac{0.16}{0.4}=0.04\text{m}/\text{sec}^2$

So, work done $\text{W}=\text{Fs}\cos\theta=\text{m}(\text{a}-\text{g})\text{s}\cos\theta$

$\Rightarrow\text{W}=30(0.04-9.8)\times2$

$\Rightarrow\text{W}=-585.5$

$\Rightarrow\text{W}=-586\text{J}$

So, $\text{W}=-586\text{J}$

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Question 443 Marks

Suppose the average mass of raindrops is 3.0 × 10^-5kg and their average terminal velocity 9m s^-1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.

Answer

Energy transferred by rain to surface of earth is kinetic energy

The velocity of rain or water is 9m/s For mass m = volume × density
= Area of base × height × p
= 1m² × 1m × 1000
= 1000kg
So, energy transferred by 100cm rainfall Click and drag to move

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Question 453 Marks

Give example of a situation in which an applied force does not result in a change in kinetic energy.

Answer

Assume a ball tied to a string and is moving in a vertical circle. Work done by tension force will be zero and hence tension force will not cause any change in KE of ball. Because at any instant of time the displacement is tangential and the force is central in nature, i.e., tension in the string and the small displacement at any instant are perpendicular to each other.

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Question 463 Marks

A body of mass 'M' at rest is struck by a body of mass 'm'. Show that the fraction of K.E. of mass m transferred to the struck particle is $\frac{4\text{mM}}{(\text{m}+\text{M})^2}.$

Answer

m₁ = m
u₁ = u
m₂= M
u₂= 0, v₂ = ?
$\text{v}_2=\frac{2\text{m}_1\text{u}_1}{\text{m}_1+\text{m}_2}+\frac{\text{m}_2+\text{u}_2}{\text{m}_1+\text{m}_2}$
$=\frac{2\text{mu}}{\text{m}+\text{M}}+0=\frac{2\text{m}\text{u}}{\text{m}+\text{M}}$
K.E of body stuck after collision,
$\text{E}_2=\frac{1}{2}\text{m}_2\text{v}_2^2=\frac{1}{2}\text{M}\Big(\frac{2\text{m}\text{u}}{\text{m}+\text{M}}\Big)^2$
$=\frac{2\text{M}\text{m}^2\text{u}^2}{(\text{m}+\text{M})^2}$
Initial K.E., $\text{E}_1=\frac{1}{2}\text{m}_1\text{u}_1^2=\frac{1}{2}\text{m}\text{u}^2$
$\therefore$ fraction of initial K.E. transferred $\frac{\text{E}_2}{\text{E}_1}=\frac{2\text{M}\text{m}^2\text{u}^2}{(\text{m}+\text{M})^2\Big(\frac{1}{2}\text{m}\text{u}^2\Big)}$
$=\frac{4\text{m}\text{M}}{(\text{m}+\text{M})^2}$

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Question 473 Marks

Consider the decay of a free neutron at rest: n → p + e^- Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus.

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e^-, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e^- + v]

Answer

The decay process of free neutron at rest is given as,
n → p + e^-
From Einstein’s mass-energy relation, we have the energy of electron as $\Delta\text{mc}^2$
Where,
$\Delta\text{m}$ = Mass defect = Mass of neutron - (Mass of proton + Mass of electron)
c = Speed of light
$\Delta\text{m}$ and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the $\beta$-decay of a neutron or a nucleus. The presence of neutrino νon the LHS of the decay correctly explains the continuous energy distribution.

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Question 483 Marks

A car of mass 1000kg travels up an incline of 1 in 25 at a constant velocity of 50km/ h. What power does the car engine have to develop if there is a resistive force of 300 N opposing the motion?

Answer

$\sin\theta=\frac{1}{25}$. force along inclination due to gravitational
Force $=\text{mg}\sin\theta$
Resistive force = 300
Net force to overcome $=300+1000\times10\times\sin\theta$
$=300+1000\times10\times\frac{1}{25}=700\text{ N}$
$\text{Power}=\vec{\text{F}}.\vec{\text{v}}=700\times50\times\frac{1000}{3600}=9722.22\text{ J/sec}$

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Question 493 Marks

Water falling from a 50m high fall is to be used for generating electric energy. If 1.8 x 10⁵kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100W lamps can be lit?

Answer

h = 50m, m = 1.8 × 10⁵ kg/hr, P = 100 watt,
P.E. = mgh = 1.8 × 10⁵ × 9.8 × 50 = 882 × 10⁵ J/hr
Because, half the potential energy is converted into electricity.
Electrical energy $=\frac{1}{2}\text{P.E.}$
= 441 × 10⁵ J/hr
So, power in watt (J/sec) is given by $=\frac{441\times10^5}{3600}$
$\therefore$ number of 100W lamps, that can be lit $=\frac{441\times10^5}{3600\times100}=122.5\approx122$

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Question 503 Marks

A particle is released from the top of an incline of height h. Does the kinetic energy of the particle at the bottom of the incline depend on the angle of incline? Do you need any more information to answer this question in Yes or No?

Answer

No, the kinetic energy of the particle at the bottom of the inclined plane does not depend on the angle of inclination. When the particle reaches the ground, all its potential energy, while at the top of the inclined plane, is converted into kinetic energy. As we know that kinetic energy depends only on the height of the particle, it will be the same for different angles of inclination.
No, we do not need any other information to answer this question.

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